Blood samples are used to screen people for certain diseases. When the disease is rare, health care workers sometimes combine or pool the blood samples of a group of individuals into one batch and then test it. If the test result of the batch is negative, no further testing is needed since none of the individuals in the group has the disease. However, if the test result of the batch is positive, each individual in the group must be tested.

Consider this hypothetical example: Suppose the probability of a person having the disease is 0.05, and a pooled sample of 15 individuals is tested. What is the probability that no further testing will be needed for the individuals in the sample? The answer to this question can be found by using what is called the binomial distribution. See Statistics Today—Revisited at the end of the chapter.

This chapter explains probability distributions in general and a specific, often used distribution called the binomial distribution. The Poisson, hypergeometric, and multinomial distributions are also explained.

Introduction

Many decisions in business, insurance, and other real-life situations are made by assigning probabilities to all possible outcomes pertaining to the situation and then evaluating the results. For example, a saleswoman can compute the probability that she will make 0, 1, 2, or 3 or more sales in a single day. An insurance company might be able to assign probabilities to the number of vehicles a family owns. A self-employed speaker might be able to compute the probabilities for giving 0, 1, 2, 3, or 4 or more speeches each week. Once these probabilities are assigned, statistics such as the mean, variance, and standard deviation can be computed for these events. With these statistics, various decisions can be made. The saleswoman will be able to compute the average number of sales she makes per week, and if she is working on commission, she will be able to approximate her weekly income over a period of time, say, monthly. The public speaker will be able to plan ahead and approximate his average income and expenses. The insurance company can use its information to design special computer forms and programs to accommodate its customers’ future needs.

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This chapter explains the concepts and applications of what is called a probability distribution. In addition, a special probability distribution, the binomial distribution, is explained.

Objective 1

Construct a probability distribution for a random variable.

5–1Probability Distributions

Before probability distribution is defined formally, the definition of a variable is reviewed. In Chapter 1 , a variable was defined as a characteristic or attribute that can assume different values. Various letters of the alphabet, such as X, Y, or Z, are used to represent variables. Since the variables in this chapter are associated with probability, they are called random variables .

For example, if a die is rolled, a letter such as X can be used to represent the outcomes. Then the value that X can assume is 1, 2, 3, 4, 5, or 6, corresponding to the outcomes of rolling a single die. If two coins are tossed, a letter, say Y, can be used to represent the number of heads, in this case 0, 1, or 2. As another example, if the temperature at 8:00 A.M. is 43° and at noon it is 53°, then the values T that the temperature assumes are said to be random, since they are due to various atmospheric conditions at the time the temperature was taken.

A random variable is a variable whose values are determined by chance.

Also recall from Chapter 1 that you can classify variables as discrete or continuous by observing the values the variable can assume. If a variable can assume only a specific number of values, such as the outcomes for the roll of a die or the outcomes for the toss of a coin, then the variable is called a discrete variable.

Discrete variables have a finite number of possible values or an infinite number of values that can be counted. The word counted means that they can be enumerated using the numbers 1, 2, 3, etc. For example, the number of joggers in Riverview Park each day and the number of phone calls received after a TV commercial airs are examples of discrete variables, since they can be counted.

Variables that can assume all values in the interval between any two given values are called continuous variables. For example, if the temperature goes from 62 to 78° in a 24-hour period, it has passed through every possible number from 62 to 78. Continuous random variables are obtained from data that can be measured rather than counted. Continuous random variables can assume an infinite number of values and can be decimal and fractional values. On a continuous scale, a person’s weight might be exactly 183.426 pounds if a scale could measure weight to the thousandths place; however, on a digital scale that measures only to tenths of pounds, the weight would be 183.4 pounds. Examples of continuous variables are heights, weights, temperatures, and time. In this chapter only discrete random variables are used; Chapter 6 explains continuous random variables.

The procedure shown here for constructing a probability distribution for a discrete random variable uses the probability experiment of tossing three coins. Recall that when three coins are tossed, the sample space is represented as TTT, TTH, THT, HTT, HHT, HTH, THH, HHH; and if X is the random variable for the number of heads, then X assumes the value 0, 1, 2, or 3.

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Probabilities for the values of X can be determined as follows:

Hence, the probability of getting no heads is , one head is , two heads is , and three heads is . From these values, a probability distribution can be constructed by listing the outcomes and assigning the probability of each outcome, as shown here.

A discrete probability distribution consists of the values a random variable can assume and the corresponding probabilities of the values. The probabilities are determined theoretically or by observation.

Discrete probability distributions can be shown by using a graph or a table. Probability distributions can also be represented by a formula. See Exercises 31–36 at the end of this section for examples.

Example 5–1

Rolling a Die

Construct a probability distribution for rolling a single die.

Solution

Since the sample space is 1, 2, 3, 4, 5, 6 and each outcome has a probability of , the distribution is as shown.

Probability distributions can be shown graphically by representing the values of X on the x axis and the probabilities P(X) on the y axis.

Example 5–2

Tossing Coins

Represent graphically the probability distribution for the sample space for tossing three coins.

Solution

The values that X assumes are located on the x axis, and the values for P(X) are located on the y axis. The graph is shown in Figure 5–1 .

Note that for visual appearances, it is not necessary to start with 0 at the origin.

Examples 5–1 and 5–2 are illustrations of theoretical probability distributions. You did not need to actually perform the experiments to compute the probabilities. In contrast, to construct actual probability distributions, you must observe the variable over a period of time. They are empirical, as shown in Example 5–3 .

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Figure 5–1

Probability Distribution for Example 5–2

Example 5–3

Baseball World Series

The baseball World Series is played by the winner of the National League and the American League. The first team to win four games wins the World Series. In other words, the series will consist of four to seven games, depending on the individual victories. The data shown consist of the number of games played in the World Series from 1965 through 2005. (There was no World Series in 1994.) The number of games played is represented by the variable X. Find the probability P(X) for each X, construct a probability distribution, and draw a graph for the data.

X	Number of times played
4	8
5	7
6	9
7	16
	40

Solution

The probability P(X) can be computed for each X by dividing the number of games X by the total.

For 4 games, = 0.200	For 6 games, = 0.225
For 5 games, = 0.175	For 7 games, = 0.400

The probability distribution is

The graph is shown in Figure 5–2 .

Figure 5–2

Probability Distribution for Example 5–3

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Speaking of Statistics

Coins, Births, and Other Random (?) Events

Examples of random events such as tossing coins are used in almost all books on probability. But is flipping a coin really a random event?

Tossing coins dates back to ancient Roman times when the coins usually consisted of the Emperor’s head on one side (i.e., heads) and another icon such as a ship on the other side (i.e., ships). Tossing coins was used in both fortune telling and ancient Roman games.

A Chinese form of divination called the I-Ching (pronounced E-Ching) is thought to be at least 4000 years old. It consists of 64 hexagrams made up of six horizontal lines. Each line is either broken or unbroken, representing the yin and the yang. These 64 hexagrams are supposed to represent all possible situations in life. To consult the I-Ching, a question is asked and then three coins are tossed six times. The way the coins fall, either heads up or heads down, determines whether the line is broken (yin) or unbroken (yang). Once the hexagon is determined, its meaning is consulted and interpreted to get the answer to the question. (Note: Another method used to determine the hexagon employs yarrow sticks.)

In the 16th century, a mathematician named Abraham DeMoivre used the outcomes of tossing coins to study what later became known as the normal distribution; however, his work at that time was not widely known.

Mathematicians usually consider the outcomes of a coin toss a random event. That is, each probability of getting a head is , and the probability of getting a tail is . Also, it is not possible to predict with 100% certainty which outcome will occur. But new studies question this theory. During World War II a South African mathematician named John Kerrich tossed a coin 10,000 times while he was interned in a German prison camp. Unfortunately, the results of his experiment were never recorded, so we don’t know the number of heads that occurred.

Several studies have shown that when a coin-tossing device is used, the probability that a coin will land on the same side on which it is placed on the coin-tossing device is about 51%. It would take about 10,000 tosses to become aware of this bias. Furthermore, researchers showed that when a coin is spun on its edge, the coin falls tails up about 80% of the time since there is more metal on the heads side of a coin. This makes the coin slightly heavier on the heads side than on the tails side.

Another assumption commonly made in probability theory is that the number of male births is equal to the number of female births and that the probability of a boy being born is and the probability of a girl being born is . We know this is not exactly true.

In the later 1700s, a French mathematician named Pierre Simon Laplace attempted to prove that more males than females are born. He used records from 1745 to 1770 in Paris and showed that the percentage of females born was about 49%. Although these percentages vary somewhat from location to location, further surveys show they are generally true worldwide. Even though there are discrepancies, we generally consider the outcomes to be 50-50 since these discrepancies are relatively small.

Based on this article, would you consider the coin toss at the beginning of a football game fair?

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Two Requirements for a Probability Distribution

1.The sum of the probabilities of all the events in the sample space must equal 1; that is, ΣP(X) = 1.

