chapter 6apstatsgrabowski.weebly.com/.../3/0/3/4/30343621/ch06_solutions.pdfthis means that about...

132 The Practice of Statistics for AP*, 4/e

Chapter 6 Section 6.1 Check Your Understanding, page 344: 1. We are looking for the probability that the student gets either an A or a B. This probability is 0.42 0.26 0.68.+ = 2. We are looking for ( )2 0.02 0.10 0.12.P X < = + = 3. This histogram is left skewed. This means that higher grades are more likely, but that there are a few lower grades.

Check Your Understanding, page 349: 1. ( ) ( ) ( ) ( )0 0.3 1 0.4 2 0.2 3 0.1 1.1.Xµ = + + + = The long-run average, over many Friday mornings, will be about 1.1 cars sold. 2. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 22 0 1.1 0.3 1 1.1 0.4 2 1.1 0.2 3 1.1 0.1 0.89.Xσ = − + − + − + − = So 0.943.Xσ = On average, the number of cars sold on a randomly selected Friday will differ from the mean (1.1) by 0.943 cars sold. Exercises, page 353: 6.1 (a) If you toss a coin 4 times, the sample space is {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT}. All of these outcomes are

equally likely and so have probability 1 0.0625.16

= To find the probability of X taking on any specific

number, count the number of outcomes with exactly this number of heads and multiply by 0.0625. For example, there are 4 ways to get exactly 3 heads so ( ) ( )3 4 0.0625 0.25.P X = = = This leads to the following distribution:

Value 0 1 2 3 4 Probability 0.0625 0.25 0.375 0.25 0.0625

Chapter 6: Random Variables 133

(b) The histogram shows that this distribution is symmetric with a center at 2.

(c) ( ) ( )3 1 4 1 0.0625 0.9375.P X P X≤ = − = = − = There is a 93.75% chance that you will get three or fewer heads on 4 tosses of a fair coin. 6.2 (a) If we roll two 6-sided dice, the sample space is {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}. All of these are equally likely, so the

probability of any one outcome is 1 .36

To find the probability of X taking on any specific number, count

the number of outcomes where the sum of the dice is exactly this number and multiply by 0.0278. For example, there are 4 ways to get a sum of 5 (the outcomes (1,4), (2,3), (3,2), and (4,1)) so

( ) 1 43 4 .36 36

P X = = =

This leads to the following distribution:

Value 2 3 4 5 6 7 8 9 10 11 12 Prob 1

36 2

36 3

36 4

36 5

36 6

36 5

36 4

36 3

36 2

36 1

36

(b) The histogram shows that the distribution is symmetric about a center of 7.

(c) ( ) ( ) 1 2 3 6 1 55 1 4 1 1 1 .36 36 36 36 6 6

P T P T ≥ = − ≤ = − + + = − = − =

This means that about five-sixths

of the time, when you roll a pair of 6-sided dice, you will have a sum of 5 or more.


6.3 (a) “At least one nonword error” is the event { }1X ≥ or { }0 .X > P(X ≥1) = 1 − P(X<1) = 1 − P(X=0) = 1 − 0.1 = 0.9. (b) The event {X ≤ 2} is “no more than two nonword errors,” or “fewer than three nonword errors.” P(X ≤ 2) = (X = 0) + P(X = 1) + P(X = 2) = 0.1 + 0.2 + 0.3 = 0.6. P(X < 2) = P(X = 0) + P(X = 1) = 0.1 + 0.2 = 0.3. 6.4 (a) “Plays with at most two toys” is the event { }2X ≤ or { }3 .X < P(X ≤ 2) = (X = 0) + P(X = 1) + P(X = 2) = 0.03 + 0.16 + 0.30 = 0.49. (b) The event {X > 3} is “the child plays with more than three toys.” ( ) ( ) ( )3 4 5 0.17 0.11 0.28.P X P X P X> = = + = = + = ( ) ( )3 1 2 1 0.49 0.51.P X P X≥ = − ≤ = − = 6.5 (a) All of the probabilities are between 0 and 1 and they sum to 1 so this is a legitimate probability distribution. (b) This is a right skewed distribution with the largest amount of probability on the digit 1.

(c) The event { }6X ≥ is the event that “the first digit in a randomly chosen record is a 6 or higher.”

( ) ( ) ( ) ( ) ( )6 6 7 8 9 0.067 0.058 0.051 0.046 0.222.P X P X P X P X P X≥ = = + = + = + = = + + + = (d) The event that “the first digit is at most 5” is the event { }5 .X ≤

( ) ( )5 1 6 1 0.222 0.778.P X P X≤ = − ≥ = − = 6.6 (a) All of the probabilities are between 0 and 1 and they sum to 1 so this is a legitimate probability distribution. (b) Most people did not work out in the last 7 days. The distribution is skewed to the right with a peak at 0 days.


