Probabilistic World

Mean and Variance of Probability Distributions

Posted on August 28, 2019 Written by The Cthaeh 13 Comments

how to solve problems involving mean and variance of probability distribution

In my previous post I introduced you to probability distributions .

In short, a probability distribution is simply taking the whole probability mass of a random variable and distributing it across its possible outcomes. Since every random variable has a total probability mass equal to 1, this just means splitting the number 1 into parts and assigning each part to some element of the variable’s sample space (informally speaking).

In this post I want to dig a little deeper into probability distributions and explore some of their properties. Namely, I want to talk about the measures of central tendency ( the mean ) and dispersion ( the variance ) of a probability distribution.

Table of Contents

Relationship to previous posts

This post is a natural continuation of my previous 5 posts. In a way, it connects all the concepts I introduced in them:

  • The Mean, The Mode, And The Median : Here I introduced the 3 most common measures of central tendency (“the three Ms”) in statistics. I showed how to calculate each of them for a collection of values, as well as their intuitive interpretation. In the current post I’m going to focus only on the mean.
  • The Law Of Large Numbers: Intuitive Introduction : This is a very important theorem in probability theory which links probabilities of outcomes to their relative frequencies of occurrence.
  • An Intuitive Explanation Of Expected Value : In this post I showed how to calculate the long-term average of a random variable by multiplying each of its possible values by their respective probabilities and summing those products.
  • The Variance: Measuring Dispersion : In this post I defined various measures of dispersion of a collection of values. In the current post I’m going to focus exclusively on variance.
  • Introduction To Probability Distributions : Finally, in this post I talked about probability distributions which are assignments of probability masses or probability densities to each possible outcome of a random variable. Probability distributions are the main protagonist of the current post as well.

Without further ado, let’s see how they all come together.

Introduction

Any finite collection of numbers has a mean and variance. In my previous posts I gave their respective formulas. Here’s how you calculate the mean if we label each value in a collection as x 1 , x 2 , x 3 , x 4 , …, x n , …, x N :

The formula for the statistical measure of central tendency called mean

If you’re not familiar with this notation, take a look at my post dedicated to the sum operator . All this formula says is that to calculate the mean of N values, you first take their sum and then divide by N (their number).

And here’s how you’d calculate the variance of the same collection:

The formula for the statistical dispersion measure called variance

So, you subtract each value from the mean of the collection and square the result. Then you add all these squared differences and divide the final sum by N. In other words, the variance is equal to the average squared difference between the values and their mean.

If you’re dealing with finite collections, this is all you need to know about calculating their mean and variance. Finite collections include populations with finite size and samples of populations. But when working with infinite populations, things are slightly different.

Let me first define the distinction between samples and populations, as well as the notion of an infinite population.

Samples versus populations

A sample is simply a subset of outcomes from a wider set of possible outcomes, coming from a population .

For example, if you’re measuring the heights of randomly selected students from some university, the sample is the subset of students you’ve chosen. The population could be all students from the same university. Or it could be all university students in the country. Or all university students in the world. The important thing is for all members of the sample to also be members of the wider population.

A big white circle with black dots inside representing a population. And smaller circles inside representing samples of the population

Samples obviously vary in size. Technically, even 1 element could be considered a sample. Whether a particular size is useful will, of course, depend on your purposes. Generally, the larger the sample is, the more representative you can expect it to be of the population it was drawn from.

The maximum size of a sample is clearly the size of the population. So, if your sample includes every member of the population, you are essentially dealing with the population itself.

It’s also important to note that whether a collection of values is a sample or a population depends on the context. For example, if you’re only interested in investigating something about students from University X, then the students of University X comprise the entirety of your population. On the other hand, if you want to learn something about all students of the country, then students from University X would be a sample of your target population.

Finite versus infinite populations

One difference between a sample and a population is that a sample is always finite in size. A population’s size, on the other hand, could be finite but it could also be infinite. An infinite population is simply one with an infinite number of members.

Where do we come across infinite populations in real life? Well, we really don’t. At any given moment, the number of any kind of entity is a fixed finite value. Even the number of atoms in the observable universe is a finite number. Infinite populations are more of a mathematical abstraction. They are born out of a hypothetical infinite repetition of a random process.

For example, if we assume that the universe will never die and our planet will manage to sustain life forever, we could consider the population of the organisms that ever existed and will ever exist to be infinite.

But where infinite populations really come into play is when we’re talking about probability distributions. A probability distribution is something you could generate arbitrarily large samples from. In fact, in a way this is the essence of a probability distribution. You will remember from my introductory post that one way to view the probability distribution of a random variable is as the theoretical limit of its relative frequency distribution (as the number of repetitions approaches infinity).

Mean and variance of infinite populations

Like I said earlier, when dealing with finite populations, you can calculate the population mean or variance just like you do for a sample of that population. Namely, by taking into account all members of the population, not just a selected subset. For instance, to calculate the mean of the population, you would sum the values of every member and divide by the total number of members.

But what if we’re dealing with a random variable which can continuously produce outcomes (like flipping a coin or rolling a die)? In this case we would have an infinite population and a sample would be any finite number of produced outcomes.

So, you can think of the population of outcomes of a random variable as an infinite sequence of outcomes produced according to its probability distribution. But how do we calculate the mean or the variance of an infinite sequence of outcomes?

The answer is actually surprisingly straightforward. Expected value to the rescue!

The equation of expected value expressed with the sum operator

In my post on expected value , I defined it to be the sum of the products of each possible value of a random variable and that value’s probability.

So, how do we use the concept of expected value to calculate the mean and variance of a probability distribution? Well, intuitively speaking, the mean and variance of a probability distribution are simply the mean and variance of a sample of the probability distribution as the sample size approaches infinity . In other words, the mean of the distribution is “the expected mean” and the variance of the distribution is “the expected variance” of a very large sample of outcomes from the distribution.

Let’s see how this actually works.

The mean of a probability distribution

Let’s say we need to calculate the mean of the collection {1, 1, 1, 3, 3, 5}.

\[ \textrm{Mean} = \frac{1+1+1+3+3+5}{6}  = \frac{14}{6} = 2.33  \]

  • Mean = 1/6 + 1/6 + 1/6 + 3/6 + 3/6 + 5/6 = 2.33
  • Mean = 3/6 * 1 + 2/6 * 3 + 1/6 * 5 = 2.33

That is, you take each unique value in the collection and multiply it by a factor of k / 6 , where k is the number of occurrences of the value.

The mean and the expected value of a distribution are the same thing

\frac{k}{6}

With this process we’re essentially creating a random variable out of the finite collection. And like all random variables, it has an infinite population of potential values, since you can keep drawing as many of them as you want. And naturally it has an underlying probability distribution.

\frac{3}{6} \cdot 1 + \frac{2}{6} \cdot 3 + \frac{1}{6} \cdot 5

  • As M approaches infinity, the mean of a sample of size M will be approaching the mean of the original collection.

But now, take a closer look at the last expression. Do you notice that it is actually equivalent to the formula for expected value? Hence, we reach an important insight!

  • The mean of a probability distribution is nothing more than its expected value.

If you remember, in my post on expected value I defined it precisely as the long-term average of a random variable. So, this should make a lot of sense.

Mean of discrete distributions

Well, here’s the general formula for the mean of any discrete probability distribution with N possible outcomes:

The formula for the statistical measure of central tendency called mean for discrete probability distributions

As you can see, this is identical to the expression for expected value. Let’s compare it to the formula for the mean of a finite collection:

x_n \cdot \frac{1}{N}

Let’s see how this works with a simulation of rolling a die. The animation below shows 250 independent die rolls. The height of each bar represents the percentage of each outcome after each roll.

A bar plot simulation of consecutive die rolls

Click on the image to start/restart the animation.

Notice how the mean is fluctuating around the expected value 3.5 and eventually starts converging to it. If the sample grows to sizes above 1 million, the sample mean would be extremely close to 3.5.

Mean of continuous distributions

In my introductory post on probability distributions, I explained the difference between discrete and continuous random variables . Let’s get a quick reminder about the latter.

In short, a continuous random variable’s sample space is on the real number line. Since its possible outcomes are real numbers, there are no gaps between them (hence the term ‘continuous’). The function underlying its probability distribution is called a probability density function .

In the post I also explained that exact outcomes always have a probability of 0 and only intervals can have non-zero probabilities. And, to calculate the probability of an interval, you take the integral of the probability density function over it.

Continuous random variables revisited

Let’s look at the pine tree height example from the same post. The plot below shows its probability density function. The shaded area is the probability of a tree having a height between 14.5 and 15.5 meters.

A curve representing the probability density function of pine tree heights. Part of the area under the curve is shaded symmetrically around the mean

Let’s use the notation f(x) for the probability density function (here x stands for height). Then the expression for the integral will be:

An integral f(x)*dx from 14.5 to 15.5

In the integrals section of my post related to 0 probabilities I said that one way to look at integrals is as the sum operator but for continuous random variables. That is, the expression above stands for the “infinite sum” of all values of f(x), where x is in the interval [14.5, 15.5]. Which happens to be approximately 0.383.

Because the total probability mass is always equal to 1, the following should also make sense:

An integral f(x)*dx from negative to positive infinity is equal to 1

In fact, this formula holds in the general case for any continuous random variable. The integral of its probability density function from negative to positive infinity should always be equal to 1, in order to be consistent with Kolmogorov’s axioms of probability.

You might be wondering why we’re integrating from negative to positive infinity. What if the possible values of the random variable are only a subset of the real numbers? For example, a tree can’t have a negative height, so negative real numbers are clearly not in the sample space. Another example would be a uniform distribution over a fixed interval like this:

The probability density function over the interval [0, 1]

Well, this is actually not a problem, since we can simply assign 0 probability density to all values outside the sample space. This way they won’t be contributing to the final value of the integral. That is, integrating from positive to negative infinity would give the same result as integrating only over the interval where the function is greater than zero.

The formula for the mean of a continuous random variable

So, after all this, it shouldn’t be too surprising when I tell you that the mean formula for continuous random variables is the following:

The formula for the statistical measure of central tendency called mean for continuous probability distributions

Notice the similarities with the discrete version of the formula:

x_n \cdot P(x_n)

And like in discrete random variables, here too the mean is equivalent to the expected value. And if we keep generating values from a probability density function, their mean will be converging to the theoretical mean of the distribution.

By the way, if you’re not familiar with integrals, don’t worry about the dx term. It means something like “an infinitesimal interval in x”. Feel free to check out my post on zero probabilities for some intuition about it.

It’s important to note that not all probability density functions have defined means. Although this topic is outside the scope of the current post, the reason is that the above integral doesn’t converge to 1 for some probability density functions (it diverges to infinity). I am going to revisit this in future posts related to such distributions.

