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Discrete Random Variables: Mean or Expected Value and Standard Deviation

Module by: Dr. Barbara Illowsky, Susan Dean

Summary: This module explores the Law of Large Numbers, the phenomenon where an experiment performed many times will yield cumulative results closer and closer to the theoretical mean over time.

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 every time you perform a particular experiment.

The mean of a random variable XX is μμ. If we do an experiment many times (for instance, flip a fair coin, as Karl Pearson did, 24,000 times and let XX = the number of heads) and record the value of XX each time, the average gets closer and closer to m as we keep repeating the experiment. This is known as the Law of Large Numbers.

Note:

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

A Step-by-Step Example

A men's soccer team plays soccer 0, 1, or 2 days a week. The probability that they play 0 days is 0.2, the probability that they play 1 day is 0.5, and the probability that they play 2 days is 0.3. Find the long-term average, μμ, or expected value of the days per week the men's soccer team plays soccer.

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

Expected Value Table
XX P(x)P(x) or P(X=x)P(X=x) x P(x)xP(x)
0 0.2 (0)(0.2) = 0
1 0.5 (1)(0.5) = 0.5
2 0.3 (2)(0.3) = 0.6

This table is called an expected value table. The table helps you calculate the expected value or long-term average, mm, of a probability distribution.

Add the last column to find the long term average or expected value: (0)(0.2)+(1)(0.5)+(2)(0.3)= 0 + 0.5(0)(0.2)+(1)(0.5)+(2)(0.3)= 0 + 0.5. 0.6 = 1.10.6 = 1.1.

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 μ=1.1μ=1.1

Example 1

Find the expected value (the expected number of times a newborn wakes its mother after midnight) for example 4-1.

The expected number of times a newborn wakes its mother after midnight is 2.1 times. You expect a newborn to wake its mother after midnight 2.1 times, on an average night.
XX P(x)P(x) or P(X=x)P(X=x) x P(x)xP(x)
0 P(X=0) = 250P(X=0)=250 (0)(250)(250) = 0
1 P(X=1) = 1150P(X=1)=1150 (1)(1150)(1150) = 11501150
2 P(X=2) = 2350P(X=2)=2350 (2)(2350)(2350) = 46504650
3 P(X=3) = 950P(X=3)=950 (3)(950)(950) = 27502750
4 P(X=4) = 450P(X=4)=450 (4)(450)(450) = 16501650
5 P(X=5) = 150P(X=5) =150 (5)(150)(150) = 550550

Note:

Add the last column to find the expected value. μμ = Expected Value = 105 50 = 2.1 105 50 =2.1

Problem 1

Go back and calculate the expected value for the number of days Nancy attends classes a week. Construct the third column to do so.

Solution 1

2.74 days a week.

Example 2

Suppose you play a game of chance in which you choose 5 numbers from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. You may choose a number more than once. You pay $2 to play and could profit $100,000 if you match all 5 numbers in order (you get your $2 back plus $100,000). Over the long term, what is your expected profit if you match all 5 numbers in order?

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

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

To win, you must get all 5 numbers correct, in order. The probability of choosing one correct number is 110110 because there are 10 numbers. You may choose a number more than once. The probability of choosing all 5 numbers correctly and in order is:

110 * 110 * 110 * 110 * 110 * = 1 * 10 -5= 0.00001110 *110 *110 *110 *110 * = 1*10 -5=0.00001(1)

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

1-0.00001=0.999991-0.00001=0.99999(2)

The expected value table is as follows.

Αdd the last column: μμ = Expected Value = −1.99998 + 1 = −0.99998
  XX P(X)P(X) XP(X)XP(X)
Loss -2 0.99999 (-2)(0.99999)=-1.99998
Profit 100,000 0.00001 (100000)(0.00001)=1

You would, on the average, expect to lose approximately one dollar 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 3

Suppose you play a game with a biased coin. You play each game by tossing the coin once. P(heads)=23P(heads)=23 and P(tails) = 13P(tails)=13. 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?

Problem 1

Define a random variable XX.

Solution 1

XX = amount of profit

Problem 2

Complete the following expected value table.

  XX ____ ____
WIN 10 1313 ____
LOSE ____ ____ -123-123

Solution 2

  XX P(x)P(x) (x)(P(x))(x)(P(x))
WIN 10 1313 103103
LOSE -6 2323 -123-123

Problem 3

What is the expected value, μμ? Do you come out ahead?

Solution 3

The expected value μ = -2 3 μ = -2 3 . You do not come out ahead.

Like data, probability distributions have standard deviations. To calculate the standard deviation, σσ, of a probability distribution, find each deviation, square it, multiply it by its probability, and add the products. To understand how to do the calculation, look at the number of days per week a men's soccer team plays soccer table again. Add the column ( x − μ ) 2 · P ( x ) ( x − μ ) 2 ·P(x).

μμ = 1.1 The sum of the last column is the variance, σ2σ2. σ2σ2 = 0.490 The standard deviation is σσ. σσ = 0.4900.490 = 0.7
XX P(x)P(x) or P(X=x)P(X=x) xP(x)xP(x) (x -μ)2 P(x)(x -μ)2P(x)
0 0.2 (0)(0.2) = 0 (0-1.1)2 (.2) = 0.242(0-1.1)2(.2)= 0.242
1 0.5 (1)(0.5) = 0.5 (1-1.1)2 (.5) = 0.005)(1-1.1)2(.5)=0.005)
2 0.3 (2)(0.3) = 0.6 (2-1.1) 2 (.3) = 0.243(2-1.1) 2(.3)=0.243

Generally for probability distributions, we use a calculator or a computer to calculate μμ and σσ to reduce roundoff error. For some of the common probability distributions, there are short-cut formulas that calculate μμ and σσ.

Glossary

Expected Value:
Expected arithmetic average when an experiment is repeated many times. (Called also mean). Notations: E(x),μE(x),μ size 12{E \( x \) ,μ} {} For discrete random variable (RV) with probability distribution function P(x)=P(X=x)P(x)=P(X=x) size 12{P \( x \) =P \( X=x \) } {} the definition also can be written in the form E(x)=μ=∑xP(x)E(x)=μ=∑xP(x) size 12{E \( x \) =μ= Sum { ital "xP" \( x \) } } {}.
Mean:
A number to measure the central tendency (average), shortening from arithmetic mean. By definition, the mean for a sample (usually denoted by XˉXˉ size 12{ { bar {X}}} {}) is Xˉ=Sum of all values in the sampleNumber of values in the sampleXˉ=Sum of all values in the sampleNumber of values in the sample size 12{ { bar {X}}= { {"Sum of all values in the sample"} over {"Number of values in the sample"} } } {}, and the mean for a population (usually denoted by mm size 12{m} {}) is m=Sum of all values in the populationNumber of values in the populationm=Sum of all values in the populationNumber of values in the population size 12{m= { {"Sum of all values in the population"} over {"Number of values in the population"} } } {}.

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