(c) The event {Y<7} is the event that the person “did not work out all 7 days.” ( ) ( )7 1 7 1 0.02 0.98.P Y P Y< = − = = − = (d) The event “worked out at least once” is the event { }1 .Y ≥

( ) ( )1 1 0 1 0.68 0.32.P Y P Y≥ = − = = − = 6.7 (a) The outcomes that make up the event A are {7, 8, 9}. ( ) ( ) ( ) ( )7 8 9 0.058 0.051 0.046 0.155.P A P X P X P X= = + = + = = + + = (b) The outcomes that make

up the event B are {1, 3, 5, 7, 9}. ( ) ( ) ( ) ( ) ( ) ( )1 3 5 7 9P B P X P X P X P X P X= = + = + = + = + = 0.301 0.125 0.079 0.058 0.046 0.609.= + + + + =

(c) The outcomes that make up the event “A or B” are {1, 3, 5, 7, 8, 9}. ( ) ( ) ( ) ( ) ( ) or 0.155 0.609 (0.058 0.046) 0.66.P A B P A B P A P B P A B= ∪ = + − ∩ = + − + = This is not

the same as ( ) ( )P A P B+ because A and B are not mutually exclusive. 6.8 (a) The outcomes that make up the event A are {1, 2, 3, 4, 5, 6, 7}. From exercise 6.6d, ( )1 0.32.P X ≥ = (b) The outcomes that make up the event B are {0, 1, 2, 3, 4}.

( ) 0.68 0.05 0.07 0.08 0.05 0.93.P B = + + + + = (c) The event “A and B” has the outcomes {1, 2, 3, 4}. So ( ) and 0.05 0.07 0.08 0.05 0.25.P A B = + + + = Note that ( ) ( ) ( )0.32 0.93 0.2976.P A P B = = These last two probabilities are not equal because A and B are not independent. 6.9 (a) The payoff is either $0, with a probability of 0.75, or $3, with a probability of 0.25.

Value $0 $3 Probability 0.75 0.25

(b) For each $1 bet, the mean payoff is Xµ = ($0)(0.75) + ($3)(0.25) = $0.75. This means that for every $1 the player bets, he only gets $0.75 back. In other words, he loses $0.25 on each bet, on average. 6.10 The company earns $300, with a probability of 0.9998, and earns $-199,700 with probability 0.0002.

Value $300 $-199,700 Probability 0.9998 0.0002

(b) The expected value of Y is ( )( ) ( )( )$300 0.9998 $ 199,700 0.0002 $260.Yµ = + − = This means that, on average, the company gains $260 per policy. 6.11 ( ) ( ) ( ) ( ) ( )0 0.1 1 0.2 2 0.3 3 0.3 4 0.1 2.1.Xµ = + + + + = On average, undergraduates make 2.1 nonword errors per 250-word essay. 6.12 ( ) ( ) ( ) ( ) ( ) ( )0 0.03 1 0.16 2 0.30 3 0.23 4 0.17 5 0.11 2.68.Xµ = + + + + + = On average, when children are given 5 toys to play with, they play with 2.68. 6.13 (a) The mean of the random variable Y is located at 5 because this distribution is symmetric and 5 is located at the center. (b) The expected value of X, the first digit following Benford’s law, is

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )1 0.301 2 0.176 3 0.125 4 0.097 5 0.079 6 0.067 7 0.058 8 0.051 9 0.046Xµ = + + + + + + + +3.441.= The average of first digits following Benford’s law is 3.441. To detect a fake expense report,

compute the sample mean of the first digits and see if it is near 5 or near 3.441. (c) Under the equally likely assumption, ( )6 0.111 0.111 0.111 0.333.P Y > = + + = Under Benford’s law

( )6 0.058 0.051 0.046 0.155.P X > = + + = So on a fake expense report, we would expect 33% of


numbers to start with digits higher than 6 whereas on a real expense report, only about 15% would start with digits higher than 6. So when looking at a suspect report, find the percentage of figures that start with numbers higher than 6. If that percentage is closer to 33% than to 15%, it is probably fake. 6.14 (a) and (b)

Death age 21 22 23 24 25 26/more Profit -$99,750 -$99,500 -$99,250 -$99,000 -$98,750 $1,250

Probability 0.00183 0.00186 0.00189 0.00191 0.00193 0.99058 (c) ( )( ) ( )( ) ( )( )$99,750 0.00183 $99,500 0.00186 $99,250 0.00189Xµ = − + − + −

( )( ) ( )( ) ( )( )$99,000 0.00191 $98,750 0.00193 $1,250 0.99058 $303.35.+ − + − + = The company makes an average of $303.35 per life insurance policy. 6.15 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 22 0 2.1 0.1 1 2.1 0.2 2 2.1 0.3 3 2.1 0.3 4 2.1 0.1 1.29.Xσ = − + − + − + − + − = So

1.29 1.1358.Xσ = = This means that, on average, the number of nonword errors in a randomly selected essay will differ from the mean (2.1) by 1.14 words. 6.16

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 22 0 2.68 0.03 1 2.68 0.16 2 2.68 0.3 3 2.68 0.23 4 2.68 0.17Xσ = − + − + − + − + −

( ) ( )25 2.68 0.11 1.7176.+ − = So 1.7176 1.3106.Xσ = = This means that, on average, the number of toys a randomly selected child will play with will differ from the mean (2.68) by 1.31 toys. 6.17 (a) ( ) ( ) ( ) ( ) ( ) ( )2 2 22 1 5 0.111 2 5 0.111 ... 9 5 0.111 6.667.Yσ = − + − + + − = So 2.58.Yσ =

(b) ( ) ( ) ( ) ( ) ( ) ( )2 2 22 1 3.441 0.301 2 3.441 0.176 ... 9 3.441 0.046 6.0605.Xσ = − + − + + − = So 2.4618.Xσ = This would not be the best way to tell the difference between a fake and a real expense report because the standard deviations are not too different from one another. 6.18 (a) The mean Xµ of the company’s “winnings” (premiums) and their “losses” (insurance claims) is about $303.35. Even though the company will lose a large amount of money on a small number of policyholders who die, it will gain a small amount from many thousands of 21-year-old men. In the long run, the insurance company can expect to make $303.35 per insurance policy. The insurance company is relying on the Law of Large Numbers. (b)

( ) ( ) ( ) ( )2 22 $99,750 303.35 0.00183 ... $1,250 303.35 0.97321 $94,236,826.64.Xσ = − − + + − = So $9,707.57.Xσ =

6.19 (a) The probability histograms are shown below. The distribution of the number of rooms is roughly symmetric for owners (graph on the left) and skewed to the right for renters (graph on the right). The center is slightly over 6 units for owners and slightly over 4 for renters. Overall, renter-occupied units tend to have fewer rooms than owner-occupied units.