2.The probability of each event in the sample space must be between or equal to 0 and 1. That is, 0 ≤ P(X) ≤ 1.

The first requirement states that the sum of the probabilities of all the events must be equal to 1. This sum cannot be less than 1 or greater than 1 since the sample space includes all possible outcomes of the probability experiment. The second requirement states that the probability of any individual event must be a value from 0 to 1. The reason (as stated in Chapter 4 ) is that the range of the probability of any individual value can be 0, 1, or any value between 0 and 1. A probability cannot be a negative number or greater than 1.

Example 5–4

Probability Distributions

Determine whether each distribution is a probability distribution.

Solution

a.Yes, it is a probability distribution.

b.No, it is not a probability distribution, since P(X) cannot be 1.5 or –1.0.

c.Yes, it is a probability distribution.

d.No, it is not, since ΣP(X) = 1.2.

Many variables in business, education, engineering, and other areas can be analyzed by using probability distributions. Section 5–2 shows methods for finding the mean and standard deviation for a probability distribution.

Applying the Concepts 5–1

Dropping College Courses

Use the following table to answer the questions.

Reason for Dropping a College Course	Frequency	Percentage
Too difficult	45
Illness	40
Change in work schedule	20
Change of major	14
Family-related problems	9
Money	7
Miscellaneous	6
No meaningful reason	3

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1.What is the variable under study? Is it a random variable?

2.How many people were in the study?

3.Complete the table.

4.From the information given, what is the probability that a student will drop a class because of illness? Money? Change of major?

5.Would you consider the information in the table to be a probability distribution?

6.Are the categories mutually exclusive?

7.Are the categories independent?

8.Are the categories exhaustive?

9.Are the two requirements for a discrete probability distribution met?

See page 295 for the answers.

Exercises 5–1

1.Define and give three examples of a random variable.

2.Explain the difference between a discrete and a continuous random variable.

3.Give three examples of a discrete random variable.

4.Give three examples of a continuous random variable.

5.What is a probability distribution? Give an example.

For Exercises 6 through 11, determine whether the distribution represents a probability distribution. If it does not, state why.

10.

11.

For Exercises 12 through 18, state whether the variable is discrete or continuous.

12.The speed of a jet airplane

13.The number of cheeseburgers a fast-food restaurant serves each day

14.The number of people who play the state lottery each day

15.The weight of a Siberian tiger

16.The time it takes to complete a marathon

17.The number of mathematics majors in your school

18.The blood pressures of all patients admitted to a hospital on a specific day

For Exercises 19 through 26, construct a probability distribution for the data and draw a graph for the distribution.

19.Medical Tests The probabilities that a patient will have 0, 1, 2, or 3 medical tests performed on entering a hospital are , and , respectively.

20.Student Volunteers The probabilities that a student volunteer hosts 1, 2, 3, or 4 prospective first-year students are 0.4, 0.3, 0.2, and 0.1, respectively.

21.Birthday Cake Sales The probabilities that a bakery has a demand for 2, 3, 5, or 7 birthday cakes on any given day are 0.35, 0.41, 0.15, and 0.09, respectively.

22.DVD Rentals The probabilities that a customer will rent 0, 1, 2, 3, or 4 DVDs on a single visit to the rental store are 0.15, 0.25, 0.3, 0.25, and 0.05, respectively.

23.Loaded Die A die is loaded in such a way that the probabilities of getting 1, 2, 3, 4, 5, and 6 are , , and , respectively.

24.Item Selection The probabilities that a customer selects 1, 2, 3, 4, and 5 items at a convenience store are 0.32, 0.12, 0.23, 0.18, and 0.15, respectively.

25.Student Classes The probabilities that a student is registered for 2, 3, 4, or 5 classes are 0.01, 0.34, 0.62, and 0.03, respectively.

26.Garage Space The probabilities that a randomly selected home has garage space for 0, 1, 2, or 3 cars are 0.22, 0.33, 0.37, and 0.08, respectively.

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27.Selecting a Monetary Bill A box contains three $1 bills, two $5 bills, five $10 bills, and one $20 bill. Construct a probability distribution for the data if x represents the value of a single bill drawn at random and then replaced.

28.Family with Children Construct a probability distribution for a family of three children. Let X represent the number of boys.

29.Drawing a Card Construct a probability distribution for drawing a card from a deck of 40 cards consisting of 10 cards numbered 1, 10 cards numbered 2, 15 cards numbered 3, and 5 cards numbered 4.

30.Rolling Two Dice Using the sample space for tossing two dice, construct a probability distribution for the sums 2 through 12.

Extending the Concepts

A probability distribution can be written in formula notation such as P(X) = 1/X, where X = 2, 3, 6. The distribution is shown as follows:

For Exercises 31 through 36, write the distribution for the formula and determine whether it is a probability distribution.

31. P(X) = X/6 for X = 1, 2, 3

32. P(X) = X for X = 0.2, 0.3, 0.5

33. P(X) = X/6 for X = 3, 4, 7

34. P(X) = X + 0.1 for X = 0.1, 0.02, 0.04

35. P(X) = X/7 for X = 1, 2, 4

36. P(X) = X/(X + 2) for X = 0, 1, 2

Objective 2

Find the mean, variance, standard deviation, and expected value for a discrete random variable.

5–2Mean, Variance, Standard Deviation, and Expectation

The mean, variance, and standard deviation for a probability distribution are computed differently from the mean, variance, and standard deviation for samples. This section explains how these measures—as well as a new measure called the expectation—are calculated for probability distributions.

Mean

In Chapter 3 , the mean for a sample or population was computed by adding the values and dividing by the total number of values, as shown in these formulas:

But how would you compute the mean of the number of spots that show on top when a die is rolled? You could try rolling the die, say, 10 times, recording the number of spots, and finding the mean; however, this answer would only approximate the true mean. What about 50 rolls or 100 rolls? Actually, the more times the die is rolled, the better the approximation. You might ask, then, How many times must the die be rolled to get the exact answer? It must be rolled an infinite number of times. Since this task is impossible, the previous formulas cannot be used because the denominators would be infinity. Hence, a new method of computing the mean is necessary. This method gives the exact theoretical value of the mean as if it were possible to roll the die an infinite number of times.

Before the formula is stated, an example will be used to explain the concept. Suppose two coins are tossed repeatedly, and the number of heads that occurred is recorded. What will be the mean of the number of heads? The sample space is

HH, HT, TH, TT

and each outcome has a probability of . Now, in the long run, you would expect two heads (HH) to occur approximately of the time, one head to occur approximately of the time (HT or TH), and no heads (TT) to occur approximately of the time. Hence, on average, you would expect the number of heads to be

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That is, if it were possible to toss the coins many times or an infinite number of times, the average of the number of heads would be 1.

Hence, to find the mean for a probability distribution, you must multiply each possible outcome by its corresponding probability and find the sum of the products.

Historical Note

A professor, Augustin Louis Cauchy (1789–1857), wrote a book on probability. While he was teaching at the Military School of Paris, one of his students was Napoleon Bonaparte.

Formula for the Mean of a Probability Distribution

The mean of a random variable with a discrete probability distribution is

µ	=	X1 · P(X1) + X2 · P(X2) + X3 · P(X3) + … + Xn · P(Xn)
	=	ΣX · P(X)

where X1, X2, X3, … , Xn are the outcomes and P(X1), P(X2), P(X3), … , P(Xn) are the corresponding probabilities.

Note: ΣX · P(X) means to sum the products.

Rounding Rule for the Mean, Variance, and Standard Deviation for a Probability Distribution The rounding rule for the mean, variance, and standard deviation for variables of a probability distribution is this: The mean, variance, and standard deviation should be rounded to one more decimal place than the outcome X. When fractions are used, they should be reduced to lowest terms.

Examples 5–5 through 5–8 illustrate the use of the formula.

Example 5–5

Rolling a Die

Find the mean of the number of spots that appear when a die is tossed.

Solution

In the toss of a die, the mean can be computed thus.

That is, when a die is tossed many times, the theoretical mean will be 3.5. Note that even though the die cannot show a 3.5, the theoretical average is 3.5.

The reason why this formula gives the theoretical mean is that in the long run, each outcome would occur approximately of the time. Hence, multiplying the outcome by its corresponding probability and finding the sum would yield the theoretical mean. In other words, outcome 1 would occur approximately of the time, outcome 2 would occur approximately of the time, etc.

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Example 5–6

Children in a Family

In a family with two children, find the mean of the number of children who will be girls.

Solution

The probability distribution is as follows:

Hence, the mean is

Example 5–7

Tossing Coins

If three coins are tossed, find the mean of the number of heads that occur. (See the table preceding Example 5–1 .)

Solution

The probability distribution is

The mean is

The value 1.5 cannot occur as an outcome. Nevertheless, it is the long-run or theoretical average.