Well, this is it for means. If there’s anything you’re not sure you understand completely, feel free to ask in the comment section below.

Now let’s take a look at the other main topic of this post: the variance.

The variance of a probability distribution

In this section I discuss the main variance formula of probability distributions. To see two useful (and insightful) alternative formulas, check out my latest post .

From the get-go, let me say that the intuition here is very similar to the one for means. The variance of a probability distribution is the theoretical limit of the variance of a sample of the distribution, as the sample’s size approaches infinity.

The variance formula for a collection with N values is:

And here’s the formula for the variance of a discrete probability distribution with N possible values:

The formula for the statistical measure of dispersion called variance for discrete probability distributions

Do you see the analogy with the mean formula? Basically, the variance is the expected value of the squared difference between each value and the mean of the distribution . In the finite case, it is simply the average squared difference.

And, to complete the picture, here’s the variance formula for continuous probability distributions:

The formula for the statistical measure of dispersion called variance for continuous probability distributions

The variance of a die roll

Let’s do this step by step.

\[ (3.5 - 1)^2 = 6.25 \]

To get an intuition about this, let’s do another simulation of die rolls. I wrote a short code that generates 250 random rolls and calculates the running relative frequency of each outcome and the variance of the sample after each roll. Click on the image below to see this simulation animated:

A bar plot simulation of consecutive die rolls

You see how the running variance keeps fluctuating around the theoretical expectation of 2.92? It doesn’t quite converge after only 250 rolls, but if we keep increasing the number of rolls, eventually it will.

The bottom line is that, as the relative frequency distribution of a sample approaches the theoretical probability distribution it was drawn from, the variance of the sample will approach the theoretical variance of the distribution.

Mean and variance of functions of random variables

This section was added to the post on the 7th of November, 2020.

To conclude this post, I want to show you something very simple and intuitive that will be useful for you in many contexts.

By now you know the general formulas for calculating the mean and variance of a probability distribution, both for discrete and continuous cases. These formulas work with the elements of the sample space associated with the distribution. To calculate the mean, you’re multiplying every element by its probability (and summing or integrating these products). Similarly, for the variance you’re multiplying the squared difference between every element and the mean by the element’s probability.

g(x) = x^2

Another die roll example

One application of what I just showed you would be in calculating the mean and variance of your expected monetary wins/losses if you’re betting on outcomes of a random variable. The association between outcomes and their monetary value would be represented by a function.

\[ g(x) = \begin{cases}x^3 & x \in \{1, 3, 5\} \\-x^2 & x \in \{2, 4, 6\}\end{cases}\]

One of my goals in this post was to show the fundamental relationship between the following concepts from probability theory:

  • Mean and variance
  • The law of large numbers
  • Expected value
  • Probability distributions

I also introduced the distinction between samples and populations. And more importantly, the difference between finite and infinite populations. I tried to give the intuition that, in a way, a probability distribution represents an infinite population of values drawn from it. And that the mean and variance of a probability distribution are essentially the mean and variance of that infinite population.

In other words, they are the theoretical expected mean and variance of a sample of the probability distribution, as the size of the sample approaches infinity.

The main takeaway from this post are the mean and variance formulas for finite collections of values compared to their variants for discrete and continuous probability distributions. I hope I managed to give you a good intuitive feel for the connection between them. Let’s take a final look at these formulas.

These are the formulas for the mean:

And here are the formulas for the variance:

Maybe take some time to compare these formulas to make sure you see the connection between them.

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November 14, 2019 at 10:49 am

THIS PRESENTATION IS VERY CLEAR. I WISH TO KNOW IF THE FOLLOWING PROCEDURE IS CORRECT. I TAKE A SET OF VARIABLES IN AN ASCENDING NUMERICAL VALUE AND I ADD THEM UP FROM THE MINIMUM TO THE MAXIMUM VALUE SO THAT I GET THE SUM OF A SUM : 1,2,3,4,5,6,7 AS (1+2)=3,(3+3=6),(6+4=10),(10+5=15),(15+6=21),(21+7=28) SO I HAVE : 1,3,6,10,15,21,28 THEN I CALCULATE THE PROBABILITY OF EACH VALUE AND TAKE THE PROBABILITY OF P(1),(P(2)-P(1)),(P(3)-P(2)) AND SO ON AS THE INDIVIDUAL PROBABILITY OF EACH NUMBER 1,2,3, . . . TO BE THEIR CORRESPONDING PROBABILITY OF OCCURENCE.AM I CORRECT IN MY APPROACH ? THANK YOU IN ADVANCE FOR YOUR CONSIDERATION ! ! !

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November 14, 2019 at 1:48 pm

Hi, Spyridon!

I’m not sure I completely understand your procedure. Could you give some more detail?

How exactly is your data being generated? Is it that you have a random variable which can take on values from the set of positive integers and you generate multiple values from it? Or do you simply have a pool of integers and you draw N of them (without replacement)? Or are the values always 1, 2, 3, 4, 5, 6, 7?

Also, once you get the cumulative sum of those values, what is your procedure (what determines) the probabilities of the sums 1, 3, 6, 10…? Since you originally operate with the actual values, couldn’t you calculate their probabilities directly?

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February 11, 2020 at 2:21 pm

Great posts. I think there’s an error in the “The mean of a probability distribution” section. The set includes 6 numbers, so the denominator should be 6 rather than 5 (including in the k/5 fraction). In the subsequent section (“The mean and the expected value of a distribution are the same thing”), 3/5+2/5+1/5 doesn’t actually equal 1, but it would if the 5 were a 6. But the posts are very helpful overall.

February 11, 2020 at 4:56 pm

Good spot, Sergey! Thank you, I corrected the mistake.

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April 19, 2020 at 10:10 pm

Hie, you guys go to great lengths to make things as clear as possible. I’m really glad I bumped into you!!!

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October 25, 2020 at 4:08 am

Very good explanation….Thank you so much. I would like to add more details on the bellow part

October 25, 2020 at 5:49 am

Hi Mansoor! Sure, feel free to add. Looks like your comment was cut in the middle?

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December 2, 2020 at 9:26 am

I am unable to unable to understand the lines—-

A probability distribution is something you could generate arbitrarily large samples from. In fact, in a way this is the essence of a probability distribution

Can you please elaborate on this.

December 2, 2020 at 4:16 pm

Hi, Karthik. If you have any finite population, you can generate samples of size less than or equal to the size of the population, right? For example, if you have a bag of 30 red balls and 70 green balls, the biggest sample of balls you could pick is 100 (the entire population). On the other hand, if every time you pick a random ball you just record its color and immediately throw it back inside the bag, then you can draw samples of arbitrary sizes (much larger than 100). This sampling with replacement is essentially equivalent to sampling from a Bernoulli distribution with parameter p = 0.3 (or 0.7, depending on which color you define as “success”). But the same holds for any probability distribution.

Let me know if this makes sense.

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February 10, 2021 at 7:48 am

Hi CTHAEH, integral calculates area uder the curve. when you calculate area under the probability density curve, what you are calculating is somewhat of a product =f(x).dx over the range of x. How can that be equal to 1? I can understand that the sum of all probablities must be euual to 1. But here it is not just the sum of probablities, but the sum of probability and corresponding x value. i.e it is not f(x) but f(x).dx where dx is an infinetesimal delta of X. so it is a bit confusing..

February 11, 2021 at 6:01 am

Hi Daraj. You can intuitively think of f(x)*dx as an “infinitely thin” rectangle whose height is the value of the function at the point x. Adding up “all” the rectangles from point A to point B gives the area under the curve in the interval [A, B]. The most trivial example of the area adding up to 1 is the uniform distribution. Imagine you have the function f(x) = 2 for all x in the interval [0, 0.5]. What is the area under the curve in this case? It’s just a rectangle whose height is 2 and whose width is 0.5, right? Then the area under the curve is simply 2 * 0.5 = 1.

Also, check out the first post on probability distributions (and its comment section) to gain a little more intuition about continuous distributions and the difference between probability masses (the familiar notion of probabilities) and probability densities. I am currently working on a series of posts related to analysis where I will explain in much more detail what integrals are and how to actually compute them.

In the meantime, let me know if my answer was a little helpful and if you need clarification.

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February 22, 2021 at 11:36 am

the variance of rolling a dice probability distribution is approximately 2.92. another example of your variance is 2725 dollar and 16 dollar(mean or expected value)

i can’t understand what we try to say here, variance mean dispersion or how value are far apart or different from each other. your example of travelling different planet and recording their temperature and calculating their mean and variance is well understood and provide good applicable use of variance.

but for rolling a dice i cant understand what actually 2.92 and 2725 dollar suggest?

it will be great help if you can clear my doubt

February 24, 2021 at 4:35 am

Hi Diwyanshu,

It’s the same idea as with the planet/temperature example. To calculate the variance of a die roll, just treat the possible outcomes as the values whose spread we’re measuring. They are 1, 2, 3, 4, 5, 6, right? Just calculate their variance as if those were temperatures on the planet.

Why does this work so straightforwardly? Because each outcome has the same probability (1/6), we can treat those values as if they were the entire population. If the probabilities were not equal, we just need to weigh each value by its corresponding probability.

Have you read my post about expected values ? I think it will give you a better intuition for why we do that. Basically think of the variance of a probability distribution as the variance of an infinite collection of numbers. Why infinite? Because we can keep generating values from a probability distribution (by sampling from it). As we continue to draw more and more samples from the distribution, the size of the collection increases. But if after each draw we keep calculating the variance, the value we’re going to obtain is going to be getting closer and closer to the theoretical variance we calculated from the formula I gave in the post.

In other words, the variance of a probability distribution is the expectation of the variance of a sample of values taken from that distribution, as the size of the sample approaches infinity .

Does this make sense?

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Mathematics LibreTexts

5.1: Basics of Probability Distributions

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  • Page ID 130249

  • Kathryn Kozak
  • Coconino Community College

As a reminder, a variable or what will be called the random variable from now on, is represented by the letter x and it represents a quantitative (numerical) variable that is measured or observed in an experiment.

Also remember there are different types of quantitative variables, called discrete or continuous. What is the difference between discrete and continuous data? Discrete data can only take on particular values in a range. Continuous data can take on any value in a range. Discrete data usually arises from counting while continuous data usually arises from measuring.

Examples of each

How tall is a plant given a new fertilizer? Continuous. This is something you measure. How many fleas are on prairie dogs in a colony? Discrete. This is something you count.

If you have a variable, and can find a probability associated with that variable, it is called a random variable . In many cases the random variable is what you are measuring, but when it comes to discrete random variables, it is usually what you are counting. So for the example of how tall is a plant given a new fertilizer, the random variable is the height of the plant given a new fertilizer. For the example of how many fleas are on prairie dogs in a colony, the random variable is the number of fleas on a prairie dog in a colony.