(b) The mean for owner-occupied units is Xµ = (1)(0.003) + (2)(0.002) + (3)(0.023) + (4)(0.104) + (5)(0.210) + (6)(0.224) + (7)(0.197) + (8)(0.149) + (9)(0.053) + (10)(0.035) = 6.284 rooms. The mean for renter-occupied units is Yµ = (1)(0.008) + (2)(0.027) + (3)(0.287) + (4)(0.363) + (5)(0.164) + (6)(0.093) + (7)(0.039) + (8)(0.013) + (9)(0.003) + (10)(0.003) = 4.187 rooms. Since the sample space for the number of rooms is the same for both owner-occupied and rented, the larger value of Xµ for owner-occupied units is consistent with the fact that the owner distribution was symmetric, rather than skewed to the right, as was the case with the renter distribution. The “center” of the owner distribution is roughly at the central peak class, 6, whereas the “center” of the renter distribution is roughly at the class 4 (closer to the left hand endpoint of the sample space). A comparison of the centers (6.284 > 4.187) matches the observation in part (a) that the number of rooms for owner-occupied units tended to be higher than the number of rooms for renter-occupied units. (c) We would expect the owner distribution to have a slightly wider spread than the renter distribution. Even though the distribution of renter-occupied units is skewed to the right, it is more concentrated (contains less variability) about the “peak” than the symmetric distribution for owner-occupied units. Thus, the average distance between a value and the mean is slightly larger for owners. The variances and standard deviations are: 2

Xσ = (1 − 6.284)20.003 + (2 − 6.284) 20.002 + (3 − 6.284) 20.023 + (4 − 6.284) 20.104 + (5 − 6.284) 20.210 + (6 − 6.284)20.224 + (7 − 6.284)20.197 + (8 − 6.284) 20.149 + (9 − 6.284) 20.053 + (10 − 6.284)20.035 = 2.68934 and

1.6399Xσ = rooms for owner-occupied units and 2Yσ = (1 − 4.187)20.008 + (2 − 4.187) 20.027 + (3 −

4.187) 20.287 + (4 − 4.187) 20.363 + (5 − 4.187)20.164 + (6 − 4.187) 20.093 + (7 − 4.187) 20.039 + (8 − 4.187)20.013 + (9 − 4.187)20.003 + (10 − 4.187) 20.003 = 1.71003 and 1.3077Yσ = rooms for renter-occupied units. 6.20 (a) Both distributions are skewed to the right. However, the event {X = 1} has a much higher probability in the household distribution. This reflects the fact that a family must consist of two or more persons. A closer look reveals that all of the values above one, except for 6, have slightly higher probabilities in the family distribution. These observations and the fact that the mean and median numbers of occupants are higher for families indicates that family sizes tend to be slightly larger than household sizes in the U.S.


(b) The means are: ( ) ( ) ( ) ( ) ( ) ( ) ( )1 0.25 2 0.32 3 0.17 4 0.15 5 0.07 6 0.03 7 0.01 2.6Xµ = + + + + + + = people for a household and ( ) ( ) ( ) ( ) ( ) ( ) ( )1 0 2 0.42 3 0.23 4 0.21 5 0.09 6 0.03 7 0.02 3.14Yµ = + + + + + + = people for a family. The family distribution has a slightly larger mean than the household distribution, matching the observation in part (a) that family sizes tend to be larger than household sizes. (c) The standard deviations are: 2

Xσ = (1 − 2.6)2×0.25 + (2 − 2.6)2× 0.32 + (3 − 2.6)2×0.17 + (4 − 2.6)2×0.15 + (5 − 2.6)2×0.07 + (6 − 2.6)2×0.03 + (7 − 2.6)2×0.01 = 2.02, and 2.02 1.421Xσ = = people for a household and

2Yσ = (1 − 3.14)2(0) + (2 − 3.14)2(0.42) + (3 − 3.14)2(0.23) + (4 − 3.14)2(0.21) + (5 − 3.14)2(0.09) + (6 −

3.14)2(0.03) + (7 − 3.14)2(0.02) =1.5604, and 1.5604 1.249Yσ = = people for a family. The standard deviation for households is only slightly larger, mainly due to the fact that a household can have only 1 person. 6.21 (a) ( )0.49 0.51.P X > = (b) ( )0.49 0.51.P X ≥ = Note: (a) and (b) are the same because there is no area under the curve at any one particular point. (c) ( ) ( ) ( )0.19 0.37 or 0.84 1.27 0.19 0.37 0.84 1.27 .P X X P X P X≤ < < ≤ = ≤ < + < ≤ Note that

