Mean or Expected Value and Standard Deviation, Discrete Random Variables | Statistics

Learning objectives

The expected value of a discrete random variable X, symbolized as E(X), is often referred to as the long-term average or mean (symbolized as μ). This means that over the long term of doing an experiment over and over, you would expect this average. For example, let X = the number of heads you get when you toss three fair coins. If you repeat this experiment (toss three fair coins) a large number of times, the expected value of X is the number of heads you expect to get for each three tosses on average.

A men's soccer team plays soccer zero, one, or two days a week. The probability that they play zero days is .2, the probability that they play one day is .5, and the probability that they play two days is .3. Find the long-term average or expected value, μ, 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), the product of the value x with the corresponding probability P(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	.2	(0)(.2) = 0
1	.5	(1)(.5) = .5
2	.3	(2)(.3) = .6

Add the last column $x * P (x)$ to get the expected value/mean of the random variable X.

E (X) = μ = \sum x P (x) = 0 + .5 + .6 = 1.1

The expected value/mean 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.

As you learned in Chapter 3, if you toss a fair coin, the probability that the result is heads is 0.5. This probability is a theoretical probability, which is what we expect to happen. This probability does not describe the short-term results of an experiment. If you flip a coin two times, the probability does not tell you that these flips will result in one head and one tail. Even if you flip a coin 10 times or 100 times, the probability does not tell you that you will get half tails and half heads. The probability 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. The relative frequency of heads is 12,012/24,000 = .5005, which is very close to the theoretical probability .5. 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). The relative frequency is also called the experimental probability, a term that means what actually happens.

In the next example, we will demonstrate how to find the expected value and standard deviation of a discrete probability distribution by using relative frequency.

Like data, probability distributions have variances and standard deviations. The variance of a probability distribution is symbolized as $σ^{2}$ and the standard deviation of a probability distribution is symbolized as σ. Both are parameters since they summarize information about a population. To find the variance $σ^{2}$ of a discrete probability distribution, find each deviation from its expected value, square it, multiply it by its probability, and add the products. To find the standard deviation σ of a probability distribution, simply take the square root of variance $σ^{2}$ . The formulas are given as below.

A researcher conducted a study to investigate how a newborn babys crying after midnight affects the sleep of the baby's mother. The researcher randomly selected 50 new mothers and asked how many times they were awakened by their newborn baby's crying after midnight per week. Two mothers were awakened zero times, 11 mothers were awakened one time, 23 mothers were awakened two times, nine mothers were awakened three times, four mothers were awakened four times, and one mother was awakened five times. Find the expected value of the number of times a newborn baby's crying wakes its mother after midnight per week. Calculate the standard deviation of the variable as well.

To do the problem, first let the random variable X = the number of times a mother is awakened by her newborns crying after midnight per week. X takes on the values 0, 1, 2, 3, 4, 5. Construct a PDF table as below. The column of P(x) gives the experimental probability of each x value. We will use the relative frequency to get the probability. For example, the probability that a mother wakes up zero times is $\frac{2}{50}$ since there are two mothers out of 50 who were awakened zero times. The third column of the table is the product of a value and its probability, xP(x).

x	P(x)	xP(x)
0	$P (x = 0) = \frac{2}{50}$	$(0) (\frac{2}{50}) = 0$
1	$P (x = 1) = \frac{11}{50}$	$(1) (\frac{11}{50}) = \frac{11}{50}$
2	$P (x = 2) = \frac{23}{50}$	$(2) (\frac{23}{50}) = \frac{46}{50}$
3	$P (x = 3) = \frac{9}{50}$	$(3) (\frac{9}{50}) = \frac{27}{50}$
4	$P (x = 4) = \frac{4}{50}$	$(4) (\frac{4}{50}) = \frac{16}{50}$
5	$P (x = 5) = \frac{1}{50}$	$(5) (\frac{1}{50}) = \frac{5}{50}$

We then add all the products in the third column to get the mean/expected value of X.

E (X) = μ = \sum x P (x) = 0 + \frac{11}{50} + \frac{46}{50} + \frac{27}{50} + \frac{16}{50} + \frac{5}{50} = \frac{105}{50} = 2.1

Therefore, we expect a newborn to wake its mother after midnight 2.1 times per week, on the average.