Now suppose you put all the values of the random variable together with the probability that that random variable would occur. You could then have a distribution like before, but now it is called a probability distribution since it involves probabilities. A probability distribution is an assignment of probabilities to the values of the random variable. The abbreviation of pdf is used for a probability distribution function.

For probability distributions, \(0 \leq P(x) \leq 1 \operatorname{and} \sum P(x)=1\)

Example \(\PageIndex{1}\): Probability Distribution

The 2010 U.S. Census found the chance of a household being a certain size. The data is in Example \(\PageIndex{1}\) ("Households by age," 2013).

In this case, the random variable is x = number of people in a household. This is a discrete random variable, since you are counting the number of people in a household.

This is a probability distribution since you have the x value and the probabilities that go with it, all of the probabilities are between zero and one, and the sum of all of the probabilities is one.

You can give a probability distribution in table form (as in Example \(\PageIndex{1}\)) or as a graph. The graph looks like a histogram. A probability distribution is basically a relative frequency distribution based on a very large sample.

Example \(\PageIndex{2}\) graphing a probability distribution

The 2010 U.S. Census found the chance of a household being a certain size. The data is in the table ("Households by age," 2013). Draw a histogram of the probability distribution.

State random variable:

x = number of people in a household

You draw a histogram, where the x values are on the horizontal axis and are the x values of the classes (for the 7 or more category, just call it 7). The probabilities are on the vertical axis.

Screenshot (73).png

Notice this graph is skewed right.

Just as with any data set, you can calculate the mean and standard deviation. In problems involving a probability distribution function (pdf), you consider the probability distribution the population even though the pdf in most cases come from repeating an experiment many times. This is because you are using the data from repeated experiments to estimate the true probability. Since a pdf is basically a population, the mean and standard deviation that are calculated are actually the population parameters and not the sample statistics. The notation used is the same as the notation for population mean and population standard deviation that was used in chapter 3.

The mean can be thought of as the expected value . It is the value you expect to get if the trials were repeated infinite number of times. The mean or expected value does not need to be a whole number, even if the possible values of x are whole numbers.

For a discrete probability distribution function,

The mean or expected value is \(\mu=\sum x P(x)\)

The variance is \(\sigma^{2}=\sum(x-\mu)^{2} P(x)\)

The standard deviation is \(\sigma=\sqrt{\sum(x-\mu)^{2} P(x)}\)

where x = the value of the random variable and P(x) = the probability corresponding to a particular x value.

Example \(\PageIndex{3}\): Calculating mean, variance, and standard deviation for a discrete probability distribution

The 2010 U.S. Census found the chance of a household being a certain size. The data is in the table ("Households by age," 2013).

  • Find the mean
  • Find the variance
  • Find the standard deviation
  • Use a TI-83/84 to calculate the mean and standard deviation
  • Using R to calculate the mean

x= number of people in a household

a. To find the mean it is easier to just use a table as shown below. Consider the category 7 or more to just be 7. The formula for the mean says to multiply the x value by the P(x) value, so add a row into the table for this calculation. Also convert all P(x) to decimal form.

Now add up the new row and you get the answer 2.525. This is the mean or the expected value, \(\mu\) = 2.525 people. This means that you expect a household in the U.S. to have 2.525 people in it. Now of course you can’t have half a person, but what this tells you is that you expect a household to have either 2 or 3 people, with a little more 3-person households than 2-person households.

b. To find the variance, again it is easier to use a table version than try to just the formula in a line. Looking at the formula, you will notice that the first operation that you should do is to subtract the mean from each x value. Then you square each of these values. Then you multiply each of these answers by the probability of each x value. Finally you add up all of these values.

Now add up the last row to find the variance, \(\sigma^{2}=2.02375 \text { people }^{2}\). (Note: try not to round your numbers too much so you aren’t creating rounding error in your answer. The numbers in the table above were rounded off because of space limitations, but the answer was calculated using many decimal places.)

c. To find the standard deviation, just take the square root of the variance, \(\sigma=\sqrt{2.023375} \approx 1.422454\) people. This means that you can expect a U.S. household to have 2.525 people in it, with a standard deviation of 1.42 people.

d. Go into the STAT menu, then the Edit menu. Type the x values into L1 and the P(x) values into L2. Then go into the STAT menu, then the CALC menu. Choose 1:1-Var Stats. This will put 1-Var Stats on the home screen. Now type in L1,L2 (there is a comma between L1 and L2) and then press ENTER. If you have the newer operating system on the TI-84, then your input will be slightly different. You will see the output in Figure \(\PageIndex{1}\) .

Screenshot (74).png

The mean is 2.525 people and the standard deviation is 1.422 people.

e. The command would be weighted.mean(x, p). So for this example, the process would look like:

x<-c(1, 2, 3, 4, 5, 6, 7) p<-c(0.267, 0.336, 0.158, 0.137, 0.063, 0.024, 0.015) weighted.mean(x, p)

Output: [1] 2.525

So the mean is 2.525.

To find the standard deviation, you would need to program the process into R. So it is easier to just do it using the formula.

Example \(\PageIndex{4}\) Calculating the expected value

In the Arizona lottery called Pick 3, a player pays $1 and then picks a three-digit number. If those three numbers are picked in that specific order the person wins $500. What is the expected value in this game?

To find the expected value, you need to first create the probability distribution. In this case, the random variable x = winnings. If you pick the right numbers in the right order, then you win $500, but you paid $1 to play, so you actually win $499. If you didn’t pick the right numbers, you lose the $1, the x value is -$1. You also need the probability of winning and losing. Since you are picking a three-digit number, and for each digit there are 10 numbers you can pick with each independent of the others, you can use the multiplication rule. To win, you have to pick the right numbers in the right order. The first digit, you pick 1 number out of 10, the second digit you pick 1 number out of 10, and the third digit you pick 1 number out of 10. The probability of picking the right number in the right order is \(\dfrac{1}{10} * \dfrac{1}{10} * \dfrac{1}{10}=\dfrac{1}{1000}=0.001\). The probability of losing (not winning) would be \(1-\dfrac{1}{1000}=\dfrac{999}{1000}=0.999\). Putting this information into a table will help to calculate the expected value.

Now add the two values together and you have the expected value. It is \(\$ 0.499+(-\$ 0.999)=-\$ 0.50\). In the long run, you will expect to lose $0.50. Since the expected value is not 0, then this game is not fair. Since you lose money, Arizona makes money, which is why they have the lottery.

The reason probability is studied in statistics is to help in making decisions in inferential statistics. To understand how that is done the concept of a rare event is needed.

Definition \(\PageIndex{1}\): Rare Event Rule for Inferential Statistics

If, under a given assumption, the probability of a particular observed event is extremely small, then you can conclude that the assumption is probably not correct.

An example of this is suppose you roll an assumed fair die 1000 times and get a six 600 times, when you should have only rolled a six around 160 times, then you should believe that your assumption about it being a fair die is untrue.

Determining if an event is unusual

If you are looking at a value of x for a discrete variable, and the P (the variable has a value of x or more) < 0.05, then you can consider the x an unusually high value. Another way to think of this is if the probability of getting such a high value is less than 0.05, then the event of getting the value x is unusual.

Similarly, if the P (the variable has a value of x or less) < 0.05, then you can consider this an unusually low value. Another way to think of this is if the probability of getting a value as small as x is less than 0.05, then the event x is considered unusual.

Why is it "x or more" or "x or less" instead of just "x" when you are determining if an event is unusual? Consider this example: you and your friend go out to lunch every day. Instead of Going Dutch (each paying for their own lunch), you decide to flip a coin, and the loser pays for both. Your friend seems to be winning more often than you'd expect, so you want to determine if this is unusual before you decide to change how you pay for lunch (or accuse your friend of cheating). The process for how to calculate these probabilities will be presented in the next section on the binomial distribution. If your friend won 6 out of 10 lunches, the probability of that happening turns out to be about 20.5%, not unusual. The probability of winning 6 or more is about 37.7%. But what happens if your friend won 501 out of 1,000 lunches? That doesn't seem so unlikely! The probability of winning 501 or more lunches is about 47.8%, and that is consistent with your hunch that this isn't so unusual. But the probability of winning exactly 501 lunches is much less, only about 2.5%. That is why the probability of getting exactly that value is not the right question to ask: you should ask the probability of getting that value or more (or that value or less on the other side).

The value 0.05 will be explained later, and it is not the only value you can use.

Example \(\PageIndex{5}\) is the event unusual

  • Is it unusual for a household to have six people in the family?
  • If you did come upon many families that had six people in the family, what would you think?
  • Is it unusual for a household to have four people in the family?
  • If you did come upon a family that has four people in it, what would you think?

a. To determine this, you need to look at probabilities. However, you cannot just look at the probability of six people. You need to look at the probability of x being six or more people or the probability of x being six or less people. The

\(\begin{aligned} P(x \leq 6) &=P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6) \\ &=26.7 \%+33.6 \%+15.8 \%+13.7 \%+6.3 \%+2.4 \% \\ &=98.5 \% \end{aligned}\)

Since this probability is more than 5%, then six is not an unusually low value. The

\(\begin{aligned} P(x \geq 6) &=P(x=6)+P(x \geq 7) \\ &=2.4 \%+1.5 \% \\ &=3.9 \% \end{aligned}\)

Since this probability is less than 5%, then six is an unusually high value. It is unusual for a household to have six people in the family.

b. Since it is unusual for a family to have six people in it, then you may think that either the size of families is increasing from what it was or that you are in a location where families are larger than in other locations.

c. To determine this, you need to look at probabilities. Again, look at the probability of x being four or more or the probability of x being four or less. The

\(\begin{aligned} P(x \geq 4) &=P(x=4)+P(x=5)+P(x=6)+P(x=7) \\ &=13.7 \%+6.3 \%+2.4 \%+1.5 \% \\ &=23.9 \% \end{aligned}\)

Since this probability is more than 5%, four is not an unusually high value. The

\(\begin{aligned} P(x \leq 4) &=P(x=1)+P(x=2)+P(x=3)+P(x=4) \\ &=26.7 \%+33.6 \%+15.8 \%+13.7 \% \\ &=89.8 \% \end{aligned}\)

Since this probability is more than 5%, four is not an unusually low value. Thus, four is not an unusual size of a family.

d. Since it is not unusual for a family to have four members, then you would not think anything is amiss.