( )0.19 0.37 0.37 0.19 0.18.P X≤ < = − = Note also that X cannot be bigger than 1, so

( ) ( )0.84 1.27 0.84 1 1 0.84 0.16.P X P X< ≤ = < ≤ = − = Therefore

( )0.19 0.37 or 0.84 1.27 0.18 0.16 0.34.P X X≤ < < ≤ = + = 6.22 (a) ( )0.4 0.4.P X ≤ = (b) ( )0.4 0.4.P X < = Note: (a) and (b) are the same because there is no area under the curve at any one particular point. (c) ( ) ( ) ( )0.1 0.15 or 0.77 0.88 0.1 0.15 0.77 0.88P Y Y P Y P Y< ≤ ≤ < = < ≤ + ≤ <

( ) ( )0.15 0.1 0.88 0.77 0.05 0.11 0.16.= − + − = + = 6.23 State: What is the probability that a randomly chosen student scores a 9 or better on the ITBS? Plan: The score X of the randomly chosen student has the N(6.8, 1.6) distribution. We want to find ( )9 .P X ≥ We’ll standardize the scores and find the area shaded in the Normal curve.


Do: The standardized score for the test is 9 6.8 1.38.1.6

z −= = ( )1.38 0.0838.P Z ≥ = Conclude: There is

about an 8% chance that the chosen student’s score is 9 or higher. 6.24 State: What is the probability that a randomly chosen student runs the mile in under 6 minutes? Plan: The time Y of the randomly chosen student has the N(7.11, 0.74) distribution. We want to find ( )6 .P Y < We’ll standardize the scores and find the area shaded in the Normal curve.

Do: The standardized time for this student is 6 7.11 1.50.0.74

z −= = − ( )1.50 0.0668.P Z < − = Conclude:

There is about a 7% chance that this student will run the mile in under 6 minutes.

6.25 (a) Since V has a Normal distribution, standardize both endpoints. 0.52 0.56 2.110.019

z −= = − and

0.60 0.56 2.11.0.19

z −= = So ( ) ( )0.52 0.60 2.11 2.11 0.9826 0.0174 0.9652P V P Z≤ ≤ = − ≤ ≤ = − = (using

Table A). (b) ( ) ( )0.72 0.560.72 8.42 0.0.19

P V P Z P Z− ≥ = ≥ = ≥ ≈

If people answered truthfully, it

would be virtually impossible to get a sample in which 72% said they had voted.

6.26 ( ) ( )8.9 9 9.1 98.9 9.1 1.33 1.33 0.9082 0.0918 0.8164.0.075 0.075

P x P Z P Z− − ≤ ≤ = ≤ ≤ = − ≤ ≤ = − =

(Using Table A.) 6.27 b 6.28 c 6.29 c 6.30 a 6.31 Yes, students did have higher scores, in general, after participating in the chess program. If we look at the differences in the scores, the mean difference was 5.38 and the median was 3. This means that at least half of the students (though less than three quarters since 1Q was negative) improved their reading scores.


6.32 No we cannot conclude that chess causes an increase in reading scores. We do not have a control group that did not participate in the chess program. This means that we have no comparison group. It may be that children of this age naturally improve their reading scores anyway and that the chess program had nothing to do with their improvement. 6.33 The equation of the linear regression model is: ( )predicted post-test 17.897 0.78301 pretest .= + 6.34 This line is appropriate. The residual plot does not show any patterns. The strength of the fit is only moderate with an 2 0.558.r = The line does not do a great job of predicting any individual score as

12.55s = so in using the line, we expect to be off by an average of 12.55 points. Given that the mean number of points is roughly 63, this is a lot to be off by. Section 6.2 Check Your Understanding, page 362: 1. ( )500 500 1.1 $550.Y Xµ µ= = = ( )500 500 0.943 $471.5Y Xσ σ= = = 2. 75 550 75 $475.T Yµ µ= − = − = $471.5.T Yσ σ= = Check Your Understanding, page 370: 1. 1.1 0.7 1.8.T X Yµ µ µ= + = + = On average, this dealership sells or leases 1.8 cars in the first hour of business on Fridays. 2. Assuming that X and Y are independent, ( ) ( )2 22 2 2 0.943 0.64 1.2988T X Yσ σ σ= + = + = so

1.2988 1.14.Tσ = = 3. The total bonus is 500 300 .B X Y= + This means that

( ) ( )500 300 500 1.1 300 0.7 $760.B X Yµ µ µ= + = + = To find the standard deviation, we must again assume that X and Y are independent and we must first calculate the variance.

( ) ( ) ( ) ( )2 2 2 22 2 2500 300 250,000 0.943 90,000 0.64 259,176.25.B X Yσ σ σ= + = + = Therefore

259,176.25 $509.09.Bσ = = Check Your Understanding, page 372: 1. 1.1 0.7 0.4.D X Yµ µ µ= − = − = On average, this dealership sells 0.4 cars more than it leases during the first hour of business on Fridays. 2. ( ) ( )2 22 2 2 0.943 0.64 1.2998D X Yσ σ σ= + = + = so 1.2998 1.14.Dσ = = 3. 500 300 .B X Y= − This means that ( ) ( )500 300 500 1.1 300 0.7 $340.B X Yµ µ µ= − = − = To find the standard deviation, we must again assume that X and Y are independent and we must first calculate the variance. ( ) ( ) ( ) ( )2 2 2 22 2 2500 300 250,000 0.943 90,000 0.64 259,176.25.B X Yσ σ σ= + = + = Therefore

259,176.25 $509.09.Bσ = = Exercises, page 378: 6.35 The relationship between the length in centimeters and the length in inches is 2.54 .Y X= So

( )2.54 2.54 1.2 3.048Y Xµ µ= = = cm and ( )2.54 2.54 0.25 0.635Y Xσ σ= = = cm.