Example 5–8

Number of Trips of Five Nights or More

The probability distribution shown represents the number of trips of five nights or more that American adults take per year. (That is, 6% do not take any trips lasting five nights or more, 70% take one trip lasting five nights or more per year, etc.) Find the mean.

Solution

µ	=	ΣX · P(X)
	=	(0)(0.06) + (1)(0.70) + (2)(0.20) + (3)(0.03) + (4)(0.01)
	=	0 + 0.70 + 0.40 + 0.09 + 0.04
	=	1.23 ≈ 1.2

Hence, the mean of the number of trips lasting five nights or more per year taken by American adults is 1.2.

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Historical Note

Fey Manufacturing Co., located in San Francisco, invented the first three-reel, automatic payout slot machine in 1895.

Variance and Standard Deviation

For a probability distribution, the mean of the random variable describes the measure of the so-called long-run or theoretical average, but it does not tell anything about the spread of the distribution. Recall from Chapter 3 that in order to measure this spread or variability, statisticians use the variance and standard deviation. These formulas were used:

These formulas cannot be used for a random variable of a probability distribution since N is infinite, so the variance and standard deviation must be computed differently.

To find the variance for the random variable of a probability distribution, subtract the theoretical mean of the random variable from each outcome and square the difference. Then multiply each difference by its corresponding probability and add the products. The formula is

Finding the variance by using this formula is somewhat tedious. So for simplified computations, a shortcut formula can be used. This formula is algebraically equivalent to the longer one and is used in the examples that follow.

Formula for the Variance of a Probability Distribution

Find the variance of a probability distribution by multiplying the square of each outcome by its corresponding probability, summing those products, and subtracting the square of the mean. The formula for the variance of a probability distribution is

σ = Σ[X2 · P(X)] – µ2

The standard deviation of a probability distribution is

Remember that the variance and standard deviation cannot be negative.

Example 5–9

Rolling a Die

Compute the variance and standard deviation for the probability distribution in Example 5–5 .

Solution

Recall that the mean is µ = 3.5, as computed in Example 5–5 . Square each outcome and multiply by the corresponding probability, sum those products, and then subtract the square of the mean.

To get the standard deviation, find the square root of the variance.

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Example 5–10

Selecting Numbered Balls

A box contains 5 balls. Two are numbered 3, one is numbered 4, and two are numbered 5. The balls are mixed and one is selected at random. After a ball is selected, its number is recorded. Then it is replaced. If the experiment is repeated many times, find the variance and standard deviation of the numbers on the balls.

Solution

Let X be the number on each ball. The probability distribution is

The mean is

The variance is

The standard deviation is

The mean, variance, and standard deviation can also be found by using vertical columns, as shown.

Find the mean by summing the ΣX · P(X) column, and find the variance by summing the X2 · P(X) column and subtracting the square of the mean.

σ2 = 16.8 – 42 = 16.8 – 16 = 0.8

and

Example 5–11

On Hold for Talk Radio

A talk radio station has four telephone lines. If the host is unable to talk (i.e., during a commercial) or is talking to a person, the other callers are placed on hold. When all lines are in use, others who are trying to call in get a busy signal. The probability that 0, 1, 2, 3, or 4 people will get through is shown in the distribution. Find the variance and standard deviation for the distribution.

Should the station have considered getting more phone lines installed?

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Solution

The mean is

µ	=	ΣX · P(X)
	=	0 · (0.18) + 1 · (0.34) + 2 · (0.23) + 3 · (0.21) + 4 · (0.04)
	=	1.6

The variance is

σ2	=	Σ[X2 · P(X)] – µ2
	=	[02 · (0.18) + 12 · (0.34) + 22 · (0.23) + 32 · (0.21) + 42 · (0.04)] – 1.62
	=	[0 + 0.34 + 0.92 + 1.89 + 0.64] – 2.56
	=	3.79 – 2.56 = 1.23
	=	1.2 (rounded)

The standard deviation is , or .

No. The mean number of people calling at any one time is 1.6. Since the standard deviation is 1.1, most callers would be accommodated by having four phone lines because µ + 2σ would be 1.6 + 2(1.1) = 1.6 + 2.2 = 3.8. Very few callers would get a busy signal since at least 75% of the callers would either get through or be put on hold. (See Chebyshev’s theorem in Section 3–2.)

Expectation

Another concept related to the mean for a probability distribution is that of expected value or expectation. Expected value is used in various types of games of chance, in insurance, and in other areas, such as decision theory.

The expected value of a discrete random variable of a probability distribution is the theoretical average of the variable. The formula is

µ = E(X) = ΣX · P(X)

The symbol E(X) is used for the expected value.

The formula for the expected value is the same as the formula for the theoretical mean. The expected value, then, is the theoretical mean of the probability distribution. That is, E(X) = µ.

When expected value problems involve money, it is customary to round the answer to the nearest cent.

Example 5–12

Winning Tickets

One thousand tickets are sold at $1 each for a color television valued at $350. What is the expected value of the gain if you purchase one ticket?

Solution

The problem can be set up as follows:

	Win	Lose
Gain X	$349	–$1
Probability P(X)

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Two things should be noted. First, for a win, the net gain is $349, since you do not get the cost of the ticket ($1) back. Second, for a loss, the gain is represented by a negative number, in this case –$1. The solution, then, is

Expected value problems of this type can also be solved by finding the overall gain (i.e., the value of the prize won or the amount of money won, not considering the cost of the ticket for the prize or the cost to play the game) and subtracting the cost of the tickets or the cost to play the game, as shown:

Here, the overall gain ($350) must be used.

Note that the expectation is –$0.65. This does not mean that you lose $0.65, since you can only win a television set valued at $350 or lose $1 on the ticket. What this expectation means is that the average of the losses is $0.65 for each of the 1000 ticket holders. Here is another way of looking at this situation: If you purchased one ticket each week over a long time, the average loss would be $0.65 per ticket, since theoretically, on average, you would win the set once for each 1000 tickets purchased.

Example 5–13

Special Die

A special six-sided die is made in which 3 sides have 6 spots, 2 sides have 4 spots, and 1 side has 1 spot. If the die is rolled, find the expected value of the number of spots that will occur.

Solution

Since there are 3 sides with 6 spots, the probability of getting a 6 is = . Since there are 2 sides with 4 spots, the probability of getting 4 spots is = . The probability of getting 1 spot is since 1 side has 1 spot.

Notice you can only get 1, 4, or 6 spots; but if you rolled the die a large number of times and found the average, it would be about .

Example 5–14

Bond Investment

A financial adviser suggests that his client select one of two types of bonds in which to invest $5000. Bond X pays a return of 4% and has a default rate of 2%. Bond Y has a % return and a default rate of 1%. Find the expected rate of return and decide which bond would be a better investment. When the bond defaults, the investor loses all the investment.

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Solution

The return on bond X is $5000 · 4% = $200. The expected return then is

E(X) = $200(0.98) – $5000(0.02) = $96

The return on bond Y is $5000 · % = $125. The expected return then is

E(X) = $125(0.99) – $5000(0.01) = $73.75

Hence, bond X would be a better investment since the expected return is higher.

In gambling games, if the expected value of the game is zero, the game is said to be fair. If the expected value of a game is positive, then the game is in favor of the player. That is, the player has a better than even chance of winning. If the expected value of the game is negative, then the game is said to be in favor of the house. That is, in the long run, the players will lose money.

In his book Probabilities in Everyday Life (Ivy Books, 1986), author John D. McGervy gives the expectations for various casino games. For keno, the house wins $0.27 on every $1.00 bet. For Chuck-a-Luck, the house wins about $0.52 on every $1.00 bet. For roulette, the house wins about $0.90 on every $1.00 bet. For craps, the house wins about $0.88 on every $1.00 bet. The bottom line here is that if you gamble long enough, sooner or later you will end up losing money.

Applying the Concepts 5–2

Expected Value

On March 28, 1979, the nuclear generating facility at Three Mile Island, Pennsylvania, began discharging radiation into the atmosphere. People exposed to even low levels of radiation can experience health problems ranging from very mild to severe, even causing death. A local newspaper reported that 11 babies were born with kidney problems in the three-county area surrounding the Three Mile Island nuclear power plant. The expected value for that problem in infants in that area was 3. Answer the following questions.

1.What does expected value mean?

2.Would you expect the exact value of 3 all the time?

3.If a news reporter stated that the number of cases of kidney problems in newborns was nearly four times as much as was usually expected, do you think pregnant mothers living in that area would be overly concerned?

4.Is it unlikely that 11 occurred by chance?

5.Are there any other statistics that could better inform the public?

6.Assume that 3 out of 2500 babies were born with kidney problems in that three-county area the year before the accident. Also assume that 11 out of 2500 babies were born with kidney problems in that three-county area the year after the accident. What is the real percent of increase in that abnormality?

7.Do you think that pregnant mothers living in that area should be overly concerned after looking at the results in terms of rates?

See page 296 for the answers.