Exercise \(\PageIndex{1}\)

Table \(\PageIndex{8}\) : Number of Days to Fix Defects a. State the random variable. b. Draw a histogram of the number of days to fix defects c. Find the mean number of days to fix defects. d. Find the variance for the number of days to fix defects. e. Find the standard deviation for the number of days to fix defects. f. Find probability that a lens will take at least 16 days to make a fix the defect. g. Is it unusual for a lens to take 16 days to fix a defect? h. If it does take 16 days for eyeglasses to be repaired, what would you think?

  • State the random variable.
  • Write the probability distribution for the number of heads.
  • Draw a histogram for the number of heads.
  • Find the mean number of heads.
  • Find the variance for the number of heads.
  • Find the standard deviation for the number of heads.
  • Find the probability of having two or more number of heads.
  • Is it unusual for to flip two heads?

1. a. See solutions, b. See solutions, c. 4.175 days, d. 8.414375 \(\text { days }^{2}\), e. 2.901 days, f. 0.004, g. See solutions, h. See solutions

how to solve problems involving mean and variance of probability distribution

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3.2.1 - expected value and variance of a discrete random variable.

By continuing with example 3-1, what value should we expect to get?  What would be the average value?     We can answer this question by finding the expected value (or mean) .

For a discrete random variable, the expected value, usually denoted as \(\mu\) or \(E(X)\), is calculated using:

\(\mu=E(X)=\sum x_if(x_i)\)

Example 3-2: Expected Value Section  

In Example 3-1 we were given the following discrete probability distribution:

What is the expected value?

\begin{align} \mu=E(X)=\sum xf(x)&=0\left(\frac{1}{5}\right)+1\left(\frac{1}{5}\right)+2\left(\frac{1}{5}\right)+3\left(\frac{1}{5}\right)+4\left(\frac{1}{5}\right)\\&=2\end{align}

For this example, the expected value was equal to a possible value of X. This may not always be the case. For example, if we flip a fair coin 9 times, how many heads should we expect? We will explain how to find this later but we should expect 4.5 heads. The expected value in this case is not a valid number of heads.

Now that we can find what value we should expect, (i.e. the expected value), it is also of interest to give a measure of the variability.

The variance of a discrete random variable is given by:

\(\sigma^2=\text{Var}(X)=\sum (x_i-\mu)^2f(x_i)\)

The formula means that we take each value of x, subtract the expected value, square that value and multiply that value by its probability. Then sum all of those values.

There is an easier form of this formula we can use.

\(\sigma^2=\text{Var}(X)=\sum x_i^2f(x_i)-E(X)^2=\sum x_i^2f(x_i)-\mu^2\)

The formula means that first, we sum the square of each value times its probability then subtract the square of the mean. We will use this form of the formula in all of our examples.

The standard deviation of a random variable, $X$, is the square root of the variance.

Example 3-3: Standard Deviation Section  

Consider the first example where we had the values 0, 1, 2, 3, 4. The PMF in tabular form was:

Find the variance and the standard deviation of X.

\(\text{Var}(X)=\left[0^2\left(\dfrac{1}{5}\right)+1^2\left(\dfrac{1}{5}\right)+2^2\left(\dfrac{1}{5}\right)+3^2\left(\dfrac{1}{5}\right)+4^2\left(\dfrac{1}{5}\right)\right]-2^2=6-4=2\)

\(\text{SD}(X)=\sqrt{2}\approx 1.4142\)

Example 3-4: Prior Convictions Section  

Click on the tab headings to see how to find the expected value, standard deviation, and variance. The last tab is a chance for you to try it.

  • Number of Prior Convictions
  • Expected Value
  • Variance and Standard Deviation

Barbed wire

Let X = number of prior convictions for prisoners at a state prison at which there are 500 prisoners. (\(x = 0,1,2,3,4\))

What is the expected value for number of prior convictions?

For this we need a weighted average since not all the outcomes have equal chance of happening (i.e. they are not equally weighted). So, we need to find our expected value of \(X\), or mean of \(X\), or \(E(X) = \Sigma f(x_i)(x_i)\). When we write this out it follows:

\(=(0.16)(0)+(0.53)(1)+(0.2)(2)+(0.08)(3)+(0.03)(4)=1.29\)

Calculate the variance and the standard deviation for the Prior Convictions example:

Using the data in our example we find that...

\begin{align} \text{Var}(X) &=[0^2(0.16)+1^2(0.53)+2^2(0.2)+3^2(0.08)+4^2(0.03)]–(1.29)^2\\ &=2.53–1.66\\ &=0.87\\ \text{SD}(X) &=\sqrt(0.87)\\ &=0.93 \end{align}

What is the expected number of prior convictions? Below is the probability distribution table for the prior conviction data. Use this table to answer the questions that follow.

3. Mean, Variance, and Standard Deviation for a Probability Distribution

Video tutorials to watch:

How to find the Mean and Standard Deviation of a Binomial Probability Distribution

Statistics Home

Introduction to Engineering Statistics

3.1) PMF, Mean, & Variance

A probability distribution is a mathematical function that describes an experiment by providing the probabilities that different possible outcomes will occur. Probability distributions are defined in terms of random variables , which are variables whose values depend on outcomes of a random phenomenon. By convention, random variables are capitalized, and their corresponding values are written in lower case. A random variable, \(Y\), describing the roll of a single die, would have six possible values, where \(y_1, y_2, y_3, y_4, y_5,\) and \(y_6\) would correspond to the die roll being a 1, 2, 3, 4, 5, or 6.

A probability distribution is specified in terms of an underlying sample space, or support , which is the set of all possible outcomes of the random phenomenon being observed. The sample space may be the set of real numbers or a set of vectors, or it may be a list of non-numerical values (eye color, political party, shoe size, etc.). For example, the sample space of a coin toss would be {heads, tails}, and if the random variable \(X\) was used to denote the outcome of a coin toss ("the experiment"), then the probability distribution of \(X\) would take the value 0.5 for \(X\) = heads, and 0.5 for \(X\) = tails because each outcome occurs with a probability of 50%.

Probability distributions are generally divided into two classes. A discrete probability distribution (applicable to the scenarios where the set of possible outcomes is discrete, such as a coin toss or a roll of dice) can be encoded by a discrete list of the probabilities of the outcomes, known as a probability mass function , or PMF . On the other hand, a continuous probability distribution (applicable to the scenarios where the set of possible outcomes can take on values in a continuous range, such as the temperature on a given day) is typically described by a probability density function , or PDF . In a continuous probability distribution, the probability of any individual outcome occurring is actually 0 (as the acceptable outcome region shrinks down to a single point, the probability shrinks to zero).

The Probability Mass Function

If an experiment has \(k\) possible distinct outcomes, then we can describe those outcomes using the discrete random variable \(X\), consisting of the values \(x_0, x_1, x_2, \ldots, x_k\). The corresponding probabilities that the outcomes occur would be given by \(p(x_0), p(x_1), p(x_2), \ldots, p(x_k)\). The function \(p(x)\) is a valid probability mass function if the following two constraints are satisfied:

\(0\lt p(x)\le 1 \hspace{20pt} \textrm{ for any } x \in \{x_1,x_2,\ldots,x_k \}\) and \(\displaystyle \sum_{i=0}^k p(x_i)=1\)

In other words, all discrete probabilities must be above 0% and no more than 100%, and the probabilities of all of the outcomes must sum to 100%. Note that an alternative notation for the probability function is \(P(X=x)\); for example, \(p(x_2)\) could be written \(P(X=x_2)\).

The Cumulative Mass Function

In some cases, we are interested in a range of possible outcomes rather than in one individual outcome.  For example, a manufacturing plant might want to know the probability of a shipment containing up to 2 defective units.  In that case, we would sum the individual probabilities for 0 defective units, 1 defective unit, and 2 defective units.  In general, the probability that a random variable, \(X\), has a value less than or equal to \(x\) is given by a Cumulative Mass Function , or CMF , defined as a sum of a portion of the PMF:

\(F(x)=P(X\le x)=\displaystyle \sum_{x_i\le x} p(x_i)\) where \(p(x_i)\) is the i th individual value of the PMF \(p(x)\).

We can find the probability of a range of outcomes by subtracting CMFs with different boundaries. For example, the probability of flipping five coins and getting 1, 2, or 3 heads would be \(F(3) - F(1)\) or \(26/32 - 1/32 = 25/32\). Similarly, we can find any individual value \(p(x_i)\) by taking \(F(x_i)\) and subtracting \(F(x_{i-1})\).

Expectation and Variance

We can restate the earlier results for the expected value and the variance in terms of probability mass functions and cumulative mass functions. The expectation, or expected value, of a random variable is the arithmetic mean of all possible results for an infinite number of trials. The expected value should closely approximate the mean result from a large series of trials following a particular probability function.

Since the expected value includes all possible results, we must know the complete probability function in order to calculate the expectation. The expected value of a random variable \(X\) is denoted by \(E(X)\) or \(\overline X\) or \(\lt X\gt\), depending on which branch of math, science, or engineering that is performing the calculation. The expected value is denoted \(\mu_x\), or simply \(\mu\), to indicate that the expected value is the mean value of the whole distribution of the random variable. The expected value is given by

Using the example in the previous section, the expected value of \(X\) (the number of heads obtained on a toss of five coins) is given by \(\mu_x = E(X) = (0)(\frac{1}{32})+(1)(\frac{5}{32})+(2)(\frac{10}{32})+(3)(\frac{10}{32})+(4)(\frac{5}{32})+(5)(\frac{1}{32}) = 2.500\)

Note that the expected value is not necessarily a possible outcome from a single trial.

A life insurance company has data indicating that the probability of a forty-year-old man surviving to age 70 is 0.995. The probability that he will die during that time is therefore 1-0.995, or 0.005. The company sells $20,000 life insurance policies for this length of time, and they want to earn an average profit of $50 per policy. How much should the company charge a forty-year-old man to buy the $20,000 policy?