6.36 The relationship between heights in inches and heights in feet is 12 .J H= So ( )12 12 5.8 69.6J Hµ µ= = = in and ( )12 12 0.24 2.88J Hσ σ= = = in.

6.37 (a) This graph is skewed to the left. Most of the time, the ferry makes $20 or $25.

(b) ( )5 5 3.87 $19.35.M Xµ µ= = = The ferry makes $19.35 per trip on average. (c)

( )5 5 1.29 $6.45.M Xσ σ= = = The individual amounts made on the ferry trips will vary by about $6.45 from the mean ($19.35) on average. 6.38 (a) This distribution is skewed to the right. Ana is more likely to get 1 or 2 tickets than 4 or 5.

(b) ( )1 1 23.8 2.3810 10T Xµ µ= = = tickets. On average, Ana will receive 2.38 tickets on a single roll. (c)

( )1 1 12.63 1.26310 10T Xσ σ= = = tickets. In individual rolls, the number of tickets that Ana wins will vary

by 1.263 from the mean (2.38) on average. 6.39 (a) The score on the test G is related to the number of questions X by the equation 10 .G X= This means that ( )10 10 7.6 76.G Xµ µ= = = (b) ( )10 10 1.32 13.2.G Xσ σ= = = (c) Since the variance of G is

the square of the standard deviation ( )22 210 100 .G X Xσ σ σ= = In other words, we need to multiply the variance of X by the square of the constant 10 to get the variance of G. 6.40 (a) The score on the test G is related to the number of questions X by the equation 10 .G X= This means that ( )median 10median 10 8.5 85.G X= = = (b) ( ) ( )10 10 1 10.G XIQR IQR= = = (c) The shape of


G’s distribution will be the same as the shape of X’s. Since the distance between the median and the minimum is much bigger than the distance between the median and the maximum, this distribution is skewed to the left. 6.41 (a) The mean of Y is 20 less than the mean of M. In other words 20.Y Mµ µ= − Thus, the total mean profit per trip is $20 less than the amount of money collected. (b) The standard deviation of Y is the same as the standard deviation of M. There is the same amount of variability in both the profit and the amount of money collected. 6.42 (a) ( ) ( )0 0.999 500 0.001 $0.50.Xµ = + = ( ) ( ) ( ) ( )2 22 0 0.50 0.999 500 0.50 0.001 249.75Xσ = − + − =

so 249.75 $15.80.Xσ = = (b) 1 0.50 1 $0.50.W Xµ µ= − = − = − $15.80.W Xσ σ= = On average, when playing this game, people will lose $0.50. Individual outcomes will vary from this amount by $15.80 on average. 6.43 ( )6 20 6 3.87 20 $3.22.Y Xµ µ= − = − = ( )6 6 1.29 $7.74.Y Xσ σ= = = 6.44 The mean and standard deviation of Y are ( )0.9 0.2 0.9 $3 0.2 $2.5Y Xµ µ= − = − = million and

( )2 2 20.9 0.9 6.35 $2.2679Y Xσ σ= = = million.

6.45 (a) ( )9 932 8.5 32 47.35 5Y Tµ µ= + = + = degrees Fahrenheit. ( )9 9 2.25 4.05

5 5Y Tσ σ= = = degrees

Fahrenheit. (b) ( ) ( )40 47.340 1.80 0.0359.4.05

P Y P Z P Z− < = < = < − =

6.46 (a) ( )28.35 273.01 28.35 9.70 273.01 1.985Y Xµ µ= − = − = grams.

( )28.35 28.35 0.03 0.8505Y Xσ σ= = = grams. (b) ( ) ( )3 1.9853 1.19 0.1170.0.8505

P Y P Z P Z− ≥ = ≥ = ≥ =

6.47 (a) The possible values of T are {1+2=3, 1+4=5, 2+2=4, 2+4=6, 5+2=7, 5+4=9}. Since X and Y are independent, ( ) ( ) ( ) and .P X x Y y P X x P Y y= = = = = So

( ) ( ) ( ) ( ) ( )3 1 and 2 1 2 0.2 0.7 0.14.P T P X Y P X P Y= = = = = = = = = The rest of the probabilities are found the same way. The probability distribution for T is

Value 3 4 5 6 7 9 Probability 0.14 0.35 0.06 0.15 0.21 0.09

(b) From the distribution of T we get ( ) ( ) ( ) ( ) ( ) ( )3 0.14 4 0.35 5 0.06 6 0.15 7 0.21 9 0.09 5.3.Tµ = + + + + + = Now using the relationship between T, X and Y we get 2.7 2.6 5.3T X Yµ µ µ= + = + = which is the same. (c) From the distribution of

T we get ( ) ( ) ( ) ( ) ( ) ( )2 2 22 3 5.3 0.14 4 5.3 0.35 ... 9 5.3 0.09 3.25.Tσ = − + − + + − = Now using the

relationship between T, X and Y we get ( ) ( )2 22 2 2 1.55 0.917 3.24T X Yσ σ σ= + = + = which is only different due to rounding error. Note that 1.80Tσ = but 1.55 0.917 2.467.X Yσ σ+ = + = These last two values are not the same.