To calculate the standard deviation σ, we add the fourth column (x-μ)² and the fifth column ${(x - μ)}^{2} ∙ P (x)$ to get the following table:

x	P(x)	xP(x)	(x-µ)²	(x-µ)²•P(x)
0	$P (x = 0) = \frac{2}{50}$	$(0) (\frac{2}{50}) = 0$	${(0 - 2.1)}^{2} = 4.41$	$4.41 • \frac{2}{50} = .1764$
1	$P (x = 1) = \frac{11}{50}$	$(1) (\frac{11}{50}) = \frac{11}{50}$	${(1 - 2.1)}^{2} = 1.21$	$1.21 • \frac{11}{50} = .2662$
2	$P (x = 2) = \frac{23}{50}$	$(2) (\frac{23}{50}) = \frac{46}{50}$	${(2 - 2.1)}^{2} = .01$	$.01 • \frac{23}{50} = .0046$
3	$P (x = 3) = \frac{9}{50}$	$(3) (\frac{9}{50}) = \frac{27}{50}$	${(3 - 2.1)}^{2} = .81$	$.81 • \frac{9}{50} = .1458$
4	$P (x = 4) = \frac{4}{50}$	$(4) (\frac{4}{50}) = \frac{16}{50}$	${(4 - 2.1)}^{2} = 3.61$	$3.61 • \frac{4}{50} = .2888$
5	$P (x = 5) = \frac{1}{50}$	$(5) (\frac{1}{50}) = \frac{5}{50}$	${(5 - 2.1)}^{2} = 8.41$	$8.41 • \frac{1}{50} = .1682$

We then add all the products in the 5^th column to get the variance of X.

σ^{2} = .1764 + .2662 + .0046 + .1458 + .2888 + .1682 = 1.05

To get the standard deviation σ, we simply take the square root of variance σ².

σ = \sqrt{σ^{2}} = \sqrt{1.05} \approx 1.0247

Try It

A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift. For a random sample of 50 patients, the following information was obtained. What is the expected value?

x	P(x)
0	P(x = 0) = $\frac{4}{50}$
1	P(x = 1) = $\frac{8}{50}$
2	P(x = 2) = $\frac{16}{50}$
3	P(x = 3) = $\frac{14}{50}$
4	P(x = 4) = $\frac{6}{50}$
5	P(x = 5) = $\frac{2}{50}$

The expected value is 2.24

(0) $\frac{4}{50}$ + (1) $\frac{4}{50}$ + (2) $\frac{16}{50}$ + (3) $\frac{14}{50}$ + (4) $\frac{6}{50}$ + (5) $\frac{2}{50}$ = 0 + $\frac{8}{50}$ + $\frac{32}{50}$ + $\frac{42}{50}$ + $\frac{24}{50}$ + $\frac{10}{50}$ = $\frac{116}{50}$ = 2.24

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 a PDF table for the amount of money you can profit.

Let X = the amount of money you profit. If your five numbers match in order, you will win the game and will get your $2 back plus $100,000. That means your profit is $100,000. If your five numbers do not match in order, you will lose the game and lose your $2. That means your profit is -$2. Therefore, X takes on the values $100,000 and –$2. That is the second column x in the PDF table below.

To win, you must get all five numbers correct, in order. The probability of choosing the correct first number is $\frac{1}{10}$ because there are 10 numbers (from zero to nine) and only one of them is correct. The probability of choosing the correct second number is also $\frac{1}{10}$ because the selection is done with replacement and there are still 10 numbers (from zero to nine) for you to choose. Due to the same reason, the probability of choosing the correct third number, the correct fourth number, and the correct fifth number are also $\frac{1}{10}$ . The selection of one number does not affect the selection of another number. That means the five selections are independent. The probability of choosing all five correct numbers and in order is equal to the product of the probabilities of choosing each number correctly.

\begin{matrix} P (choosing all five numbers correctly) • P (choosing 1st number correctly) • \\ P (choosing 2nd number correctly) • P (choosing 5th number correctly) \\ = (\frac{1}{10}) • (\frac{1}{10}) • (\frac{1}{10}) • (\frac{1}{10}) • (\frac{1}{10}) \\ = .00001 \end{matrix}

Therefore, the probability of winning is .00001 and the probability of losing is 1 − .00001 = .99999. That is how we get the third column P(x) in the PDF table below.