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The variance of a frequency distribution is the average of the squared deviations from the mean: \(\sum_{i=1}^N (x_i-\mu_x)^2 / N\). Since the mean is also the expected value, the variance of a discrete random variable can be expressed as \begin{align}%\label{} \nonumber \sigma_x^2 &= E(x-\mu_x)^2\\ \textrm{ } \\ &= \sum_{i} (x_i-\mu_x)^2 P(x_i)\\ \textrm{ } \\ &\textrm{ or }\\ \textrm{ } \\ \sigma_x^2 &=E(X^2)-(\;E(X)\;)^2 \hspace{20pt}\textrm{ as described when introducing descriptive measures} \end{align}

The standard deviation is always the square root of the variance \begin{align}%\label{} \nonumber \sigma_x &= \sqrt{E(\;(x-\mu_x)^2\;)}\\ \textrm{ } \\ &\textrm{ or }\\ \textrm{ } \\ \sigma_x &=\sqrt{E(X^2)-(\;E(X)\;)^2} \end{align}

Referring to the example for five coin tosses, we can find the standard deviation using the two expected values. \(E(X^2) = (0^2)(\frac{1}{32})+(1^2)(\frac{5}{32})+(2^2)(\frac{10}{32})+(3^2)(\frac{10}{32})+(4^2)(\frac{5}{32})+(5^2)(\frac{1}{32}) = 7.500\)

and as shown in a previous calculation \(E(X) = (0)(\frac{1}{32})+(1)(\frac{5}{32})+(2)(\frac{10}{32})+(3)(\frac{10}{32})+(4)(\frac{5}{32})+(5)(\frac{1}{32}) = 2.500\)

Thus \(\sigma_x =\sqrt{E(X^2)-(\;E(X)\;)^2} = \sqrt{7.500 - (2.500)^2} = \sqrt{1.25} = 1.118\)

A probability function is defined as \(p(0) = 0.3164\), \(p(1)=0.4219\), \(p(2)=0.2109\), \(p(3)=0.0469\), and \(p(4)=0.0039\). Find its mean and variance.

Three items are selected at random without replacement from a box containing ten items, of which four are defective.

a) Calculate the probability distribution for the number of defective items chosen.

b) Calculate the expected number of defective items chosen.

c) Calculate the variance and standard deviation of the number of defective items chosen.

Probability is the ratio of intended outcomes to total outcomes. The total outcomes would be the number of possible ways to choose 3 items from a pool of 10, or \(_{10}C_3 = 120\).

We need to treat the intended outcomes separately for each case in the distribution, and we need to count both defective and non-defective items:

0 defective and 3 non-defective: \((_4C_0)(_6C_3) = (1)(20) = 20\)

1 defective and 2 non-defective: \((_4C_1)(_6C_2) = (4)(15) = 60\)

2 defective and 1 non-defective: \((_4C_2)(_6C_1) = (6)(6) = 36\)

3 defective and 0 non-defective: \((_4C_3)(_6C_0) = (4)(1) = 4\)

a) If we use a random variable \(X\) to denote the number of defective items, where \(X\) = 0, 1, 2, or 3. Then \(P(X)\) = 20/120, 60/120, 36/120, or 4/120. Reducing the fractions gives the following probability mass function: P(0) = 1/6 P(1) = 1/2 P(2) = 3/10 P(3) = 1/30

Check: \(P(0)+P(1)+P(2)+P(3) = 100\%\;\;\;\;\;\;\;\)✔

b) The expected value is \begin{align}%\label{} \nonumber E(X) &= (0)(P(0)) + (1)(P(1)) + (2)(P(2)) + (3)(P(3)) \\ \textrm{ } \\ &= 0 + 1/2 + 3/5 + 1/10 \\ \textrm{ } \\ &= 1.2 \end{align} c) The variance is \(E(X^2)-(\;E(X)\;)^2\) \begin{align}%\label{} \nonumber E(X^2) &= (0^2)(P(0)) + (1^2)(P(1)) + (2^2)(P(2)) + (3^2)(P(3)) \\ \textrm{ } \\ &= 0 + 1/2 + 12/10 + 9/30 = 2 \end{align}

System Reliability

System diagrams may be used to create a probability mass function for the system using the rules for series and parallel connections. In the example system below, the total system output rate is determined by whether components A, B, C, and D operate.

The probabilities that components A, B, C, and D fail are as follows: \(P(A) = 0.30\),\( P(B) = 0.40\),\( P(C) = 0.10\), and \( P(D) = 0.15\). The reliabilities are then 0.70, 0.60, 0.90, and 0.85 respectively.

If no components fail, then the system output is 100. If component D fails, or if component C and either A or B fail, the system output is zero. If only component C fails, the system output is 80. If either component A or B fails, the system output is 20. Find the expected value and standard deviation of the output.

If only component C fails (so components A, B, and D do not fail), then the output is 80. The probability that A, B, and D do not fail is the product of the three reliabilities: \((0.70)(0.60)(0.85) = 0.357\). \begin{align*} P(80) &= P(\textrm{ C fails })\times P(\textrm{ A, B and D don't fail }) \\ \textrm{ } \\ &= (0.10)(0.357) \\ \textrm{ } \\ &= 0.0357 \end{align*}

Finally, we get 100 output if no component fails \begin{align*} P(100) &= (0.70)(0.60)(0.90)(0.85) \\ \textrm{ } \\ &= 0.3213 \end{align*}

So, our probability mass function is: \begin{align*} P(0) &= 0.1993 \\ \textrm{ } \\ P(20) &= 0.4437 \\ \textrm{ } \\ P(80) &= 0.0357 \\ \textrm{ } \\ P(100) &= 0.3213 \end{align*}

The expected value of the output is \begin{align*} E(\textrm{ Output }) &= (0)(P(0)) + (20)(P(20)) + (80)(P(80)) + (100)(P(100))\\ \textrm{ } \\ &= 0 + (20)(0.4437) + (80)(0.0357) + (100)(0.3213)\\ \textrm{ } \\ &= 43.86 \end{align*}

The second term in the variance formula is \(E(\textrm{ Output}^2)\) which is \begin{align*} E(\textrm{ Output}^2) &= (0^2)(P(0)) + (20^2)(P(20)) + (80^2)(P(80)) + (100^2)(P(100))\\ \textrm{ } \\ &= 0 + (400)(0.4437) + (6400)(0.0357) + (10000)(0.3213)\\ \textrm{ } \\ &= 3618.96 \end{align*}

The variance and standard deviation are then given by \begin{align*} \textrm{ variance } = \sigma^2 &= E(\textrm{ Output}^2) - (\;E(\textrm{Output})\;)^2 \\ \textrm{ } \\ &= 3618.96 - 43.86^2 \\ \textrm{ } \\ &= 1695.26\\ \textrm{ } \\ \\ \textrm{ } \\ \sigma &= \sqrt{1695.26} \\ \textrm{ } \\ &= 41.17 \end{align*}

In some situations, it is easier to use a spreadsheet to make a table of all of the possible outcomes and mark the output values for each one. The probability for each outcome is then the product of the A, B, C, and D factors, and the probability for each output level can be found by summing the rows with matching output levels.

Example Involving Bayes' Rule

In some cases, Bayes' Rule must be used to reverse the direction of a dependent probability when constructing the PMF and finding the expected value. The example below involves two related probability trees: one for the chance of a flood warning, and the other for the chance of a flood occurring.

A flood forecaster issues a flood warning under two conditions only: i) Winter snowfall exceeds 20 cm, regardless of fall rainfall ii) Fall rainfall exceeds 10 cm and winter snowfall is between 15 and 20 cm

The probability of winter snowfall exceeding 20cm is 0.05. The probability of winter snowfall of between 15 and 20 cm is 0.10. The probability of fall rainfall exceeding 10 cm is 0.10.

The probability that flooding occurs is 0.75 for condition (i) above, 0.60 for condition (ii) above, and 0.05 for conditions where no flooding was anticipated.

The cost of a flood after a warning is $100,000. The cost of a flood with no warning is $1,000,000. The cost of a warning without a flood is $50,000. The cost with no flood and no warning is $0.

a) What is the probability that a flood warning will be issued? b) Given that a warning was issued, what is the probability that winter snowfall was greater than 20 cm? c) What is the expected cost in any given year?

b) Bayes' Rule says P(snow>20 | warning) = P(snow>20 AND warning) / P(warning) = (0.05)(1.00) / (0.06) = 0.83

c) To find the expected cost, we need to find the probabilities for each flooding outcome. P(type (i) and flood) = (0.05)(1.00)(0.75) = 0.0375 P(type (i) no flood) = (0.05)(1.00)(0.25) = 0.0125 P(type (ii) and flood) = (0.10)(0.10)(0.60) = 0.006 P(type (ii) no flood) = (0.10)(0.10)(0.40) = 0.004 P(no warning and flood) = ((0.10)(0.90)+(0.85)(1.00))(0.05) = 0.047 P(no warning no flood) = ((0.10)(0.90)+(0.85)(1.00))(0.95) = 0.893

E(cost) = (100,000)(0.0375) + (50,000)(0.0125) + (100,000)(0.006) + (50,000)(0.004) + (1,000,000)(0.047) + (0)(0.893) E(cost) = $52,175

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5.3: Expectation, Variance and Standard Deviation

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The expected value is often referred to as the "long-term" average or mean. This means that over the long term of doing an experiment over and over, you would expect this average.

You toss a coin and record the result. What is the probability that the result is heads? If you flip a coin two times, does probability tell you that these flips will result in one heads and one tail? You might toss a fair coin ten times and record nine heads. As you learned in Chapter 3, probability does not describe the short-term results of an experiment. It gives information about what can be expected in the long term. To demonstrate this, Karl Pearson once tossed a fair coin 24,000 times! He recorded the results of each toss, obtaining heads 12,012 times. In his experiment, Pearson illustrated the Law of Large Numbers.

The Law of Large Numbers states that, as the number of trials in a probability experiment increases, the difference between the theoretical probability of an event and the relative frequency approaches zero (the theoretical probability and the relative frequency get closer and closer together). When evaluating the long-term results of statistical experiments, we often want to know the “average” outcome. This “long-term average” is known as the mean or expected value of the experiment and is denoted by the Greek letter \(\mu\). In other words, after conducting many trials of an experiment, you would expect this average value.

To find the expected value or long term average, \(\mu\), simply multiply each value of the random variable by its probability and add the products.

Example \(\PageIndex{1}\)

A men's soccer team plays soccer zero, one, or two days a week. The probability that they play zero days is 0.2, the probability that they play one day is 0.5, and the probability that they play two days is 0.3. Find the long-term average or expected value, \(\mu\), of the number of days per week the men's soccer team plays soccer.

To do the problem, first let the random variable \(X =\) the number of days the men's soccer team plays soccer per week. \(X\) takes on the values 0, 1, 2. Construct a PDF table adding a column \(x*P(x)\). In this column, you will multiply each \(x\) value by its probability.

Add the last column \(x*P(x)\) to find the long term average or expected value:

\[(0)(0.2) + (1)(0.5) + (2)(0.3) = 0 + 0.5 + 0.6 = 1.1. \nonumber\]

The expected value is 1.1. The men's soccer team would, on the average, expect to play soccer 1.1 days per week. The number 1.1 is the long-term average or expected value if the men's soccer team plays soccer week after week after week. We say \(\mu = 1.1\).

Example \(\PageIndex{2}\)

Find the expected value of the number of times a newborn baby's crying wakes its mother after midnight. The expected value is the expected number of times per week a newborn baby's crying wakes its mother after midnight. Calculate the standard deviation of the variable as well.