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Exercises 5–2

1.Defective Transistors From past experience, a company has found that in cartons of transistors, 92% contain no defective transistors, 3% contain one defective transistor, 3% contain two defective transistors, and 2% contain three defective transistors. Find the mean, variance, and standard deviation for the defective transistors.

About how many extra transistors per day would the company need to replace the defective ones if it used 10 cartons per day?

2.Suit Sales The number of suits sold per day at a retail store is shown in the table, with the corresponding probabilities. Find the mean, variance, and standard deviation of the distribution.

If the manager of the retail store wants to be sure that he has enough suits for the next 5 days, how many should the manager purchase?

3.Number of Credit Cards A bank vice president feels that each savings account customer has, on average, three credit cards. The following distribution represents the number of credit cards people own. Find the mean, variance, and standard deviation. Is the vice president correct?

4.Trivia Quiz The probabilities that a player will get 5 to 10 questions right on a trivia quiz are shown below. Find the mean, variance, and standard deviation for the distribution.

5.Cellular Phone Sales The probability that a cellular phone company kiosk sells X number of new phone contracts per day is shown below. Find the mean, variance, and standard deviation for this probability distribution.

What is the probability that they will sell 6 or more contracts three days in a row?

6.Animal Shelter Adoptions The local animal shelter adopts out cats and dogs each week with the following probabilities.

Find the mean, variance, and standard deviation for the number of animals adopted each week. What is the probability that they find homes for more than 5 animals in a given week?

7.Commercials During Children’s TV Programs A concerned parents group determined the number of commercials shown in each of five children’s programs over a period of time. Find the mean, variance, and standard deviation for the distribution shown.

8.Number of Televisions per Household A study conducted by a TV station showed the number of televisions per household and the corresponding probabilities for each. Find the mean, variance, and standard deviation.

If you were taking a survey on the programs that were watched on television, how many program diaries would you send to each household in the survey?

9.Students Using the Math Lab The number of students using the Math Lab per day is found in the distribution below. Find the mean, variance, and standard deviation for this probability distribution.

What is the probability that fewer than 8 or more than 12 use the lab in a given day?

10.Pizza Deliveries A pizza shop owner determines the number of pizzas that are delivered each day. Find the mean, variance, and standard deviation for the distribution shown. If the manager stated that 45 pizzas were delivered on one day, do you think that this is a believable claim?

11.Insurance An insurance company insures a person’s antique coin collection worth $20,000 for an annual premium of $300. If the company figures that the probability of the collection being stolen is 0.002, what will be the company’s expected profit?

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12.Job Bids A landscape contractor bids on jobs where he can make $3000 profit. The probabilities of getting 1, 2, 3, or 4 jobs per month are shown.

Find the contractor’s expected profit per month.

13.Rolling Dice If a person rolls doubles when she tosses two dice, she wins $5. For the game to be fair, how much should she pay to play the game?

14.Dice Game A person pays $2 to play a certain game by rolling a single die once. If a 1 or a 2 comes up, the person wins nothing. If, however, the player rolls a 3, 4, 5, or 6, he or she wins the difference between the number rolled and $2. Find the expectation for this game. Is the game fair?

15.Lottery Prizes A lottery offers one $1000 prize, one $500 prize, and five $100 prizes. One thousand tickets are sold at $3 each. Find the expectation if a person buys one ticket.

16.In Exercise 15, find the expectation if a person buys two tickets. Assume that the player’s ticket is replaced after each draw and that the same ticket can win more than one prize.

17.Winning the Lottery For a daily lottery, a person selects a three-digit number. If the person plays for $1, she can win $500. Find the expectation. In the same daily lottery, if a person boxes a number, she will win $80. Find the expectation if the number 123 is played for $1 and boxed. (When a number is “boxed,” it can win when the digits occur in any order.)

18.Life Insurance A 35-year-old woman purchases a $100,000 term life insurance policy for an annual payment of $360. Based on a period life table for the U.S. government, the probability that she will survive the year is 0.999057. Find the expected value of the policy for the insurance company.

19.Raffle Ticket Sales A civic group sells 1000 raffle tickets to raise $2500 for its namesake charity. First prize is $1000, second prize is $300, and third prize is $200. How much should the group charge for each ticket?

Extending the Concepts

20.Rolling Dice Construct a probability distribution for the sum shown on the faces when two dice are rolled. Find the mean, variance, and standard deviation of the distribution.

21.Rolling a Die When one die is rolled, the expected value of the number of spots is 3.5. In Exercise 20, the mean number of spots was found for rolling two dice. What is the mean number of spots if three dice are rolled?

22.The formula for finding the variance for a probability distribution is

σ = Σ[(X – µ)2 · P(X)]

Verify algebraically that this formula gives the same result as the shortcut formula shown in this section.

23.Rolling a Die Roll a die 100 times. Compute the mean and standard deviation. How does the result compare with the theoretical results of Example 5–5 ?

24.Rolling Two Dice Roll two dice 100 times and find the mean, variance, and standard deviation of the sum of the spots. Compare the result with the theoretical results obtained in Exercise 20.

25.Extracurricular Activities Conduct a survey of the number of extracurricular activities your classmates are enrolled in. Construct a probability distribution and find the mean, variance, and standard deviation.

26.Promotional Campaign In a recent promotional campaign, a company offered these prizes and the corresponding probabilities. Find the expected value of winning. The tickets are free.

Number of prizes	Amount	Probability
1	$100,000
2	10,000
5	1,000
10	100

If the winner has to mail in the winning ticket to claim the prize, what will be the expectation if the cost of the stamp is considered? Use the current cost of a stamp for a first-class letter.

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Speaking of Statistics

This study shows that a part of the brain reacts to the impact of losing, and it might explain why people tend to increase their bets after losing when gambling. Explain how this type of split decision making may influence fighter pilots, firefighters, or police officers, as the article states.

Source: Psychology Today, August 2002, p. 22 . Used with permission.

Technology Step by Step

TI-83 Plus or TI-84 Plus

Step by Step

To calculate the mean and variance for a discrete random variable by using the formulas:

1.Enter the x values into L1 and the probabilities into L2.

2.Move the cursor to the top of the L3 column so that L3 is highlighted.

3.Type L1 multiplied by L2, then press ENTER.

4.Move the cursor to the top of the L4 column so that L4 is highlighted.

5.Type L1 followed by the x2 key multiplied by L2, then press ENTER.

6.Type 2nd QUIT to return to the home screen.

7.Type 2nd LIST, move the cursor to MATH, type 5 for sum, then type L3, then press ENTER.

8.Type 2nd ENTER, move the cursor to L3, type L4, then press ENTER.

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Using the data from Example 5–10 gives the following:

To calculate the mean and standard deviation for a discrete random variable without using the formulas, modify the procedure to calculate the mean and standard deviation from grouped data ( Chapter 3 ) by entering the x values into L1 and the probabilities into L2.

Objective 3

Find the exact probability for X successes in n trials of a binomial experiment.

5–3The Binomial Distribution

Many types of probability problems have only two outcomes or can be reduced to two outcomes. For example, when a coin is tossed, it can land heads or tails. When a baby is born, it will be either male or female. In a basketball game, a team either wins or loses. A true/false item can be answered in only two ways, true or false. Other situations can be reduced to two outcomes. For example, a medical treatment can be classified as effective or ineffective, depending on the results. A person can be classified as having normal or abnormal blood pressure, depending on the measure of the blood pressure gauge. A multiple-choice question, even though there are four or five answer choices, can be classified as correct or incorrect. Situations like these are called binomial experiments.

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Historical Note

In 1653, Blaise Pascal created a triangle of numbers called Pascal’s triangle that can be used in the binomial distribution.

A binomial experiment is a probability experiment that satisfies the following four requirements:

1.There must be a fixed number of trials.

2.Each trial can have only two outcomes or outcomes that can be reduced to two outcomes. These outcomes can be considered as either success or failure.

3.The outcomes of each trial must be independent of one another.

4.The probability of a success must remain the same for each trial.

A binomial experiment and its results give rise to a special probability distribution called the binomial distribution.

The outcomes of a binomial experiment and the corresponding probabilities of these outcomes are called a binomial distribution.

In binomial experiments, the outcomes are usually classified as successes or failures. For example, the correct answer to a multiple-choice item can be classified as a success, but any of the other choices would be incorrect and hence classified as a failure. The notation that is commonly used for binomial experiments and the binomial distribution is defined now.

Notation for the Binomial Distribution

P(S)	The symbol for the probability of success
P(F)	The symbol for the probability of failure
p	The numerical probability of a success
q	The numerical probability of a failure
	P(S) = p and P(F) = 1 – p = q
n	The number of trials
X	The number of successes in n trials

Note that 0 ≤ X ≤ n and X = 0, 1, 2, 3, … , n.

The probability of a success in a binomial experiment can be computed with this formula.

Binomial Probability Formula

In a binomial experiment, the probability of exactly X successes in n trials is

An explanation of why the formula works is given following Example 5–15 .

