<< Chapter < Page Chapter >> Page >
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.

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

To find the expected value or long term average, μ , 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, μ , or expected value of the 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 xP(x) . In this column, you will multiply each x value by its probability.

This table is called an expected value table. The table helps you calculate the expected value or long-term average.
Expected value table
x P(x) x P(x)
0 0.2 (0)(0.2) = 0
1 0.5 (1)(0.5) = 0.5
2 0.3 (2)(0.3) = 0.6

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 + 0.6 = 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=1.1

Find the expected value for the example about the number of times a newborn baby's crying wakes its mother after midnight. The expected value is the expected number of times a newborn wakes its mother after midnight.

You expect a newborn to wake its mother after midnight 2.1 times, on the average.
x P(X) x P(X)
0 P(x=0) = 2 50 (0) ( 2 50 ) = 0
1 P(x=1) = 11 50 (1) ( 11 50 ) = 11 50
2 P(x=2) = 23 50 (2) ( 23 50 ) = 46 50
3 P(x=3) = 9 50 (3) ( 9 50 ) = 27 50
4 P(x=4) = 4 50 (4) ( 4 50 ) = 16 50
5 P(x=5) = 1 50 (5) ( 1 50 ) = 5 50

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

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

2.74 days a week.

Got questions? Get instant answers now!
Got questions? Get instant answers now!

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 0 to 9 with replacement. 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 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 5 numbers correct, in order. The probability of choosing one correct number is 1 10 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:

1 10 * 1 10 * 1 10 * 1 10 * 1 10 * = 1 * 10 -5 = 0.00001

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

1 - 0.00001 = 0.99999

The expected value table is as follows.

Αdd the last column. -1.99998 + 1 = -0.99998
x P(x) x P(x)
Loss -2 0.99999 (-2)(0.99999)=-1.99998
Profit 100,000 0.00001 (100000)(0.00001)=1

Since -0.99998 is about -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 theaverage or expected LOSS per game after playing this game over and over.

Got questions? Get instant answers now!

Suppose you play a game with a biased coin. You play each game by tossing the coin once. P(heads) = 2 3 and P(tails) = 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 .

X = amount of profit

Got questions? Get instant answers now!

Complete the following expected value table.

x ____ ____
WIN 10 1 3 ____
LOSE ____ ____ -12 3
x P(x) x P(x)
WIN 10 1 3 10 3
LOSE -6 2 3 -12 3
Got questions? Get instant answers now!

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

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

Got questions? Get instant answers now!
Got questions? Get instant answers now!

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 thenumber of days per week a men's soccer team plays soccer. To find the standard deviation, add the entries in the column labeled ( x μ ) 2 · P ( x ) and take the square root.

x P(x) x P(x) (x -μ) 2 P(x)
0 0.2 (0)(0.2) = 0 ( 0 - 1.1 ) 2 ( .2 ) = 0.242
1 0.5 (1)(0.5) = 0.5 ( 1 - 1.1 ) 2 ( .5 ) = 0.005
2 0.3 (2)(0.3) = 0.6 ( 2 - 1.1 ) 2 ( .3 ) = 0.243

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 . σ = 0.49 = 0.7

Generally for probability distributions, we use a calculator or a computer to calculate μ and σ to reduce roundoff error. For some probability distributions, there are short-cut formulas that calculate μ and σ .

Questions & Answers

a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
how did I we'll learn this
Noor Reply
f(x)= 2|x+5| find f(-6)
Prince Reply
f(n)= 2n + 1
Samantha Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Got questions? Join the online conversation and get instant answers!
QuizOver.com Reply
Practice Key Terms 2

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, Collaborative statistics. OpenStax CNX. Jul 03, 2012 Download for free at http://cnx.org/content/col10522/1.40
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Collaborative statistics' conversation and receive update notifications?

Ask