Add the values in the third column of the table to find the expected value of \(X\):

\[\mu = \text{Expected Value} = \dfrac{105}{50} = 2.1 \nonumber\]

Use \(\mu\) to complete the table. The fourth column of this table will provide the values you need to calculate the standard deviation. For each value \(x\), multiply the square of its deviation by its probability. (Each deviation has the format \(x – \mu\).

Add the values in the fourth column of the table:

\[0.1764 + 0.2662 + 0.0046 + 0.1458 + 0.2888 + 0.1682 = 1.05 \nonumber\]

The standard deviation of \(X\) is the square root of this sum: \(\sigma = \sqrt{1.05} \approx 1.0247\)

The mean, μ , of a discrete probability function is the expected value.

\[μ=∑(x∙P(x))\nonumber\]

The standard deviation, Σ, of the PDF is the square root of the variance.

\[σ=\sqrt{∑[(x – μ)2 ∙ P(x)]}\nonumber\]

When all outcomes in the probability distribution are equally likely, these formulas coincide with the mean and standard deviation of the set of possible outcomes.

Suppose you play a game of chance in which five numbers are chosen from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. A computer randomly selects five numbers from zero to nine with replacement. You pay $2 to play and could profit $100,000 if you match all five numbers in order (you get your $2 back plus $100,000). Over the long term, what is your expected profit of playing the game?

To do this problem, set up an expected value table for the amount of money you can profit.

Let \(X =\) the amount of money you profit. The values of \(x\) are not 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Since you are interested in your profit (or loss), the values of \(x\) are 100,000 dollars and −2 dollars.

To win, you must get all five numbers correct, in order. The probability of choosing one correct number is \(\dfrac{1}{10}\) because there are ten numbers. You may choose a number more than once. The probability of choosing all five numbers correctly and in order is

\[\begin{align*} \left(\dfrac{1}{10}\right) \left(\dfrac{1}{10}\right) \left(\dfrac{1}{10}\right) \left(\dfrac{1}{10}\right) \left(\dfrac{1}{10}\right) &= (1)(10^{-5}) \\[5pt] &= 0.00001. \end{align*}\]

Therefore, the probability of winning is 0.00001 and the probability of losing is

\[1−0.00001=0.99999.1−0.00001 = 0.99999.\nonumber\]

The expected value table is as follows:

Since –0.99998 is about –1, you would, on average, expect to lose approximately $1 for each game you play. However, each time you play, you either lose $2 or profit $100,000. The $1 is the average or expected LOSS per game after playing this game over and over.

Example \(\PageIndex{4}\)

Suppose you play a game with a biased coin. You play each game by tossing the coin once. \(P(\text{heads}) = \dfrac{2}{3}\) and \(P(\text{tails}) = \dfrac{1}{3}\). If you toss a head, you pay $6. If you toss a tail, you win $10. If you play this game many times, will you come out ahead?

  • Define a random variable \(X\).
  • Complete the following expected value table.
  • What is the expected value, \(\mu\)? Do you come out ahead?

\(X\) = amount of profit

Add the last column of the table. The expected value \(\mu = \dfrac{-2}{3}\). You lose, on average, about 67 cents each time you play the game so you do not come out ahead.

Like data, probability distributions have standard deviations. To calculate the standard deviation ( σ ) of a probability distribution, find each deviation from its expected value, square it, multiply it by its probability, add the products, and take the square root. To understand how to do the calculation, look at the table for the number of days per week a men's soccer team plays soccer. To find the standard deviation, add the entries in the column labeled \((x) – \mu^{2}P(x)\) and take the square root.

Add the last column in the table. \(0.242 + 0.005 + 0.243 = 0.490\). The standard deviation is the square root of 0.49, or \(\sigma = \sqrt{0.49} = 0.7\)

Generally for probability distributions, we use a calculator or a computer to calculate \(\mu\) and \(\sigma\) to reduce roundoff error. For some probability distributions, there are short-cut formulas for calculating \(\mu\) and \(\sigma\).

Example \(\PageIndex{5}\)

Toss a fair, six-sided die twice. Let \(X\) = the number of faces that show an even number. Construct a table like Table and calculate the mean \(\mu\) and standard deviation \(\sigma\) of \(X\).

Tossing one fair six-sided die twice has the same sample space as tossing two fair six-sided dice. The sample space has 36 outcomes:

Use the sample space to complete the following table:

Add the values in the third column to find the expected value: \(\mu\) = \(\dfrac{36}{36}\) = 1. Use this value to complete the fourth column.

Add the values in the fourth column and take the square root of the sum:

\[\sigma = \sqrt{\dfrac{18}{36}} \approx 0.7071.\]

Example \(\PageIndex{6}\)

On May 11, 2013 at 9:30 PM, the probability that moderate seismic activity (one moderate earthquake) would occur in the next 48 hours in Iran was about 21.42%. Suppose you make a bet that a moderate earthquake will occur in Iran during this period. If you win the bet, you win $50. If you lose the bet, you pay $20. Let X = the amount of profit from a bet.

\(P(\text{win}) = P(\text{one moderate earthquake will occur}) = 21.42%\)

\(P(\text{loss}) = P(\text{one moderate earthquake will not occur}) = 100% – 21.42%\)

If you bet many times, will you come out ahead? Explain your answer in a complete sentence using numbers. What is the standard deviation of \(X\)? Construct a table similar to Table and Table to help you answer these questions.

Mean = Expected Value = 10.71 + (–15.716) = –5.006.

If you make this bet many times under the same conditions, your long term outcome will be an average loss of $5.01 per bet.

Standard Deviation \(= \sqrt{648.0964+176.6636} \approx 28.7186\)

Some of the more common discrete probability functions are binomial, geometric, hypergeometric, and Poisson. Most elementary courses do not cover the geometric, hypergeometric, and Poisson. Your instructor will let you know if he or she wishes to cover these distributions.

A probability distribution function is a pattern. You try to fit a probability problem into a pattern or distribution in order to perform the necessary calculations. These distributions are tools to make solving probability problems easier. Each distribution has its own special characteristics. Learning the characteristics enables you to distinguish among the different distributions.

The expected value, or mean, of a discrete random variable predicts the long-term results of a statistical experiment that has been repeated many times. The standard deviation of a probability distribution is used to measure the variability of possible outcomes.

Formula Review

  • Mean or Expected Value: \(\mu = \sum_{x \in X}xP(x)\)
  • Standard Deviation: \(\sigma = \sqrt{\sum_{x \in X}(x - \mu)^{2}P(x)}\)

WeBWorK Problems

Query \(\PageIndex{1}\)

Query \(\PageIndex{2}\)

Query \(\PageIndex{3}\)

Query \(\PageIndex{4}\)

Query \(\PageIndex{5}\)

Query \(\PageIndex{6}\)

Query \(\PageIndex{7}\)

Query \(\PageIndex{8}\)

Query \(\PageIndex{9}\)

Query \(\PageIndex{10}\)

Query \(\PageIndex{11}\)

Query \(\PageIndex{12}\)

  • Class Catalogue at the Florida State University. Available online at apps.oti.fsu.edu/RegistrarCo...archFormLegacy (accessed May 15, 2013).
  • “World Earthquakes: Live Earthquake News and Highlights,” World Earthquakes, 2012. www.world-earthquakes.com/ind...thq_prediction (accessed May 15, 2013).

Contributors and Attributions

Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] .

Normal Distribution Problems with Solutions

Problems and applications on normal distributions are presented. The solutions to these problems are at the bottom of the page. An online normal probability calculator and an inverse normal probability calculator may be useful to check your answers.

Problems with Solutions

  • X is a normally distributed variable with mean = 30 and standard deviation = 4. Find the probabilities a) P(X < 40) b) P(X > 21) c) P(30 < X < 35)  
  • A radar unit is used to measure the speeds of cars on a motorway. The speeds are normally distributed with a mean of 90 km/hr and a standard deviation of 10 km/hr. What is the probability that a car picked at random is traveling at more than 100 km/hr?  
  • For certain types of computers, the length of time between charges of the battery is normally distributed with a mean of 50 hours and a standard deviation of 15 hours. John owns one of these computers and wants to know the probability that the length of time will be between 50 and 70 hours.  
  • Entry to a certain University is determined by a national test. The scores on this test are normally distributed with a mean of 500 and a standard deviation of 100. Tom wants to be admitted to this university and he knows that he must score better than at least 70% of the students who took the test. Tom takes the test and scores 585. Will he be admitted to this university?  
  • The length of similar components produced by a company is approximated by a normal distribution model with a mean of 5 cm and a standard deviation of 0.02 cm. If a component is chosen at random a) what is the probability that the length of this component is between 4.98 and 5.02 cm? b) What is the probability that the length of this component is between 4.96 and 5.04 cm?  
  • The length of life of an instrument produced by a machine has a normal distribution with a mean of 12 months and a standard deviation of 2 months. Find the probability that an instrument produced by this machine will last a) less than 7 months. b) between 7 and 12 months.  
  • The time taken to assemble a car in a certain plant is a random variable having a normal distribution of 20 hours and a standard deviation of 2 hours. What is the probability that a car can be assembled at this plant in a period of time a) less than 19.5 hours? b) between 20 and 22 hours?  
  • A large group of students took a test in Physics and the final grades have a mean of 70 and a standard deviation of 10. If we can approximate the distribution of these grades by a normal distribution, what percent of the students a) scored higher than 80? b) Should pass the test (grades60)? c) Should fail the test (grades<60)?  
  • The annual salaries of employees in a large company are approximately normally distributed with a mean of $50,000 and a standard deviation of $20,000. a) What percent of people earn less than $40,000? b) What percent of people earn between $45,000 and $65,000? c) What percent of people earn more than $70,000?  