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Example 5–15

Tossing Coins

A coin is tossed 3 times. Find the probability of getting exactly two heads.

Solution

This problem can be solved by looking at the sample space. There are three ways to get two heads.

HHH, HHT, HTH, THH, TTH, THT, HTT, TTT

The answer is , or 0.375.

Looking at the problem in Example 5–15 from the standpoint of a binomial experiment, one can show that it meets the four requirements.

1.There are a fixed number of trials (three).

2.There are only two outcomes for each trial, heads or tails.

3.The outcomes are independent of one another (the outcome of one toss in no way affects the outcome of another toss).

4.The probability of a success (heads) is in each case.

In this case, n = 3, X = 2, p = , and q = . Hence, substituting in the formula gives

which is the same answer obtained by using the sample space.

The same example can be used to explain the formula. First, note that there are three ways to get exactly two heads and one tail from a possible eight ways. They are HHT, HTH, and THH. In this case, then, the number of ways of obtaining two heads from three coin tosses is 3C2, or 3, as shown in Chapter 4 . In general, the number of ways to get X successes from n trials without regard to order is

This is the first part of the binomial formula. (Some calculators can be used for this.)

Next, each success has a probability of and can occur twice. Likewise, each failure has a probability of and can occur once, giving the part of the formula. To generalize, then, each success has a probability of p and can occur X times, and each failure has a probability of q and can occur n – X times. Putting it all together yields the binomial probability formula.

Example 5–16

Survey on Doctor Visits

A survey found that one out of five Americans say he or she has visited a doctor in any given month. If 10 people are selected at random, find the probability that exactly 3 will have visited a doctor last month.

Source: Reader’s Digest.

Solution

In this case, n = 10, X = 3, p = , and q = . Hence,

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Example 5–17

Survey on Employment

A survey from Teenage Research Unlimited (Northbrook, Illinois) found that 30% of teenage consumers receive their spending money from part-time jobs. If 5 teenagers are selected at random, find the probability that at least 3 of them will have part-time jobs.

Solution

To find the probability that at least 3 have part-time jobs, it is necessary to find the individual probabilities for 3, or 4, or 5, and then add them to get the total probability.

Hence,

P(at least three teenagers have part-time jobs)

= 0.132 + 0.028 + 0.002 = 0.162

Computing probabilities by using the binomial probability formula can be quite tedious at times, so tables have been developed for selected values of n and p. Table B in Appendix C gives the probabilities for individual events. Example 5–18 shows how to use Table B to compute probabilities for binomial experiments.

Example 5–18

Tossing Coins

Solve the problem in Example 5–15 by using Table B .

Solution

Since n = 3, X = 2, and p = 0.5, the value 0.375 is found as shown in Figure 5–3 .

Figure 5–3

Using Table B for Example 5–18

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Example 5–19

Survey on Fear of Being Home Alone at Night

Public Opinion reported that 5% of Americans are afraid of being alone in a house at night. If a random sample of 20 Americans is selected, find these probabilities by using the binomial table.

a.There are exactly 5 people in the sample who are afraid of being alone at night.

b.There are at most 3 people in the sample who are afraid of being alone at night.

c.There are at least 3 people in the sample who are afraid of being alone at night.

Source: 100% American by Daniel Evan Weiss.

Solution

a.n = 20, p = 0.05, and X = 5. From the table, we get 0.002.

b.n = 20 and p = 0.05. “At most 3 people” means 0, or 1, or 2, or 3.

Hence, the solution is

P(0) + P(1) + P(2) + P(3)	=	0.358 + 0.377 + 0.189 + 0.060
	=	0.984

c.n = 20 and p = 0.05. “At least 3 people” means 3, 4, 5, … , 20. This problem can best be solved by finding P(0) + P(1) + P(2) and subtracting from 1.

P(0) + P(1) + P(2)	=	0.358 + 0.377 + 0.189 = 0.924
1 – 0.924	=	0.076

Example 5–20

Driving While Intoxicated

A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at night on weekends involve an intoxicated driver. If a sample of 15 single-vehicle traffic fatalities that occur at night on a weekend is selected, find the probability that exactly 12 involve a driver who is intoxicated.

Source: 100% American by Daniel Evan Weiss.

Solution

Now, n = 15, p = 0.70, and X = 12. From Table B , P(12) = 0.170. Hence, the probability is 0.17.

Remember that in the use of the binomial distribution, the outcomes must be independent. For example, in the selection of components from a batch to be tested, each component must be replaced before the next one is selected. Otherwise, the outcomes are not independent. However, a dilemma arises because there is a chance that the same component could be selected again. This situation can be avoided by not replacing the component and using a distribution called the hypergeometric distribution to calculate the probabilities. The hypergeometric distribution is beyond the scope of this book.

Objective 4

Find the mean, variance, and standard deviation for the variable of a binomial distribution.

Mean, Variance, and Standard Deviation for the Binomial Distribution

The mean, variance, and standard deviation of a variable that has the binomial distribution can be found by using the following formulas.

Mean: µ = n · p

Variance: σ2 = n · p · q

Standard deviation: σ =

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These formulas are algebraically equivalent to the formulas for the mean, variance, and standard deviation of the variables for probability distributions, but because they are for variables of the binomial distribution, they have been simplified by using algebra. The algebraic derivation is omitted here, but their equivalence is shown in Example 5–21 .

Example 5–21

Tossing a Coin

A coin is tossed 4 times. Find the mean, variance, and standard deviation of the number of heads that will be obtained.

Solution

With the formulas for the binomial distribution and n = 4, p = , and q = , the results are

From Example 5–21 , when four coins are tossed many, many times, the average of the number of heads that appear is 2, and the standard deviation of the number of heads is 1. Note that these are theoretical values.

As stated previously, this problem can be solved by using the formulas for expected value. The distribution is shown.

Hence, the simplified binomial formulas give the same results.

Example 5–22

Rolling a Die

A die is rolled 360 times. Find the mean, variance, and standard deviation of the number of 4s that will be rolled.

Solution

This is a binomial experiment since getting a 4 is a success and not getting a 4 is considered a failure. Hence n = 360, p = , and q = .

On average, sixty 4s will be rolled. The standard deviation is 7.07.

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Example 5–23

Likelihood of Twins

The Statistical Bulletin published by Metropolitan Life Insurance Co. reported that 2% of all American births result in twins. If a random sample of 8000 births is taken, find the mean, variance, and standard deviation of the number of births that would result in twins.

Source: 100% American by Daniel Evan Weiss.

Solution

This is a binomial situation, since a birth can result in either twins or not twins (i.e., two outcomes).

For the sample, the average number of births that would result in twins is 160, the variance is 156.8, or 157, and the standard deviation is 12.5, or 13 if rounded.

Applying the Concepts 5–3

Unsanitary Restaurants

Health officials routinely check sanitary conditions of restaurants. Assume you visit a popular tourist spot and read in the newspaper that in 3 out of every 7 restaurants checked, there were unsatisfactory health conditions found. Assuming you are planning to eat out 10 times while you are there on vacation, answer the following questions.

1.How likely is it that you will eat at three restaurants with unsanitary conditions?

2.How likely is it that you will eat at four or five restaurants with unsanitary conditions?

3.Explain how you would compute the probability of eating in at least one restaurant with unsanitary conditions. Could you use the complement to solve this problem?

4.What is the most likely number to occur in this experiment?

5.How variable will the data be around the most likely number?

6.Is this a binomial distribution?

7.If it is a binomial distribution, does that mean that the likelihood of a success is always 50% since there are only two possible outcomes?

Check your answers by using the following computer-generated table.

Mean = 4.3

Std. dev. = 1.56557

X	P(X)	*Cum. Prob.*
0	0.00362	0.00362
1	0.02731	0.03093
2	0.09272	0.12365
3	0.18651	0.31016
4	0.24623	0.55639
5	0.22291	0.77930
6	0.14013	0.91943
7	0.06041	0.97983
8	0.01709	0.99692
9	0.00286	0.99979
10	0.00022	1.00000

See page 296 for the answers.

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Exercises 5–3

1.Which of the following are binomial experiments or can be reduced to binomial experiments?

a.Surveying 100 people to determine if they like Sudsy Soap

b.Tossing a coin 100 times to see how many heads occur

c.Drawing a card with replacement from a deck and getting a heart

d.Asking 1000 people which brand of cigarettes they smoke

e.Testing four different brands of aspirin to see which brands are effective

f.Testing one brand of aspirin by using 10 people to determine whether it is effective

g.Asking 100 people if they smoke

h.Checking 1000 applicants to see whether they were admitted to White Oak College

i.Surveying 300 prisoners to see how many different crimes they were convicted of

j.Surveying 300 prisoners to see whether this is their first offense

2.(ans) Compute the probability of X successes, using Table B in Appendix C .

a.n = 2, p = 0.30, X = 1

b.n = 4, p = 0.60, X = 3

c.n = 5, p = 0.10, X = 0

d.n = 10, p = 0.40, X = 4

e.n = 12, p = 0.90, X = 2

f.n = 15, p = 0.80, X = 12

g.n = 17, p = 0.05, X = 0

h.n = 20, p = 0.50, X = 10

i.n = 16, p = 0.20, X = 3

3.Compute the probability of X successes, using the binomial formula.

a.n = 6, X = 3, p = 0.03

b.n = 4, X = 2, p = 0.18

c.n = 5, X = 3, p = 0.63

d.n = 9, X = 0, p = 0.42

e.n = 10, X = 5, p = 0.37

For Exercises 4 through 13, assume all variables are binomial. (Note: If values are not found in Table B of Appendix C , use the binomial formula.)