6.48 (a) The possible values for D are {1-2= -1, 1-4= -3, 2-2=0, 2-4= -2, 5-2=3, 5-4=1}. Since X and Y are independent, ( ) ( ) ( ) and .P X x Y y P X x P Y y= = = = = So

( ) ( ) ( ) ( ) ( )1 1 and 2 1 2 0.2 0.7 0.14.P D P X Y P X P Y= − = = = = = = = = The rest of the probabilities are found the same way. The probability distribution for D is

Value -3 -2 -1 0 1 3 Probability 0.06 0.15 0.14 0.35 0.09 0.21

(b) From the distribution of D we get ( ) ( ) ( ) ( ) ( ) ( )3 0.06 2 0.15 1 0.14 0 0.35 1 0.09 3 0.21 0.1.Dµ = − + − + − + + + = Now using the relationship between D, X and Y we get 2.7 2.6 0.1D X Yµ µ µ= − = − = which is the same. (c) From the distribution of

D we get ( ) ( ) ( ) ( ) ( ) ( )2 2 22 3 0.1 0.06 2 0.1 0.15 ... 3 0.1 0.21 3.25.Dσ = − − + − − + + − = Now using the

relationship between D, X and Y we get ( ) ( )2 22 2 2 1.55 0.917 3.24D X Yσ σ σ= + = + = which is only different due to rounding error. 6.49 (a) Dependent: since the cards are being drawn from the deck without replacement, the nature of the third card (and thus the value of Y) will depend upon the nature of the first two cards that were drawn (which determine the value of X). (b) Independent: X relates to the outcome of the first roll, Y to the outcome of the second roll, and individual dice rolls are independent (the dice have no memory). 6.50 (a) Independent: Weather conditions a year apart should be independent. (b) Not independent: Weather patterns tend to persist for several days; today’s weather tells us something about tomorrow’s. (c) Not independent: The two locations are very close together, and would likely have similar weather conditions. 6.51 (a) Yes. The mean of a sum is always

equal to the sum of the means. (b) No. The variance of the sum is not equal to the sum of the variances, because it is not reasonable to assume that X and Y are independent.

6.52 (a) Randomly selected student’s scores would presumably be unrelated. (b) The mean of the difference F M F Mµ µ µ− = − = 120 – 105 = 15 points. The variance of the difference is

2 2 2 2 228 35 2009F M F Mσ σ σ− = + = + = , so the standard deviation of the difference is 2009 44.8219F Mσ − = = points. (c) We cannot find the probability based on only the mean and standard

deviation. Many different distributions have the same mean and standard deviation. Many students will assume normality and do the calculation, but we are not given any information about the distributions of the scores. 6.53 (a) Let T represent the total payoff and X represent the first day’s payoff and Y the second day’s payoff. Then T X Y= + and 0.50 0.50 $1.00.T X Yµ µ µ= + = + = This also means that

( ) ( )2 22 2 2 15.80 15.80 499.28T X Yσ σ σ= + = + = so 499.28 $22.34.Tσ = = (b) Let W represent total winnings. Then 2W T= − since you had to pay $1 each of the two days for the tickets. So

2 1 2 $1.00.W Tµ µ= − = − = − Since we are only subtracting a constant, $22.34.W Tσ σ= =

6.54 (a) Let T represent the total payoff and iX represent the payoff on the ith day. Then 365

1i

iT X

=

=∑ so

365

10.50 0.50 ... 0.50

iT Xi

µ µ=

= = + + +∑ (365 times). So ( )365 0.50 $182.50.Tµ = = To find the standard


deviation, note that we add the variances for the individual days. So

( )365

22 2

1365 15.80 91,118.60.

iT Xi

σ σ=

= = =∑ So 91,118.6 $301.86.Tσ = = (b) So, over the course of many

years, you would expect to have a total payoff of $182.50, on average. But individual years will vary from this by $301.86, on average. 6.55 The mean of the difference Y−X is Y X Y Xµ µ µ− = − = 2.001 – 2.000 = 0.001g. The variance of the difference is 2 2 2

Y X Y Xσ σ σ− = + = 0.0022 + 0.0012 = 0.000005 so Y Xσ − = 0.002236g. On average, the difference between the readings of the two scales will be 0.001 g. Individual differences will vary from this by 0.002236 g on average. 6.56 (a) 1.0 2.1 1.1.Y X Y Xµ µ µ− = − = − = − ( ) ( )2 22 2 2 1.0 1.136 2.2905Y X Y Xσ σ σ− = + = + = so

2.2905 1.51.Y Xσ − = = On average, students make 1.1 more nonword errors than they do word errors. Individual differences will vary from this by 1.51 errors on average. (b) Let .D Y X= − If we want to find the probability that there are more word errors than nonword errors, then we are asking for ( )0 .P D > The outcomes that make up this event are {1-0=1, 2-0=2, 2-1=1, 3-0=3, 3-1=2, 3-2=1}. If we

assume that X and Y are independent, then we can calculate the probabilities of these of these events. For example ( ) ( ) ( ) ( )1 1 0 2 1 3 2P D P Y X Y X Y X = = = ∩ = ∪ = ∩ = ∪ = ∩ =