To get the fourth column xP(x) in the table, we simply multiply the value x with the corresponding probability P(x).

The PDF table is as follows:

	x	P(x)	x*P(x)
Loss	–2	.99999	(–2)(.99999) = –1.99998
Profit	100,000	.00001	(100000)(.00001) = 1

We then add all the products in the last column to get the mean/expected value of X.

E (X) = μ = \sum x P (x) = - 1.99998 + 1 = - .9998 .

Since –.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.

Try It

You are playing a game of chance in which four cards are drawn from a standard deck of 52 cards. You guess the suit of each card before it is drawn. The cards are replaced in the deck on each draw. You pay $1 to play. If you guess the right suit every time, you get your money back and $256. What is your expected profit of playing the game over the long term?

Let X = the amount of money you profit. The x-values are –$1 and $256.

The probability of guessing the right suit each time is $(\frac{1}{4}) (\frac{1}{4}) (\frac{1}{4}) (\frac{1}{4}) = \frac{1}{256}$ = 0.0039

The probability of losing is 1 – $\frac{1}{256}$ = $\frac{255}{256}$ = 0.9961

(0.0039)256 + (0.9961)(–1) = 0.9984 + (–0.9961) = 0.0023 or 0.23 cents.

Suppose you play a game with a biased coin. You play each game by tossing the coin once. P(heads) = $\frac{2}{3}$ and P(tails) = $\frac{1}{3}$ . 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?

a. Define a random variable X.

a. X = amount of profit

b. Complete the following expected value table.

	x	____	____
WIN	10	$\frac{1}{3}$	____
LOSE	____	____	$\frac{–12}{3}$

	x	P(x)	xP(x)
WIN	10	$\frac{1}{3}$	$\frac{10}{3}$
LOSE	–6	$\frac{2}{3}$	$\frac{–12}{3}$

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

c. Add the last column of the table. The expected value $E (X) = μ = \frac{10}{3} + (- \frac{12}{3}) = - \frac{2}{3} \approx - .67$ . You lose, on average, about 67 cents each time you play the game, so you do not come out ahead.

Try It

Suppose you play a game with a spinner. You play each game by spinning the spinner once. P(red) = $\frac{2}{5}$ , P(blue) = $\frac{2}{5}$ , and P(green) = $\frac{1}{5}$ . If you land on red, you pay $10. If you land on blue, you don't pay or win anything. If you land on green, you win $10. Complete the following expected value table.

	x	P(x)
Red			$– \frac{20}{5}$
Blue		$\frac{2}{5}$
Green	10

	x	P(x)	x*P(x)
Red	–10	$\frac{2}{5}$	$- \frac{20}{5}$
Blue	0	$\frac{2}{5}$	$\frac{0}{5}$
Green	10	$\frac{1}{5}$	$\frac{10}{5}$

Generally for probability distributions, we use a calculator or a computer to calculate μ and σ to reduce rounding errors. For some probability distributions, there are shortcut formulas for calculating μ and σ.

Toss a fair, six-sided die twice. Let X = the number of faces that show an even number. Construct a table like fs-idm71508016 and calculate the mean μ and standard deviation σ 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.

(1, 1)	(1, 2)	(1, 3)	(1, 4)	(1, 5)	(1, 6)
(2, 1)	(2, 2)	(2, 3)	(2, 4)	(2, 5)	(2, 6)
(3, 1)	(3, 2)	(3, 3)	(3, 4)	(3, 5)	(3, 6)
(4, 1)	(4, 2)	(4, 3)	(4, 4)	(4, 5)	(4, 6)
(5, 1)	(5, 2)	(5, 3)	(5, 4)	(5, 5)	(5, 6)
(6, 1)	(6, 2)	(6, 3)	(6, 4)	(6, 5)	(6, 6)

Use the sample space to complete the following table.

Calculating μ and σ.

x	P(x)	xP(x)	(x – μ)² $\cdot$ P(x)
0	$\frac{9}{36}$	0	(0 – 1)² ⋅ $\frac{9}{36}$ = $\frac{9}{36}$
1	$\frac{18}{36}$	$\frac{18}{36}$	(1 – 1)² ⋅ $\frac{18}{36}$ = 0
2	$\frac{9}{36}$	$\frac{18}{36}$	(2 – 1)² ⋅ $\frac{9}{36}$ = $\frac{9}{36}$

Add the values in the third column to find the expected value: μ = $\frac{36}{36}$ = 1. Use this value to complete the fourth column.

