# Mathematics

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

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

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