Solutions to the Above Problems

  • Note: What is meant here by area is the area under the standard normal curve. a) For x = 40, the z-value z = (40 - 30) / 4 = 2.5 Hence P(x < 40) = P(z < 2.5) = [area to the left of 2.5] = 0.9938 b) For x = 21, z = (21 - 30) / 4 = -2.25 Hence P(x > 21) = P(z > -2.25) = [total area] - [area to the left of -2.25] = 1 - 0.0122 = 0.9878 c) For x = 30 , z = (30 - 30) / 4 = 0 and for x = 35, z = (35 - 30) / 4 = 1.25 Hence P(30 < x < 35) = P(0 < z < 1.25) = [area to the left of z = 1.25] - [area to the left of 0] = 0.8944 - 0.5 = 0.3944  
  • Let x be the random variable that represents the speed of cars. x has = 90 and = 10. We have to find the probability that x is higher than 100 or P(x > 100) For x = 100 , z = (100 - 90) / 10 = 1 P(x > 90) = P(z > 1) = [total area] - [area to the left of z = 1] = 1 - 0.8413 = 0.1587 The probability that a car selected at random has a speed greater than 100 km/hr is equal to 0.1587  
  • Let x be the random variable that represents the length of time. It has a mean of 50 and a standard deviation of 15. We have to find the probability that x is between 50 and 70 or P( 50< x < 70) For x = 50 , z = (50 - 50) / 15 = 0 For x = 70 , z = (70 - 50) / 15 = 1.33 (rounded to 2 decimal places) P( 50< x < 70) = P( 0< z < 1.33) = [area to the left of z = 1.33] - [area to the left of z = 0] = 0.9082 - 0.5 = 0.4082 The probability that John's computer has a length of time between 50 and 70 hours is equal to 0.4082.  
  • Let x be the random variable that represents the scores. x is normally distributed with a mean of 500 and a standard deviation of 100. The total area under the normal curve represents the total number of students who took the test. If we multiply the values of the areas under the curve by 100, we obtain percentages. For x = 585 , z = (585 - 500) / 100 = 0.85 The proportion P of students who scored below 585 is given by P = [area to the left of z = 0.85] = 0.8023 = 80.23% Tom scored better than 80.23% of the students who took the test and he will be admitted to this University.  
  • a) P(4.98 < x < 5.02) = P(-1 < z < 1) = 0.6826 b) P(4.96 < x < 5.04) = P(-2 < z < 2) = 0.9544  
  • a) P(x < 7) = P(z < -2.5) = 0.0062 b) P(7 < x < 12) = P(-2.5 < z < 0) = 0.4938  
  • a) P(x < 19.5) = P(z < -0.25) = 0.4013 b) P(20 < x < 22) = P(0 < z < 1) = 0.3413  
  • a) For x = 80, z = 1 Area to the right (higher than) z = 1 is equal to 0.1586 = 15.87% scored more that 80. b) For x = 60, z = -1 The area to the right of z = -1 is equal to 0.8413 = 84.13% should pass the test. c) 100% - 84.13% = 15.87% should fail the test.  
  • a) For x = 40000, z = -0.5 The area to the left (less than) of z = -0.5 is equal to 0.3085 = 30.85% earn less than $40,000. b) For x = 45000 , z = -0.25 and for x = 65000, z = 0.75 The area between z = -0.25 and z = 0.75 is equal to 0.3720 = 37.20 earning between $45,000 and $65,000. c) For x = 70000, z = 1 The area to the right (higher) of z = 1 is equal to 0.1586 = 15.86% earning more than $70,000.

More References and links

  • Normal Distribution Definition
  • Elementary statistics and probabilities

Statology

Statistics Made Easy

How to Find the Mean of a Probability Distribution (With Examples)

A probability distribution tells us the probability that a random variable takes on certain values.

For example, the following probability distribution tells us the probability that a certain soccer team scores a certain number of goals in a given game:

how to solve problems involving mean and variance of probability distribution

Note: The probabilities in a valid probability distribution will always add up to 1. We can confirm that this probability distribution is valid: 0.18 + 0.34 + 0.35 + 0.11 + 0.02 = 1.

To find the  mean (sometimes called the “expected value”) of any probability distribution, we can use the following formula:

For example, consider our probability distribution for the soccer team:

The mean number of goals for the soccer team would be calculated as:

μ = 0*0.18  +  1*0.34  +  2*0.35  +  3*0.11  +  4*0.02  =   1.45 goals.

The following examples show how to calculate the mean of a probability distribution in a few other scenarios.

Example 1: Mean Number of Vehicle Failures

The following probability distribution tells us the probability that a given vehicle experiences a certain number of battery failures during a 10-year span:

Example of finding the mean of a probability distribution

Question: What is the mean number of expected failures for this vehicle?

Solution: The mean number of expected failures is calculated as:

μ = 0*0.24  +  1*0.57  +  2*0.16  +  3*0.03 =  0.98 failures.

Example 2: Mean Number of Wins

The following probability distribution tells us the probability that a given basketball team wins a certain number of games in a tournament:

Mean of probability distribution

Question: What is the mean number of expected wins for this team?

Solution: The mean number of expected wins is calculated as:

μ = 0*.06  +  1*.15  +  2*0.17  +  3*0.24  +  4*.23  +  5*.09  +  6*.06  =  2.94 wins.

Example 3: Mean Number of Sales

The following probability distribution tells us the probability that a given salesman will make a certain number of sales in the upcoming month:

how to solve problems involving mean and variance of probability distribution

Question: What is the mean number of expected sales for this salesman in the upcoming month?

Solution: The mean number of expected sales is calculated as:

μ = 10*.24  +  20*.31  +  30*0.39  +  40*0.06  =  22.7 sales.

Bonus: Probability Distribution calculator

You can use this calculator to automatically calculate the mean of any probability distribution.

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Probability Distribution Mean (Expectation), Variance :: Problems

© Krishbhavara ♣

Random Variables: Mean, Variance and Standard Deviation

A Random Variable is a set of possible values from a random experiment.

Example: Tossing a coin: we could get Heads or Tails.

Let's give them the values Heads=0 and Tails=1 and we have a Random Variable "X":

  • We have an experiment (like tossing a coin)
  • We give values to each event
  • The set of values is a Random Variable

Learn more at Random Variables .

Mean, Variance and Standard Deviation

single die

Example: Tossing a single unfair die

For fun, imagine a weighted die (cheating!) so we have these probabilities:

Mean or Expected Value: μ

When we know the probability p of every value x we can calculate the Expected Value (Mean) of X:

μ = Σxp

Note: Σ is Sigma Notation , and means to sum up.

To calculate the Expected Value:

  • multiply each value by its probability
  • sum them up

Example continued:

μ = Σxp = 0.1+0.2+0.3+0.4+0.5+3 = 4.5

The expected value is 4.5

Note: this is a weighted mean : values with higher probability have higher contribution to the mean.

Variance: Var(X)

The Variance is:

Var(X) = Σx 2 p − μ 2

To calculate the Variance :

  • square each value and multiply by its probability
  • sum them up and we get Σx 2 p
  • then subtract the square of the Expected Value μ 2

Σx 2 p = 0.1+0.4+0.9+1.6+2.5+18 = 23.5

Var(X) = Σx 2 p − μ 2 = 23.5 - 4.5 2 = 3.25

The variance is 3.25

Standard Deviation: σ

The Standard Deviation is the square root of the Variance:

σ = √Var(X)

σ = √Var(X) = √3.25 = 1.803...

The Standard Deviation is 1.803...

Let's have another example!

(Note that we run the table downwards instead of along this time.)

fried chicken

You plan to open a new McDougals Fried Chicken, and found these stats for similar restaurants:

Using that as probabilities for your new restaurant's profit, what is the Expected Value and Standard Deviation?

The Random Variable is X = 'possible profit'.

Sum up xp and x 2 p :

μ = Σxp = 25

Var(X) = Σx 2 p − μ 2 = 3750 − 25 2 = 3750 − 625 = 3125

σ = √3125 = 56 (to nearest whole number)

But remember these are in thousands of dollars, so:

  • μ = $25,000
  • σ = $56,000

So you might expect to make $25,000, but with a very wide deviation possible.

Let's try that again, but with a much higher probability for $50,000:

Example (continued):

Now with different probabilities (the $50,000 value has a high probability of 0.7 now):

μ = Σxp = 45

Var(X) = Σx 2 p − μ 2 = 4250 − 45 2 = 4250 − 2025 = 2225

σ = √2225 = 47 (to nearest whole number)

In thousands of dollars:

  • μ = $45,000
  • σ = $47,000

The mean is now much closer to the most probable value.

And the standard deviation is a little smaller (showing that the values are more central.)

Random Variables can be either Discrete or Continuous :

  • Discrete Data can only take certain values (such as 1,2,3,4,5)
  • Continuous Data can take any value within a range (such as a person's height)

Here we looked only at discrete data, as finding the Mean, Variance and Standard Deviation of continuous data needs Integration .

  • A Random Variable is a variable whose possible values are numerical outcomes of a random experiment.
  • The Mean (Expected Value) is: μ = Σxp
  • The Variance is: Var(X) = Σx 2 p − μ 2
  • The Standard Deviation is: σ = √Var(X)

Solutions to Mean and Variance Problems

Practice problems, solutions to practice problems, solution problem 1, solution problem 2, solution problem 3, solution problem 4, solution problem 5, solution problem 6, solution problem 7.

In previous sections of this guide, you learned how to calculate measures of central tendency and variability . Specifically, we showed you the formulas for mean and variance, how to interpret these measures and why they are used. In this section, we’ll put what you’ve learned to the test with various practice problems meant to challenge and reinforce the knowledge you’ve built.

Paolo

As a brief overview of the mean, recall that it is one of the most common measures of central tendency. The mean is defined as an average of a variable or data set and is typically used when there are no extreme values. This is because the mean is sensitive to outliers. Below, you can find the formulas for the various means we’ve explained throughout this guide on descriptive statistics.

The variance, unlike the mean, is a measure of variability. Meaning, it captures the amount of spread within a variable or data set. Measures of variability are often reported alongside measures of central tendency. Meaning, the variance is typically taken with the mean or the median, as this helps form a greater picture of what the data set looks like. Below are the formulas for the variance.

Below, you’ll find practice problems for the mean and variance. Don’t worry if you don’t get the answer right away, you can always compare it to the solutions below.

Find the group mean of the data below.

Find the final grade of this student given the following data.

Find the geometric mean of the following series.

Given the following information, find the population variance .

Find the variance of the following data.

Calculate the variance of the data sets given the following information.

Interpret the variance and mean given.

Below are the solutions to the practice problems above. Make sure to first try them on your own before checking for the solution. If some of the answers are still slightly confusing to you, make sure to review the other sections of our guide dealing with measures of central tendency and variability.

To find the group mean, you simply need to follow the formula,

x_{group} = \frac{\Sigma(f_{i}*x_{m})}{n}

Plugging this into the formula , we get the group mean as,

\dfrac{2464.5}{61} = 40.4

Looking at the chart below, you can see the placement of the mean within the data set.

how to solve problems involving mean and variance of probability distribution

To find the final grade of this student, we have to find the weighted mean , following the formula,

\bar{x}_{weighted} = \frac{\Sigma x_{i}* w_{i}}{\Sigma w_{i}}

Plugging this into the formula, we get,

\dfrac{98.6}{9.8} = 10.1

In order to find the geometric mean of the geometric series , we must use the following formula,

\sqrt[n]{a_{1}*a_{2}*\dotsm*a_{n}}

\sqrt[6]{3*9*27*81*243*729} = \sqrt[6]{10460353203} = 46.8

Where we can see the mean plotted in the graph below.

how to solve problems involving mean and variance of probability distribution

Which may be more clear when plotted on one axis.

how to solve problems involving mean and variance of probability distribution

We were given information on the population mean and population standard deviation. To find the population variance, simply follow the steps in the table below.