Add the values in the fourth column and take the square root of the sum: σ = $\sqrt{\frac{18}{36}}$ ≈ .7071.

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.

References

Florida State University. (n.d.). Class catalogue at the Florida State University. Retrieved from https://apps.oti.fsu.edu/RegistrarCourseLookup/SearchFormLegacy

World Earthquakes. (2012). World earthquakes: Live earthquake news and highlights. Retrieved from http://www.worldearthquakes.com/index.php?option=ethq_prediction

Chapter Review

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: $μ = \underset{x \in X}{\sum^{}} x P (x)$

Standard Deviation: $σ = \sqrt{\underset{x \in X}{\sum^{}} {(x - μ)}^{2} P (x)}$

Complete the expected value table.

x	P(x)	x*P(x)
0	.2
1	.2
2	.4
3	.2

Find the expected value from the expected value table.

x	P(x)	x*P(x)
2	.1	2(.1) = .2
4	.3	4(.3) = 1.2
6	.4	6(.4) = 2.4
8	.2	8(.2) = 1.6

.2 + 1.2 + 2.4 + 1.6 = 5.4

Find the standard deviation.

x	P(x)	x*P(x)	(x – μ)²P(x)
2	0.1	2(.1) = .2	(2–5.4)²(.1) = 1.156
4	0.3	4(.3) = 1.2	(4–5.4)²(.3) = .588
6	0.4	6(.4) = 2.4	(6–5.4)²(.4) = .144
8	0.2	8(.2) = 1.6	(8–5.4)²(.2) = 1.352

Identify the mistake in the probability distribution table.

x	P(x)	x*P(x)
1	.15	.15
2	.25	.50
3	.30	.90
4	.20	.80
5	.15	.75

The values of P(x) do not sum to one.

Identify the mistake in the probability distribution table.

x	P(x)	x*P(x)
1	.15	.15
2	.25	.40
3	.25	.65
4	.20	.85
5	.15	1

Use the following information to answer the next five exercises: A physics professor wants to know what percent of physics majors will spend the next several years doing postgraduate research. He has the following probability distribution:

x	P(x)	x*P(x)
1	.35
2	.20
3	.15
4
5	.10
6	.05

Define the random variable X.

Let X = the number of years a physics major will spend doing postgraduate research.

Define P(x), or the probability of x.

Find the probability that a physics major will do postgraduate research for four years. P(x = 4) = ________

1 – .35 – .20 – .15 – .10 – .05 = .15

Find the probability that a physics major will do postgraduate research for at most three years. P(x ≤ 3) = ________

On average, how many years would you expect a physics major to spend doing postgraduate research?

1(.35) + 2(.20) + 3(.15) + 4(.15) + 5(.10) + 6(.05) = .35 + .40 + .45 + .60 + .50 + .30 = 2.6 years

Use the following information to answer the next seven exercises: A ballet instructor is interested in knowing what percent of each year's class will continue on to the next so that she can plan what classes to offer. Over the years, she has established the following probability distribution:

Let X = the number of years a student will study ballet with the teacher.
Let P(x) = the probability that a student will study ballet x years.

Complete M02_Ch04_tbl021 using the data provided.

x	P(x)	x*P(x)
1	.10
2	.05
3	.10
4
5	.30
6	.20
7	.10

In words, define the random variable X.

X is the number of years a student studies ballet with the teacher.

P(x = 4) = ________

P(x < 4) = ________

.10 + .05 + .10 = .25

On average, how many years would you expect a child to study ballet with this teacher?

What does the column P(x) sum to and why?

The sum of the probabilities sum to one because it is a probability distribution.

What does the column x*P(x) sum to and why?

You are playing a game by drawing a card from a standard deck and replacing it. If the card is a face card, you win $30. If it is not a face card, you pay $2. There are 12 face cards in a deck of 52 cards. What is the expected value of playing the game?

$- 2 (\frac{40}{52}) + 30 (\frac{12}{52}) = - 1.54 + 6.92 = 5.38$

HOMEWORK

A theater group holds a fund-raiser. It sells 100 raffle tickets for $5 apiece. Suppose you purchase four tickets. The prize is two passes to a Broadway show, worth a total of $150.