( )( ) ( )( ) ( )( )0.3 0.1 0.2 0.2 0.1 0.3 0.10.= + + = Similarly

( ) ( ) ( ) ( )( ) ( )( )2 2 0 3 1 0.2 0.1 0.1 0.2 0.04P D P Y X Y X = = = ∩ = ∪ = ∩ = = + = and

( ) ( ) ( )( )3 3 0 0.1 0.1 0.01.P D P Y X= = = ∩ = = = So there is a 15% chance that a randomly chosen student will have more word errors than nonword errors. 6.57 Using properties of means, the mean of W is 0.5 0.5W X Yµ µ µ= + = 0.5×$303.35 + 0.5×$303.35 = $303.35. Using properties of variances, the variance of W is

( )2 2 20.25 0.25 0.5 94,236,915.3 47,118,457.65W X Yσ σ σ= + = = so the standard deviation is

47,118,457.65 $6,864.29Wσ = = . 6.58 For 4 men, the expected value of the average income is

1 2 3 40.25 0.25 0.25 0.25V X X X Xµ µ µ µ µ= + + + = $303.35; the same as it was for one policy and two

policies. The variance of the average income is

1 2 3 4 1

2 2 2 2 2 20.0625 0.0625 0.0625 0.0625 0.25 23,559,228.83V X X X X Xσ σ σ σ σ σ= + + + = = , so the standard

deviation is 23,559,228.83 $4853.78Vσ = = (smaller by a factor of 12

).

6.59 (a) The total mean is 11 + 20 = 31 seconds. The standard deviation for the total time to position and

attach the part is 2 22 4 4.4721+ = seconds. (b) ( ) ( )30 3130 0.22 0.4129.4.4721

P T P Z P Z− < = < = < − =

6.60 (a) The total resistance 1 2T R R= + is Normal with mean 100 + 250 = 350 ohms and standard

deviation 2 22.5 2.8 3.7537+ = ohms. (b) The probability is ( )345 355P T≤ ≤ =


345 350 355 3503.7537 3.7537

P Z− − ≤ ≤

= ( )1.332 1.332P Z− ≤ ≤ = 0.9086 − 0.0914 = 0.8172 (Table A gives

0.9082 − 0.0918 = 0.8164). 6.61 State: What is the probability that the difference between the NOX levels for two randomly selected cars is at least 0.80 g/mi? Plan: Since both cars have NOX levels that follow a N(1.4, 0.3) distribution,

the distribution of the difference X – Y will be N(0, 2 20.3 0.3+ ) ≈ N(0, 0.4243). Use this to find ( )0.8 or 0.8 .P X Y X Y− > − < − Do:

( ) ( ) ( )0.8 or 0.8 0.8 0.8P X Y X Y P X Y P X Y− > − < − = − > + − < −

( ) ( )0.8 0 0.8 0 1.89 1.89 0.0294 0.0294 0.0588.0.4243 0.4243

P Z P Z P Z P Z− − − = > + < = > + < − = + =

Conclude: There is only about a 6% chance of randomly finding two cars with as big a difference in NOX levels as the attendant found. 6.62 State: What is the probability that the scores differ by 5 or more points? Plan: Let L and F denote the respective scores of Leona and Fred. The difference L − F has a Normal distribution with mean

L Fµ − = 24 – 24 = 0 points and standard deviation 2 22 2 2.8284L Fσ − = + points. Use this to find ( )5 or 5 .P L F L F− < − − > Do: ( ) ( ) ( )5 or 5 5 5P L F L F P L F P L F− < − − > = − < − + − >

( ) ( )5 0 5 0 1.77 1.77 0.0384 0.0384 0.0768.2.8284 2.8284

P Z P Z P Z P Z− − − = < + > = < − + > = + =

Conclude: There is about an 8% chance that one of the two friends will have to buy the other one a pizza. 6.63 State: What is the probability that the total team swim time is less than 220 seconds. Plan: Let T stand for the total team swim time and 1X be Wendy’s time, 2X be Jill’s time, 3X be Carmen’s time, and

4X be Latrice’s time. Then 1 2 3 4.T X X X X= + + + The mean and variance are 1 2 3 4 55.2 58.0 56.3 54.7 224.2Tµ µ µ µ µ= + + + = + + + = seconds and

( ) ( ) ( ) ( )1 2 3 4

2 2 2 22 2 2 2 2 2.8 3.0 2.6 2.7 30.89.T X X X Xσ σ σ σ σ= + + + = + + + = This means that

30.89 5.56Tσ = = seconds. Since each individual’s time is approximately Normally distributed, that means that T is approximately Normally distributed. Putting this all together we have that T has a N(224.2, 5.56) distribution. Use this to find ( )220 .P T <

Do: ( ) ( )220 224.2220 0.76 0.2236.5.56

P T P Z P Z− < = < = < − =

Conclude: There is approximately a

22% chance that the team’s swim time will be less than 220 seconds. 6.64 State: What is the probability that Ken uses more than 0.85 ounces of toothpaste if he brushes his teeth six times? Plan: Let T be the total amount of toothpaste and iX be the amount used the ith time he brushes his teeth. 1 2 3 4 5 6.T X X X X X X= + + + + + Each of the iX has a N(0.13, 0.02) distribution and we will assume that the iX are independent. So ( )1 2 3 4 5 6 6 0.13 0.78Tµ µ µ µ µ µ µ= + + + + + = = ounces

and ( )6

22 2

16 0.02 0.0024.

iT Xi

σ σ=

= = =∑ Since each of the iX are Normally distributed, then so is T.