To find the variance of the data, simply follow the formula,

s^2 = \frac{\Sigma(x_{i}-\bar{x})^2}{n-1}

Where the mean is,

\bar{x} = \dfrac{224}{7} = 32

\dfrac{304}{7-1} = 50.7

Looking at the graph below, we can see the mean and standard deviation plotted.

how to solve problems involving mean and variance of probability distribution

Which may be more clear when plotted on a single axis.

dotplot line plot 2

As in the previous example, we can calculate the sample variance by taking the square of the SD.

Find a sample interpretation below.

Mean: The centre is located at the point 24.

Variance: The spread of the data is relatively small, meaning that the data points are clustered closely around the mean.

This would look something like the following.

spread 1

A measure of variability is important to report with the mean because it indicates the spread of the data. Take a look at the following data points, which also have a mean of 24 but have a variance of 110.

spread 2

As you can see, the variance is a great indicator for how close or how far the data points are located about the mean. The general rule is, the larger the variance, the more spread out the data points are. Note that a bigger variance isn’t inherently bad .

For example, if these data points were test scores, as a teacher you would probably prefer the data points with a higher variance. This is because it would be easier to pinpoint which questions and concepts students didn’t understand by taking a look at the students with lower than average test scores, rather than trying to find frequently missed concepts for a test with a cluster of the same grades.

Did you like this article? Rate it!

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Frequency Distribution

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A competition was held at Richfield with students selected to participate in a game of quiz

questions. Each question was rated as a pass or fail. What is the probability that a student

passed more than 5 questions given that there are 15 questions asked and the probability of

success for answering one question correct is 0.65? Show all your working in this question. (5)

3.2 It is known that 38 % of all students coming at Richfield are registering for Bsc IT

qualification. A random sample of 25 students were interviewed during registration time.

Calculate the following and show all your working.

a) What is the probability that greeter than 3 and less than 24 students are not registering

for BSC IT qualification? (5)

b) Calculate the mean, variance and standard deviation of this probability distribution (5)

3.3 What is the probability that if you pick 2 cards from the deck of cards without replacing them,

the first card will be a King and the second card will be an 8. (5)

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  1. SOLVING PROBLEMS INVOLVING MEAN AND VARIANCE OF PROBABILITY

    how to solve problems involving mean and variance of probability distribution

  2. SOLVING PROBLEMS INVOLVING MEAN AND VARIANCE OF PROBABILITY

    how to solve problems involving mean and variance of probability distribution

  3. SOLUTION: Solving problems involving mean and variance of probability

    how to solve problems involving mean and variance of probability distribution

  4. How To Calculate Variance In 4 Simple Steps

    how to solve problems involving mean and variance of probability distribution

  5. Problems Involving Mean and Variance of Discrete Random Variable

    how to solve problems involving mean and variance of probability distribution

  6. Lesson 9: Solving Problems Involving the Mean and the Variance of

    how to solve problems involving mean and variance of probability distribution

VIDEO

  1. Solving Problems Involving Mean and Variance of Probability Distribution

  2. Statistics & Probability

  3. Probability JEE Main

  4. Practice Combining Normally Distributed Random Variables

  5. JEE Mains: Probability 09

  6. SOLVING PROBLEMS INVOLVING MEAN AND VARIANCE OF PROBABILITY DISTRIBUTION

COMMENTS

  1. Mean and Variance of Probability Distributions

    And more importantly, the difference between finite and infinite populations. I tried to give the intuition that, in a way, a probability distribution represents an infinite population of values drawn from it. And that the mean and variance of a probability distribution are essentially the mean and variance of that infinite population.

  2. How to Calculate the Variance of a Probability Distribution

    To find the variance of a probability distribution, we can use the following formula: σ2 = Σ (xi-μ)2 * P (xi) where: For example, consider our probability distribution for the soccer team: The mean number of goals for the soccer team would be calculated as: μ = 0*0.18 + 1*0.34 + 2*0.35 + 3*0.11 + 4*0.02 = 1.45 goals.

  3. SOLVING PROBLEMS INVOLVING MEAN AND VARIANCE OF PROBABILITY ...

    ‼️statistics and probability‼️🟣 grade 11: solving problems involving mean and variance of probability distributions‼️shs mathematics playlist‼️general mathe...

  4. Lesson 9: Solving Problems Involving the Mean and the Variance of

    This is a video lesson which is a supplementary material for MELC 9-Lesson 9 in Statistics and Probability. The teacher explains the lesson from the Learning...

  5. 6.E: Sampling Distributions (Exercises)

    In one study it was found that 86% 86 % of all homes have a functional smoke detector. Suppose this proportion is valid for all homes. Find the probability that in a random sample of 600 600 homes, between 80% 80 % and 90% 90 % will have a functional smoke detector. You may assume that the normal distribution applies.

  6. 5.1: Basics of Probability Distributions

    Just as with any data set, you can calculate the mean and standard deviation. In problems involving a probability distribution function (pdf), you consider the probability distribution the population even though the pdf in most cases come from repeating an experiment many times. ... Calculating mean, variance, and standard deviation for a ...

  7. INTERPRETING THE MEAN AND VARIANCE OF A PROBABILITY ...

    ‼️statistics and probability‼️🟣 grade 11: interpreting the mean and variance of a probability distributions‼️shs mathematics playlist‼️general mathematicsfi...

  8. 3.2.1

    The variance of a discrete random variable is given by: σ 2 = Var ( X) = ∑ ( x i − μ) 2 f ( x i) The formula means that we take each value of x, subtract the expected value, square that value and multiply that value by its probability. Then sum all of those values. There is an easier form of this formula we can use.

  9. 3. Mean, Variance, and Standard Deviation for a Probability Distribution

    Math with Melissa - 3. Mean, Variance, and Standard Deviation for a Probability Distribution. 3. Mean, Variance, and Standard Deviation for a Probability Distribution. Parameters for Probability Distributions. Video tutorials to watch:

  10. Probability Distribution

    A probability distribution is an idealized frequency distribution. A frequency distribution describes a specific sample or dataset. It's the number of times each possible value of a variable occurs in the dataset. The number of times a value occurs in a sample is determined by its probability of occurrence. Probability is a number between 0 ...

  11. 9 Common Probability Distributions with Mean & Variance ...

    Bernoulli distribution, Binomial distribution, Geometric distribution, Negative Binomial distribution, Hypergeometric distribution, Poisson distribution 2. Continuous Probability Distributions

  12. 3.1) PMF, Mean, & Variance

    3.1) PMF, Mean, & Variance. A probability distribution is a mathematical function that describes an experiment by providing the probabilities that different possible outcomes will occur. Probability distributions are defined in terms of random variables, which are variables whose values depend on outcomes of a random phenomenon.

  13. 5.3: Expectation, Variance and Standard Deviation

    Like data, probability distributions have standard deviations. To calculate the standard deviation (σ) of a probability distribution, find each deviation from its expected value, square it, multiply it by its probability, add the products, and take the square root. To understand how to do the calculation, look at the table for the number of ...

  14. Normal Distribution Problems with Solutions

    Problems with Solutions. X is a normally distributed variable with mean μ = 30 and standard deviation σ = 4. Find the probabilities. a) P (X < 40) b) P (X > 21) c) P (30 < X < 35) A radar unit is used to measure the speeds of cars on a motorway. The speeds are normally distributed with a mean of 90 km/hr and a standard deviation of 10 km/hr.

  15. How to Find the Mean of a Probability Distribution (With Examples

    For example, consider our probability distribution for the soccer team: The mean number of goals for the soccer team would be calculated as: μ = 0*0.18 + 1*0.34 + 2*0.35 + 3*0.11 + 4*0.02 = 1.45 goals. The following examples show how to calculate the mean of a probability distribution in a few other scenarios.

  16. Computing for the Mean, Variance and Standard Deviation of a

    This video will teach you a method for solving the mean, variance and standard deviation of a probability distribution.

  17. Mean (Expectation), Variance from a Probability Distribution of a

    Find also the mean and variance of the distribution Solution [Expectation: 3.46; Variance: 4.0284 ; Standard Deviation : +2.007] 04. The monthly demand for radios is known to have the following probability distribution

  18. Finding the mean and variance of probability distributions

    problems involving mean and variance of probability distribution - Free download as Powerpoint Presentation (.ppt / .pptx), PDF File (.pdf), Text File (.txt) or view presentation slides online. This document contains examples and objectives for calculating mean and variance of probability distributions. It includes examples of finding the expected gain from lottery tickets where one ticket ...

  19. Random Variables

    A Random Variable is a variable whose possible values are numerical outcomes of a random experiment. The Mean (Expected Value) is: μ = Σxp. The Variance is: Var (X) = Σx2p − μ2. The Standard Deviation is: σ = √Var (X) Mathopolis: Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10. Local popup: Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10.

  20. Solutions to Mean and Variance Problems

    Solution Problem 4. We were given information on the population mean and population standard deviation. To find the population variance, simply follow the steps in the table below. Measure. Description. 1. Remember that the variance is simply the square of the standard deviation. \ [. (\sqrt {\sigma^2}) = \sigma.

  21. Problems Involving Mean and Variance of Probability Distributions

    PROBLEMS INVOLVING MEAN AND VARIANCE OF PROBABILITY DISTRIBUTIONS - Free download as Powerpoint Presentation (.ppt / .pptx), PDF File (.pdf), Text File (.txt) or view presentation slides online.

  22. Hurricane Electric Mathematics

    Calculate the following and show all your working. a) What is the probability that greeter than 3 and less than 24 students are not registering ... Calculate the mean, variance and standard deviation of this probability distribution (5) 3.3 What is the probability that if you pick 2 cards from the deck of cards without replacing them,

  23. SOLVING PROBLEMS INVOLVING MEAN AND VARIANCE OF PROBABILITY ...

    SOLVING PROBLEMS INVOLVING MEAN AND VARIANCE OF PROBABILITY DISTRIBUTION

  24. Greenhouse Gas Emissions Standards for Heavy-Duty Vehicles-Phase 3

    Concerns about uncertainties relating to supporting infrastructure included: limited nature of today's HD charging infrastructure, the magnitude of buildout of electrical distribution systems necessary to support (BEVs especially in the early model years of the program), the cost and length of time needed for infrastructure buildout, a chicken ...

  25. Problems Involving Mean and Variance of Discrete Random ...

    Statistics and Probability by @ProfD Solving problems involving mean and variance of discrete random variableGeneral Mathematics Playlisthttps://www.youtube....