4.Burglar Alarms A burglar alarm system has six fail-safe components. The probability of each failing is 0.05. Find these probabilities.

a.Exactly three will fail.

b.Fewer than two will fail.

c.None will fail.

d.Compare the answers for parts a, b, and c, and explain why the results are reasonable.

5.True/False Exam A student takes a 20-question, true/false exam and guesses on each question. Find the probability of passing if the lowest passing grade is 15 correct out of 20. Would you consider this event likely to occur? Explain your answer.

6.Multiple-Choice Exam A student takes a 20-question, multiple-choice exam with five choices for each question and guesses on each question. Find the probability of guessing at least 15 out of 20 correctly. Would you consider this event likely or unlikely to occur? Explain your answer.

7.Driving to Work Alone It is reported that 77% of workers aged 16 and over drive to work alone. Choose 8 workers at random. Find the probability that

a.All drive to work alone

b.More than one-half drive to work alone

c.Exactly 3 drive to work alone

Source: www.factfinder.census.gov

8.High School Dropouts Approximately 10.3% of American high school students drop out of school before graduation. Choose 10 students entering high school at random. Find the probability that

a.No more than two drop out

b.At least 6 graduate

c.All 10 stay in school and graduate

Source: www.infoplease.com

9.Survey on Concern for Criminals In a survey, 3 of 4 students said the courts show “too much concern” for criminals. Find the probability that at most 3 out of 7 randomly selected students will agree with this statement.

Source: Harper’s Index.

10.Labor Force Couples The percentage of couples where both parties are in the labor force is 52.1. Choose 5 couples at random. Find the probability that

a.None of the couples have both persons working

b.More than 3 of the couples have both persons in the labor force

c.Fewer than 2 of the couples have both parties working

Source: www.bls.gov

11.College Education and Business World Success R. H. Bruskin Associates Market Research found that 40% of Americans do not think that having a college education is important to succeed in the business world. If a random sample of five Americans is selected, find these probabilities.

a.Exactly 2 people will agree with that statement.

b.At most 3 people will agree with that statement.

c.At least 2 people will agree with that statement.

d.Fewer than 3 people will agree with that statement.

Source: 100% American by Daniel Evans Weiss.

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12.Destination Weddings Twenty-six percent of couples who plan to marry this year are planning destination weddings. In a random sample of 12 couples who plan to marry, find the probability that

a.Exactly 6 couples will have a destination wedding

b.At least 6 couples will have a destination wedding

c.Fewer than 5 couples will have a destination wedding

Source: Time magazine.

13.People Who Have Some College Education Fifty-three percent of all persons in the U.S. population have at least some college education. Choose 10 persons at random. Find the probability that

a.Exactly one-half have some college education

b.At least 5 do not have any college education

c.Fewer than 5 have some college education

Source: New York Times Almanac.

14.(ans) Find the mean, variance, and standard deviation for each of the values of n and p when the conditions for the binomial distribution are met.

a.n = 100, p = 0.75

b.n = 300, p = 0.3

c.n = 20, p = 0.5

d.n = 10, p = 0.8

e.n = 1000, p = 0.1

f.n = 500, p = 0.25

g.n = 50, p =

h.n = 36, p =

15.Social Security Recipients A study found that 1% of Social Security recipients are too young to vote. If 800 Social Security recipients are randomly selected, find the mean, variance, and standard deviation of the number of recipients who are too young to vote.

Source: Harper’s Index.

16.Find the mean, variance, and standard deviation for the number of heads when 20 coins are tossed.

17.Defective Calculators If 3% of calculators are defective, find the mean, variance, and standard deviation of a lot of 300 calculators.

18.Federal Government Employee E-mail Use It has been reported that 83% of federal government employees use e-mail. If a sample of 200 federal government employees is selected, find the mean, variance, and standard deviation of the number who use e-mail.

Source: USA TODAY.

19.Watching Fireworks A survey found that 21% of Americans watch fireworks on television on July 4. Find the mean, variance, and standard deviation of the number of individuals who watch fireworks on television on July 4 if a random sample of 1000 Americans is selected.

Source: USA Snapshot, USA TODAY.

20.Alternate Sources of Fuel Eighty-five percent of Americans favor spending government money to develop alternative sources of fuel for automobiles. For a random sample of 120 Americans, find the mean, variance, and standard deviation for the number who favor government spending for alternative fuels.

Source: www.pollingreport.com

21.Survey on Bathing Pets A survey found that 25% of pet owners had their pets bathed professionally rather than do it themselves. If 18 pet owners are randomly selected, find the probability that exactly 5 people have their pets bathed professionally.

Source: USA Snapshot, USA TODAY.

22.Survey on Answering Machine Ownership In a survey, 63% of Americans said they own an answering machine. If 14 Americans are selected at random, find the probability that exactly 9 own an answering machine.

Source: USA Snapshot, USA TODAY.

23.Poverty and the Federal Government One out of every three Americans believes that the U.S. government should take “primary responsibility” for eliminating poverty in the United States. If 10 Americans are selected, find the probability that at most 3 will believe that the U.S. government should take primary responsibility for eliminating poverty.

Source: Harper’s Index.

24.Internet Purchases Thirty-two percent of adult Internet users have purchased products or services online. For a random sample of 200 adult Internet users, find the mean, variance, and standard deviation for the number who have purchased goods or services online.

Source: www.infoplease.com

25.Survey on Internet Awareness In a survey, 58% of American adults said they had never heard of the Internet. If 20 American adults are selected at random, find the probability that exactly 12 will say they have never heard of the Internet.

Source: Harper’s Index.

26.Job Elimination In the past year, 13% of businesses have eliminated jobs. If 5 businesses are selected at random, find the probability that at least 3 have eliminated jobs during the last year.

Source: USA TODAY.

27.Survey of High School Seniors Of graduating high school seniors, 14% said that their generation will be remembered for their social concerns. If 7 graduating seniors are selected at random, find the probability that either 2 or 3 will agree with that statement.

Source: USA TODAY.

28.Is this a binomial distribution? Explain.

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Extending the Concepts

29.Children in a Family The graph shown here represents the probability distribution for the number of girls in a family of three children. From this graph, construct a probability distribution

30.Construct a binomial distribution graph for the number of defective computer chips in a lot of 4 if p = 0.3.

Technology Step by Step

MINITAB

Step by Step

The Binomial Distribution

Calculate a Binomial Probability

From Example 5–19 , it is known that 5% of the population is afraid of being alone at night. If a random sample of 20 Americans is selected, what is the probability that exactly 5 of them are afraid?

n = 20 p = 0.05 (5%) and X = 5 (5 out of 20)

No data need to be entered in the worksheet.

1.Select Calc>Probability Distributions>Binomial.

2.Click the option for Probability.

3.Click in the text box for Number of trials:.

4.Type in 20, then Tab to Probability of success, then type .05.

5.Click the option for Input constant, then type in 5. Leave the text box for Optional storage empty. If the name of a constant such as K1 is entered here, the results are stored but not displayed in the session window.

6.Click [OK]. The results are visible in the session window.

Probability Density Function

Binomial with n = 20 and p = 0.05

x f(x)

5 0.0022446

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Construct a Binomial Distribution

These instructions will use n = 20 and p = 0.05.

1.Select Calc>Make Patterned Data>Simple Set of Numbers.

2.You must enter three items:

a)Enter X in the box for Store patterned data in:. MINITAB will use the first empty column of the active worksheet and name it X.

b)Press Tab. Enter the value of 0 for the first value. Press Tab.

c)Enter 20 for the last value. This value should be n. In steps of:, the value should be 1.

3.Click [OK].

4.Select Calc>Probability Distributions>Binomial.

5.In the dialog box you must enter five items.

a)Click the button for Probability.

b)In the box for Number of trials enter 20.

c)Enter .05 in the Probability of success.

d)Check the button for Input columns, then type the column name, X, in the text box.

e)Click in the box for Optional storage, then type Px.

6.Click [OK]. The first available column will be named Px, and the calculated probabilities will be stored in it.

7.To view the completed table, click the worksheet icon on the toolbar.

Graph a Binomial Distribution

The table must be available in the worksheet.