What are you interested in here?
In words, define the random variable X.
List the values that X may take on.
Construct a PDF.
If this fund-raiser is repeated often and you always purchase four tickets, what would be your expected average winnings per raffle?

A game involves selecting a card from a regular 52-card deck and tossing a coin. The coin is a fair coin and is equally likely to land on heads or tails.

If the card is a face card, and the coin lands on heads, you win $6.
If the card is a face card, and the coin lands on tails, you win $2.
If the card is not a face card, you lose $2, no matter what the coin shows.

Find the expected value for this game (expected net gain or loss).
Explain what your calculations indicate about your long-term average profits and losses on this game.
Should you play this game to win money?

The variable of interest is X, or the gain or loss, in dollars.

The face cards jack, queen, and king. There are (3)(4) = 12 face cards and 52 – 12 = 40 cards that are not face cards.

We first need to construct the probability distribution for X. We use the card and coin events to determine the probability for each outcome, but we use the monetary value of X to determine the expected value.

Card Event	X net gain/loss	P(X)
Face Card and Heads	6	$(\frac{12}{52}) (\frac{1}{2}) = (\frac{6}{52})$
Face Card and Tails	2	$(\frac{12}{52}) (\frac{1}{2}) = (\frac{6}{52})$
(Not Face Card) and (H or T)	–2	$(\frac{40}{52}) (1) = (\frac{40}{52})$

$Expected value = (6) (\frac{6}{52}) + (2) (\frac{6}{52}) + (- 2) (\frac{40}{52}) = - \frac{32}{52}$
Expected value = –$0.62, rounded to the nearest cent
If you play this game repeatedly, over a long string of games, you would expect to lose 62 cents per game, on average.
You should not play this game to win money because the expected value indicates an expected average loss.

You buy a ticket to a raffle that costs $10 per ticket. There are only 100 tickets available to be sold in this raffle. In this raffle there are one $500 prize, two $100 prizes, and four $25 prizes. Find your expected gain or loss.

Complete the PDF and answer the questions.

x	P(x)	xP(x)
0	.3
1	.2
2
3	.4

Find the probability that x = 2.
Find the expected value.

Suppose that you are offered the following deal: You roll a die. If you roll a six, you win $10. If you roll a four or five, you win $5. If you roll a one, two, or three, you pay $6.

What are you ultimately interested in here (the value of the roll or the money you win)?
In words, define the random variable X.
List the values that X may take on.
Construct a PDF.
Over the long run of playing this game, what are your expected average winnings per game?
Based on numerical values, should you take the deal? Explain your decision in complete sentences.

A venture capitalist, willing to invest $1,000,000, has three investments to choose from: The first investment, a software company, has a 10 percent chance of returning $5,000,000 profit, a 30 percent chance of returning $1,000,000 profit, and a 60 percent chance of losing the million dollars. The second company, a hardware company, has a 20 percent chance of returning $3,000,000 profit, a 40 percent chance of returning $1,000,000 profit, and a 40 percent chance of losing the million dollars. The third company, a biotech firm, has a 10 percent chance of returning $6,000,000 profit, a 70 percent of no profit or loss, and a 20 percent chance of losing the million dollars.

Construct a PDF for each investment.
Find the expected value for each investment.
Which is the safest investment? Why do you think so?
Which is the riskiest investment? Why do you think so?
Which investment has the highest expected return, on average?

Software Company

x P(x)

5,000,000 .10

1,000,000 .30

–1,000,000 .60

Hardware Company

x P(x)

3,000,000 .20

1,000,000 .40

–1,000,00 .40

Biotech Firm

x P(x)

6,000,000 .10

0 .70

–1,000,000 .20
$200,000; $600,000; $400,000
third investment because it has the lowest probability of loss
first investment because it has the highest probability of loss
second investment

Software Company
5,000,000	.10
1,000,000	.30
–1,000,000	.60

Hardware Company
3,000,000	.20
1,000,000	.40
–1,000,00	.40

Biotech Firm
6,000,000	.10
0	.70
–1,000,000	.20

Suppose that 20,000 married adults in the United States were randomly surveyed as to the number of children they have. The results are compiled and are used as theoretical probabilities. Let X = the number of children married people have.

x	P(x)	xP(x)
0	.10
1	.20
2	.30
3
4	.10
5	.05
6 (or more)	.05

Find the probability that a married adult has three children.
In words, what does the expected value in this example represent?
Find the expected value.
Is it more likely that a married adult will have two to three children or four to six children? How do you know?

