Solutions

.10 + .05 = .15

1

.35 + .40 + .10 = .85

1(.15) + 2(.35) + 3(.40) + 4(.10) = .15 + .70 + 1.20 + .40 = 2.45

Let X = the number of events Javier volunteers for each month.

1 – .05 = .95

.2 + 1.2 + 2.4 + 1.6 = 5.4

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

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

1 – .35 – .20 – .15 – .10 – .05 = .15

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

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

.10 + .05 + .10 = .25

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

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

X = the number that reply yes

0, 1, 2, 3, 4, 5, 6, 7, 8

5.7

.4151

X = the number of freshmen selected from the study until one replied yes to the law that was passed.

1,2,…

1.4

X = the number of business majors in the sample.

2, 3, 4, 5, 6, 7, 8, 9

6.26

0, 1, 2, 3, 4, …

.0485

.0214

X = the number of United States teens who die from motor vehicle injuries per day.

0, 1, 2, 3, 4, ...

no

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

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

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

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

1. .1
2. 1.6

1. Software Company
   x	P(x)
   5,000,000	.10
   1,000,000	.30
   –1,000,000	.60

   Table 4.42

   Hardware Company
   x	P(x)
   3,000,000	.20
   1,000,000	.40
   –1,000,00	.40

   Table 4.43

   Biotech Firm
   x	P(x)
   6,000,000	.10
   0	.70
   –1,000,000	.20

   Table 4.44
2. $200,000; $600,000; $400,000
3. third investment because it has the lowest probability of loss
4. first investment because it has the highest probability of loss
5. second investment

4.85 years

b

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

Calculate the expected value of X.

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

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

X = the number of patients calling in claiming to have the flu, who actually have the flu.

X = 0, 1, 2, ...25

.0165

1. X = the number of DVDs a Video to Go customer rents
2. .12
3. .11
4. .77

d. 4.43

c

• X = number of questions answered correctly
• X ~ B$\left(\text{32, }\frac{\text{1}}{\text{3}}\right)$
• We are interested in MORE THAN 75 percent of 32 questions correct. 75 percent of 32 is 24. We want to find P(x > 24). The event more than 24 is the complement of less than or equal to 24.
• Using your calculator's distribution menu: 1 – binomcdf$\left(\text{32, }\frac{\text{1}}{\text{3}},\text{ 24}\right)$
• P(x > 24) = 0
• The probability of getting more than 75 percent of the 32 questions correct when randomly guessing is very small and practically zero.

1. X = the number of college and universities that offer online offerings.
2. 0, 1, 2, …, 13
3. X ~ B(13, 0.96)
4. 12.48
5. .0135
6. P(x = 12) = .3186 P(x = 13) = 0.5882 More likely to get 13.

1. X = the number of fencers who do not use the foil as their main weapon
2. 0, 1, 2, 3,... 25
3. X ~ B(25,.40)
4. 10
5. .0442
6. The probability that all 25 not use the foil is almost zero. Therefore, it would be very surprising.

1. X = the number of audits in a 20-year period
2. 0, 1, 2, …, 20
3. X ~ B(20, .02)
4. .4
5. .6676
6. .0071

1. X = the number of matches
2. 0, 1, 2, 3
3. X ~ B$\left(3,\frac{1}{6}\right)$
4. In dollars: −1, 1, 2, 3
5. $\frac{1}{2}$
6. Multiply each Y value by the corresponding X probability from the PDF table. The answer is −.0787. You lose about eight cents, on average, per game.
7. The house has the advantage.

1. X ~ B(15, .281)

   Figure 4.10
2. 1. Mean = μ = np = 15(.281) = 4.215
   2. Standard Deviation = σ = $\sqrt{npq}$ = $\sqrt{15(.281)(.719)}$ = 1.7409
3. P(x > 5) = 1 – P(x ≤ 5) = 1 – binomcdf(15, .281, 5) = 1 – 0.7754 = .2246
   P(x = 3) = binompdf(15, .281, 3) = .1927
   P(x = 4) = binompdf(15, .281, 4) = .2259
   It is more likely that four people are literate than three people are.

1. X = the number of adults in America who are surveyed until one says he or she will watch the Super Bowl.
2. X ~ G(.40)
3. 2.5
4. .0187
5. .2304

1. X = the number of pages that advertise footwear
2. X takes on the values 0, 1, 2, ..., 20
3. X ~ B(20, $\frac{29}{192}$)
4. 3.02
5. no
6. .9997
7. X = the number of pages we must survey until we find one that advertises footwear. X ~ G($\frac{29}{192}$)
8. .3881
9. 6.6207 pages

0, 1, 2, and 3

1. X ~ G(.25)
2. 1. mean = μ = $\frac{1}{p}$ = $\frac{1}{0.25}$ = 4
   2. standard deviation = σ = $\sqrt{\frac{1-p}{p^2}}$ = $\sqrt{\frac{1-\text{.25}}{{.25}^2}}$ ≈ 3.4641
3. P(x = 10) = geometpdf(.25, 10) = .0188
4. P(x = 20) = geometpdf(.25, 20) = .0011
5. P(x ≤ 5) = geometcdf(.25, 5) = .7627

1. X = the number of pages that advertise footwear
2. 0, 1, 2, 3, ..., 20
3. X ~ H(29, 163, 20), r = 29, b = 163, n = 20
4. 3.03
5. 1.5197

1. X = the number of Patriots picked
2. 0, 1, 2, 3, 4
3. X ~ H(4, 8, 9)
4. without replacement

1. X ~ P(5.5); μ = 5.5; $\sigma \text{ = }\sqrt{5.5}$ ≈ 2.3452
2. P(x ≤ 6) = poissoncdf(5.5, 6) ≈ .6860
3. There is a 15.7 percent probability that the law staff will receive more calls than they can handle.
4. P(x > 8) = 1 – P(x ≤ 8) = 1 – poissoncdf(5.5, 8) ≈ 1 – .8944 = .1056

Let X = the number of defective bulbs in a string.

Using the Poisson distribution:

• μ = np = 100(.03) = 3
• X ~ P(3)
• P(x ≤ 4) = poissoncdf(3, 4) ≈ .8153

Using the binomial distribution

• X ~ B(100, .03)
• P(x ≤ 4) = binomcdf(100, .03, 4) ≈ .8179

The Poisson approximation is very good—the difference between the probabilities is only .0026.

1. X = the number of children for a Spanish woman
2. 0, 1, 2, 3,...
3. X ~ P(1.47)
4. .2299
5. .5679
6. .4321

1. X = the number of fortune cookies that have an extra fortune
2. 0, 1, 2, 3,... 144
3. X ~ B(144, .03) or P(4.32)
4. 4.32
5. .0124 or .0133
6. .6300 or .6264
7. As n gets larger, the probabilities get closer together.

1. X = the number of people audited in one year
2. 0, 1, 2, ..., 100
3. X ~ P(2)
4. 2
5. .1353
6. .3233

1. X = the number of shell pieces in one cake
2. 0, 1, 2, 3,...
3. X ~ P(1.5)
4. 1.5
5. .2231
6. .0001
7. yes

d

1. You can use randInt (0,1,5) to generate five trials of the experiment. Count the number of 1’s generated to determine the number of successes.
2. Student answers may vary.
3. Student answers may vary.
4. The theoretical mean is $\left(5\right)\left(.5\right)=2.5$. The theoretical standard deviation is $\sqrt{\left(5\right)\left(.5\right)\left(0.5\right)}=\sqrt{1.25}$.