1.Select Graph>Scatterplot, then Simple.

a)Double-click on C2 Px for the Y variable and C1 X for the X variable.

b)Click [Data view], then Project lines, then [OK]. Deselect any other type of display that may be selected in this list.

c)Click on [Labels], then Title/Footnotes.

d)Type an appropriate title, such as Binomial Distribution n = 20, p = .05.

e)Press Tab to the Subtitle 1, then type in Your Name.

f)Optional: Click [Scales] then [Gridlines] then check the box for Y major ticks.

g)Click [OK] twice.

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The graph will be displayed in a window. Right-click the control box to save, print, or close the graph.

TI-83 Plus or TI-84 Plus

Step by Step

Binomial Random Variables

To find the probability for a binomial variable:

Press 2nd [DISTR] then 0 for binomial pdf( (Note: On the TI-84 Plus Use A)

The form is binompdf(n,p,X).

Example: n = 20, X = 5, p = .05. ( Example 5–19a from the text) binompdf(20,.05,5)

Example: n = 20, X = 0, 1, 2, 3, p = .05. ( Example 5–19b from the text) binompdf(20,.05,{0,1,2,3})

The calculator will display the probabilities in a list. Use the arrow keys to view entire display.

To find the cumulative probability for a binomial random variable:

Press 2nd [DISTR] then A (ALPHA MATH) for binomcdf( (Note: On the TI-84 Plus Use B) The form is binomcdf(n,p,X). This will calculate the cumulative probability for values from 0 to X.

Example: n = 20, X = 0, 1, 2, 3, p = .05 ( Example 5–19b from the text) binomcdf(20,.05,3)

To construct a binomial probability table:

1.Enter the X values 0 through n into L1.

2.Move the cursor to the top of the L2 column so that L2 is highlighted.

3.Type the command binompdf(n,p,L1), then press ENTER.

Example: n = 20, p = .05 ( Example 5–19 from the text)

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Excel

Step by Step

Creating a Binomial Distribution and Graph

These instructions will demonstrate how Excel can be used to construct a binomial distribution table for n = 20 and p = 0.35.

1.Type X for the binomial variable label in cell A1 of an Excel worksheet.

2.Type P(X) for the corresponding probabilities in cell B1.

3.Enter the integers from 0 to 20 in column A starting at cell A2. Select the Data tab from the toolbar. Then select Data Analysis. Under Analysis Tools, select Random Number Generation and click [OK].

4.In the Random Number Generation dialog box, enter the following:

a)Number of Variables: 1

b)Distribution: Patterned

c)Parameters: From 0 to 20 in steps of 1, repeating each number: 1 times and repeating each sequence 1 times

d)Output range: A2:A21

5.Then click [OK].

6.To determine the probability corresponding to the first value of the binomial random variable, select cell B2 and type: =BINOMDIST(0,20,.35,FALSE). This will give the probability of obtaining 0 successes in 20 trials of a binomial experiment for which the probability of success is 0.35.

7.Repeat step 6, changing the first parameter, for each of the values of the random variable from column A.

Note: If you wish to obtain the cumulative probabilities for each of the values in column A, you can type: =BINOMDIST(0,20,.35,TRUE) and repeat for each of the values in column A.

To create the graph:

1.Select the Insert tab from the toolbar and the Column Chart.

2.Select the Clustered Column (the first column chart under the 2-D Column selections).

3.You will need to edit the data for the chart.

a)Right-click the mouse on any location of the chart. Click the Select Data option. The Select Data Source dialog box will appear.

b)Click X in the Legend Entries box and click Remove.

c)Click the Edit button under Horizontal Axis Labels to insert a range for the variable X.

d)When the Axis Labels box appears, highlight cells A2 to A21 on the worksheet, then click [OK].

4.To change the title of the chart:

a)Left-click once on the current title.

b)Type a new title for the chart, for example, Binomial Distribution (20, .35, .65).

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Summary

Many variables have special probability distributions. This chapter presented several of the most common probability distributions, including the binomial distribution, the multinomial distribution, the Poisson distribution, and the hypergeometric distribution.

The binomial distribution is used when there are only two outcomes for an experiment, there are a fixed number of trials, the probability is the same for each trial, and the outcomes are independent of one another.

A probability distribution can be graphed, and the mean, variance, and standard deviation can be found. The mathematical expectation can also be calculated for a probability distribution. Expectation is used in insurance and games of chance.

Important Terms

binomial distribution

binomial experiment

discrete probability distribution

expected value

random variable

Important Formulas

Formula for the mean of a probability distribution:

µ = ΣX · P(X)

Formulas for the variance and standard deviation of a probability distribution:

Formula for expected value:

E(X) = ΣX · P(X)

Binomial probability formula:

Formula for the mean of the binomial distribution:

µ = n · p

Formulas for the variance and standard deviation of the binomial distribution:

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Review Exercises

For Exercises 1 through 3, determine whether the distribution represents a probability distribution. If it does not, state why.

4.Emergency Calls The number of emergency calls a local police department receives per 24-hour period is distributed as shown here. Construct a graph for the data.

5.Credit Cards A large retail company encourages its employees to get customers to apply for the store credit card. Below is the distribution for the number of credit card applications received per employee for an 8-hour shift.

a. What is the probability that an employee will get 2 or 3 applications during any given shift?

b. Find the mean, variance, and standard deviation for this probability distribution.

6.Coins in a Box A box contains 5 pennies, 3 dimes, 1 quarter, and 1 half-dollar. Construct a probability distribution and draw a graph for the data.

7. Tie Purchases At Tyler’s Tie Shop, Tyler found the probabilities that a customer will buy 0, 1, 2, 3, or 4 ties, as shown. Construct a graph for the distribution.

8.Customers in a Bank A bank has a drive-through service. The number of customers arriving during a 15-minute period is distributed as shown. Find the mean, variance, and standard deviation for the distribution.

9. Museum Visitors At a small community museum, the number of visitors per hour during the day has the distribution shown here. Find the mean, variance, and standard deviation for the data.

10.Cans of Paint Purchased During a recent paint sale at Corner Hardware, the number of cans of paint purchased was distributed as shown. Find the mean, variance, and standard deviation of the distribution.

11. Inquiries Received The number of inquiries received per day for a college catalog is distributed as shown. Find the mean, variance, and standard deviation for the data.

12.Outdoor Regatta A producer plans an outdoor regatta for May 3. The cost of the regatta is $8000. This includes advertising, security, printing tickets, entertainment, etc. The producer plans to make $15,000 profit if all goes well. However, if it rains, the regatta will have to be canceled. According to the weather report, the probability of rain is 0.3. Find the producer’s expected profit.

13. Card Game A game is set up as follows: All the diamonds are removed from a deck of cards, and these 13 cards are placed in a bag. The cards are mixed up, and then one card is chosen at random (and then replaced). The player wins according to the following rules.

If the ace is drawn, the player loses $20.

If a face card is drawn, the player wins $10.

If any other card (2–10) is drawn, the player wins $2.

How much should be charged to play this game in order for it to be fair?

14.Using Exercise 13, how much should be charged if instead of winning $2 for drawing a 2–10, the player wins the amount shown on the card in dollars?

15.Let x be a binomial random variable with n = 12 and p = 0.3. Find the following:

a. P(X = 8)

b. P(X < 5)

c. P(X ≥ 10)

d. P(4 < X ≤ 9)

16.Internet Access via Cell Phone Fourteen percent of cell phone users use their cell phones to access the Internet. In a random sample of 10 cell phone users, what is the probability that exactly 2 have used their phones to access the Internet? More than 2?

Source: www.infoplease.com

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17. Drug Calculation Test If 75% of nursing students are able to pass a drug calculation test, find the mean, variance, and standard deviation of the number of students who pass the test in a sample of 180 nursing students.

18.Flu Shots It has been reported that 63% of adults aged 65 and over got their flu shots last year. In a random sample of 300 adults aged 65 and over, find the mean, variance, and standard deviation for the number who got their flu shots.

Source: U.S. Center for Disease Control and Prevention.

19. U.S. Police Chiefs and the Death Penalty The chance that a U.S. police chief believes the death penalty “significantly reduces the number of homicides” is 1 in 4. If a random sample of 8 police chiefs is selected, find the probability that at most 3 believe that the death penalty significantly reduces the number of homicides.

Source: Harper’s Index.

20.Household Wood Burning American Energy Review reported that 27% of American households burn wood. If a random sample of 500 American households is selected, find the mean, variance, and standard deviation of the number of households that burn wood.

Source: 100% American by Daniel Evan Weiss.

21. Pizza for Breakfast Three out of four American adults under age 35 have eaten pizza for breakfast. If a random sample of 20 adults under age 35 is selected, find the probability that exactly 16 have eaten pizza for breakfast.

Source: Harper’s Index.

22.Unmarried Women According to survey records, 75.4% of women aged 20–24 have never been married. In a random sample of 250 young women aged 20–24, find the mean, variance, and standard deviation for the number who are or who have been married.