Suppose that the PDF for the number of years it takes to earn a bachelor of science (B.S.) degree is given as in id488jhjhj72479.

x	P(x)
3	.05
4	.40
5	.30
6	.15
7	.10

On average, how many years do you expect it to take for an individual to earn a B.S.?

4.85 years

People visiting video rental stores often rent more than one DVD at a time. The probability distribution for DVD rentals per customer at Video to Go is given in the following table. There is a five-video limit per customer at this store, so nobody ever rents more than five DVDs.

x	P(x)
0	.03
1	.50
2	.24
3
4	.07
5	.04

Describe the random variable X in words.
Find the probability that a customer rents three DVDs.
Find the probability that a customer rents at least four DVDs.
Find the probability that a customer rents at most two DVDs.
Another shop, Entertainment Headquarters, rents DVDs and video games. The probability distribution for DVD rentals per customer at this shop is given as follows. They also have a five-DVD limit per customer.

x P(x)

0 .35

1 .25

2 .20

3 .10

4 .05

5 .05
At which store is the expected number of DVDs rented per customer higher?
If Video to Go estimates that they will have 300 customers next week, how many DVDs do they expect to rent next week? Answer in sentence form.
If Video to Go expects 300 customers next week, and Entertainment Headquarters projects that they will have 420 customers, for which store is the expected number of DVD rentals for next week higher? Explain.
Which of the two video stores experiences more variation in the number of DVD rentals per customer? How do you know that?

x	P(x)
0	.35
1	.25
2	.20
3	.10
4	.05
5	.05

A “friend” offers you the following deal: For a $10 fee, you may pick an envelope from a box containing 100 seemingly identical envelopes. However, each envelope contains a coupon for a free gift.

Ten of the coupons are for a free gift worth $6.
Eighty of the coupons are for a free gift worth $8.
Six of the coupons are for a free gift worth $12.
Four of the coupons are for a free gift worth $40.

Based upon the financial gain or loss over the long run, should you play the game?

Yes, I expect to come out ahead in money.
No, I expect to come out behind in money.
It doesn’t matter. I expect to break even.

A university has 14 statistics classes scheduled for its Summer 2013 term. One class has space available for 30 students, eight classes have space for 60 students, one class has space for 70 students, and four classes have space for 100 students.

What is the average class size assuming each class is filled to capacity?
Space is available for 980 students. Suppose that each class is filled to capacity and select a statistics student at random. Let the random variable X equal the size of the student’s class. Define the PDF for X.
Find the mean of X.
Find the standard deviation of X.

In a raffle, there are 250 prizes of $5, 50 prizes of $25, and 10 prizes of $100. Assuming that 10,000 tickets are to be issued and sold, what is a fair price to charge to break even?

Let X = the amount of money to be won on a ticket. The following table shows the PDF for X:

x	P(x)
0	.969
5	$\frac{250}{10,000}$ = .025
25	$\frac{50}{10,000}$ = .005
100	$\frac{10}{10,000}$ = .001

Calculate the expected value of X.

0(.969) + 5(.025) + 25(.005) + 100(.001) = .35

A fair price for a ticket is $0.35. Any price over $0.35 will enable the lottery to raise money.

Glossary

expected value: expected arithmetic average when an experiment is repeated many times; also called the mean; notations μ; for a discrete random variable (RV) with probability distribution function P(x),the definition can also be written in the form μ = $\sum$ xP(x)
mean: a number that measures the central tendency; a common name for mean is average
The term mean is a shortened form of arithmetic mean. By definition, the mean for a sample (denoted by $\bar{x}$ ) is $\bar{x} = \frac{Sum of all values in the sample}{Number of values in the sample}$ and the mean for a population (denoted by μ) is μ = $\frac{Sum of all values in the population}{Number of values in the population}$ .
mean of a probability distribution: the long-term average of many trials of a statistical experiment
standard deviation of a probability distribution: a number that measures how far the outcomes of a statistical experiment are from the mean of the distribution
the law of large numbers: as the number of trials in a probability experiment increases, the difference between the theoretical probability of an event and the relative frequency probability approaches zero

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