Source: www.infoplease.com

Statistics Today

Is Pooling Worthwhile?—Revisited

In the case of the pooled sample, the probability that only one test will be needed can be determined by using the binomial distribution. The question being asked is, In a sample of 15 individuals, what is the probability that no individual will have the disease? Hence, n = 15, p = 0.05, and X = 0. From Table B in Appendix C , the probability is 0.463, or 46% of the time, only one test will be needed. For screening purposes, then, pooling samples in this case would save considerable time, money, and effort as opposed to testing every individual in the population.

Chapter Quiz

Determine whether each statement is true or false. If the statement is false, explain why.

1. The expected value of a random variable can be thought of as a long-run average.

2. The number of courses a student is taking this semester is an example of a continuous random variable.

3. When the binomial distribution is used, the outcomes must be dependent.

4. A binomial experiment has a fixed number of trials.

Complete these statements with the best answer.

5. Random variable values are determined by _______.

6. The mean for a binomial variable can be found by using the formula _______.

7. One requirement for a probability distribution is that the sum of all the events in the sample space must equal ______.

Select the best answer.

8. What is the sum of the probabilities of all outcomes in a probability distribution?

a.0

c.1

d.It cannot be determined.

9. How many outcomes are there in a binomial experiment?

a.0

b.1

c.2

d.It varies.

10. The number of trials for a binomial experiment

a.Can be infinite

b.Is unchanged

c.Is unlimited

d.Must be fixed

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For questions 11 through 14, determine if the distribution represents a probability distribution. If not, state why.

15. Calls for a Fire Company The number of fire calls the Conestoga Valley Fire Company receives per day is distributed as follows:

Construct a graph for the data.

16. Telephones per Household A study was conducted to determine the number of telephones each household has. The data are shown here.

Construct a probability distribution and draw a graph for the data.

17. CD Purchases During a recent CD sale at Matt’s Music Store, the number of CDs customers purchased was distributed as follows:

Find the mean, variance, and standard deviation of the distribution.

18. Calls for a Crisis Hot Line The number of calls received per day at a crisis hot line is distributed as follows:

Find the mean, variance, and standard deviation of the distribution.

19. Selecting a Card There are 6 playing cards placed face down in a box. They are the 4 of diamonds, the 5 of hearts, the 2 of clubs, the 10 of spades, the 3 of diamonds, and the 7 of hearts. A person selects a card. Find the expected value of the draw.

20. Selecting a Card A person selects a card from an ordinary deck of cards. If it is a black card, she wins $2. If it is a red card between or including 3 and 7, she wins $10. If it is a red face card, she wins $25; and if it is a black jack, she wins an extra $100. Find the expectation of the game.

21. Carpooling If 40% of all commuters ride to work in carpools, find the probability that if 8 workers are selected, 5 will ride in carpools.

22.Employed Women If 60% of all women are employed outside the home, find the probability that in a sample of 20 women,

a. Exactly 15 are employed

b. At least 10 are employed

c. At most 5 are not employed outside the home

23. Driver’s Exam If 80% of the applicants are able to pass a driver’s proficiency road test, find the mean, variance, and standard deviation of the number of people who pass the test in a sample of 300 applicants.

24. Meeting Attendance A history class has 75 members. If there is a 12% absentee rate per class meeting, find the mean, variance, and standard deviation of the number of students who will be absent from each class.

Critical Thinking Challenges

1.Lottery Numbers Pennsylvania has a lottery entitled “Big 4.” To win, a player must correctly match four digits from a daily lottery in which four digits are selected. Find the probability of winning.

2.Lottery Numbers In the Big 4 lottery, for a bet of $100, the payoff is $5000. What is the expected value of winning? Is it worth it?

3.Lottery Numbers If you played the same four-digit number every day (or any four-digit number for that matter) in the Big 4, how often (in years) would you win, assuming you have average luck?

4.Chuck-a-Luck In the game Chuck-a-Luck, three dice are rolled. A player bets a certain amount (say $1.00) on a number from 1 to 6. If the number appears on 1 die, the person wins $1.00. If it appears on 2 dice, the person wins $2.00, and if it appears on all 3 dice, the person wins $3.00. What are the chances of winning $1.00? $2.00? $3.00?

5.Chuck-a-Luck What is the expected value of the game of Chuck-a-Luck if a player bets $1.00 on one number?

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Data Projects

1.Business and Finance Assume that a life insurance company would like to make a profit of $250 on a $100,000 policy sold to a person whose probability of surviving the year is 0.9985. What premium should the company charge the customer? If the company would like to make a $250 profit on a $100,000 policy at a premium of $500, what is the lowest life expectancy it should accept for a customer?

2.Sports and Leisure Baseball, hockey, and basketball all use a seven-game series to determine their championship. Find the probability that with two evenly matched teams a champion will be found in 4 games. Repeat for 5, 6, and 7 games. Look at the historical results for the three sports. How do the actual results compare to the theoretical?

3.Technology Use your most recent itemized phone bill for the data in this problem. Assume that incoming and outgoing calls are equal in the population (why is this a reasonable assumption?). This means assume p = 0.5. For the number of calls you made last month, what would be the mean number of outgoing calls in a random selection of calls? Also, compute the standard deviation. Was the number of outgoing calls you made an unusual amount given the above? In a selection of 12 calls, what is the probability that less than 3 were outgoing?

4.Health and Wellness Use Red Cross data to determine the percentage of the population with an Rh factor that is positive (A+, B+, AB+ , or O+ blood types). Use that value for p. How many students in your class have a positive Rh factor? Is this an unusual amount?

5.Politics and Economics Find out what percentage of citizens in your state is registered to vote. Assuming that this is a binomial variable, what would be the mean number of registered voters in a random group of citizens with a sample size equal to the number of students in your class? Also determine the standard deviation. How many students in your class are registered to vote? Is this an unusual number, given the above?

6.Your Class Have each student in class toss 4 coins on her or his desk, and note how many heads are showing. Create a frequency table displaying the results. Compare the frequency table to the theoretical probability distribution for the outcome when 4 coins are tossed. Find the mean for the frequency table. How does it compare with the mean for the probability distribution?

Answers to Applying the Concepts

Section 5–1 Dropping College Courses

1.The random variable under study is the reason for dropping a college course.

2.There were a total of 144 people in the study.

3.The complete table is as follows:

Reason for dropping a college course	Frequency	Percentage
Too difficult	45	31.25
Illness	40	27.78
Change in work schedule	20	13.89
Change of major	14	9.72
Family-related problems	9	6.25
Money	7	4.86
Miscellaneous	6	4.17
No meaningful reason	3	2.08

4.The probability that a student will drop a class because of illness is about 28%. The probability that a student will drop a class because of money is about 5%. The probability that a student will drop a class because of a change of major is about 10%.

5.The information is not itself a probability distribution, but it can be used as one.

6.The categories are not necessarily mutually exclusive, but we treated them as such in computing the probabilities.

7.The categories are not independent.

8.The categories are exhaustive.

9.Since all the probabilities are between 0 and 1, inclusive, and the probabilities sum to 1, the requirements for a discrete probability distribution are met.

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Section 5–2 Expected Value

1.The expected value is the mean in a discrete probability distribution.

2.We would expect variation from the expected value of 3.

3.Answers will vary. One possible answer is that pregnant mothers in that area might be overly concerned upon hearing that the number of cases of kidney problems in newborns was nearly 4 times what was usually expected. Other mothers (particularly those who had taken a statistics course!) might ask for more information about the claim.

4.Answers will vary. One possible answer is that it does seem unlikely to have 11 newborns with kidney problems when we expect only 3 newborns to have kidney problems.

5.The public might better be informed by percentages or rates (e.g., rate per 1000 newborns).

6.The increase of 8 babies born with kidney problems represents a 0.32% increase (less than %).

7.Answers will vary. One possible answer is that the percentage increase does not seem to be something to be overly concerned about.

Section 5–3 Unsanitary Restaurants

1.The probability of eating at 3 restaurants with unsanitary conditions out of the 10 restaurants is 0.18651.

2.The probability of eating at 4 or 5 restaurants with unsanitary conditions out of the 10 restaurants is 0.24623 + 0.22291 = 0.46914.

3.To find this probability, you could add the probabilities for eating at 1, 2, … , 10 unsanitary restaurants. An easier way to compute the probability is to subtract the probability of eating at no unsanitary restaurants from 1 (using the complement rule).

4.The highest probability for this distribution is 4, but the expected number of unsanitary restaurants that you would eat at is 10 • = 4.3.

5.The standard deviation for this distribution is = 1.56.

6.This is a binomial distribution. We have two possible outcomes: “success” is eating in an unsanitary restaurant; “failure” is eating in a sanitary restaurant. The probability that one restaurant is unsanitary is independent of the probability that any other restaurant is unsanitary. The probability that a restaurant is unsanitary remains constant at . And we are looking at the number of unsanitary restaurants that we eat at out of 10 “trials.”

7.The likelihood of success will vary from situation to situation. Just because we have two possible outcomes, this does not mean that each outcome occurs with probability 0.50.