Putting all of this together, T has a N(0.78, 0.049) distribution. Use this to find ( )0.85 .P T > Do:


( ) ( )0.85 0.780.85 1.43 0.0764.0.049

P T P Z P Z− > = > = > =

Conclude: There is about an 8% chance that

Ken will use the entire tube of toothpaste on his trip. 6.65 c. 6.66 d. 6.67 (a) Fidelity Technology Fund is more closely tied to the stock market as a whole, because its correlation is larger. (b) No: correlations tell nothing about the absolute size of the variables, only the relative sizes (above/below average).

6.68 (a) After one day the stock is either worth $1300 (gain of 30%) or $750 (loss of 25%), each with probability 0.5. If the stock goes up the first day, then it is either worth $1690 (30% gain) or $975 (25% loss) after the second day. If the stock goes down the first day, then it is either worth $975 (30% gain) or $562.5 (25% loss) after the second day. Since gain or loss is equally likely both days, the four different outcomes each have probability 0.25. Note, however, that there are two ways to get $975. So the probability distribution is

Amount $562.50 $975 $1690 Probability 0.25 0.50 0.25

The probability that the stock is worth more than the $1000 paid for it, after two days, is 0.25 (the probability of it being worth $1690). (b) The mean value after two days is

( ) ( ) ( )562.5 0.25 975 0.5 1690 0.25 $1,050.625.µ = + + = Section 6.3 Check Your Understanding, page 385: 1. Check the BINS: Binary? “Success” = get an ace. “Failure” = don’t get an ace. Independent? Because you are replacing the card in the deck and shuffling each time, the result of one trial does not affect the outcome of any other trial. Number? The number of trials is set at 10 in advance. Success?

The probability of success is 452

for each trial. This is a binomial setting. Since X counts the number of

successes, it is a binomial random variable with 10n = and 4 .52

p =

2. Check the BINS: Binary? “Success” = over 6 feet. “Failure” = not over 6 feet. Independent? Because we are selecting without replacement from a small number of students, the observations are not independent. Number? The number of trials is set to 3 in advance. Success? The probability of success will not change from person to person. This is not a binomial setting. 3. Check the BINS: Binary? “Success” = roll a 5. “Failure” = don’t roll a 5. Independent? Because you are rolling a die, the outcome of any one trial does not affect the outcome of any other trial. Number? The number of trials is set to 5 in advance. Success? The probability of success will change depending on whether a 6-sided or an 8-sided die is used. This is not a binomial setting. Check Your Understanding, page 390: 1. Check the BINS: Binary? “Success” = question answered “correctly.” “Failure” = question not answered “correctly.” Independent? The computer randomly assigned correct answers to the questions, so one trial should not affect any other trial. Number? There were 10 questions. Success? The


probability of success is 0.20 for each trial. This is a binomial setting. Since X counts the number of successes, it is a binomial random variable with 10n = and 0.20.p =

2. ( ) ( ) ( )3 7103 0.2 0.8 0.2013.

3P X

= = =

There is a 20% chance that Patti will answer exactly 3

questions correctly. 3. ( ) ( ) ( )6 1 6 1 5 .P X P X P X≥ = − < = − ≤ Using binomcdf(10,0.2,5) we get 0.9936. So

( )6 1 0.9936 0.0064.P X ≥ = − = There is only a 0.64% chance that a student would pass so we would be quite surprised if Patti was able to pass. Check Your Understanding, page 393: 1. ( )10 0.20 2.X npµ = = = We would expect the average student to get 2 answers correct on such a quiz. 2. ( ) ( )( )1 10 0.20 0.80 1.265.X np pσ = − = = We would expect individual students’ scores to vary from a mean of 2 correct answers by an average of 1.265 correct answers. 3. Two standard deviations above the mean is ( )2 2 1.265 4.53.+ = Since Patti can only score an integer number of questions correctly, we are asked to find ( ) ( )5 1 4 .P X P X≥ = − ≤ Using binomcdf(10,0.2,4) we get 0.9672. So ( )5 1 0.9672 0.0328.P X ≥ = − = There is about a 3% chance that Patti will score more than two standard deviations above the mean. Check Your Understanding, page 401: 1. Check the BITS: Binary? “Success” = roll doubles. “Failure” = don’t roll doubles. Independent: Rolling dice is an independent process. Trials: We are counting trials up to and including the first

success. Success? The probability of success is 16

for each roll (there are 6 ways to get doubles out of a

total of 36 possible rolls). This is a geometric setting. Since X is measuring the number of trials to get

the first success, it is a geometric random variable with 1 .6

p =

2. ( )25 13 0.1157.

6 6P T = = =

There is about a 12% chance that you will get the first set of doubles

on the third roll of the dice. 3.

( ) ( ) ( ) ( ) 1 5 13 1 2 3 0.1157 0.1667 0.1389 0.1157 0.4213.6 6 6

P T P T P T P T ≤ = = + = + = = + + = + + =

Exercises, page 403: 6.69 Binary? “Success” = seed germinates. “Failure” = seed does not germinate. Independent? Possibly although it is possible that if one seed does not germinate, it’s more likely that others in the packet will not grow either. Number? 20 seeds. Success? The probability that each seed germinates is 85%, assuming the advertised percentage is true. Assuming independence does hold, this is a binomial setting and X would have a binomial distribution. 6.70 Binary? “Success” = name has more than 6 letters. “Failure” = name has 6 letters or less. Independent? Since we are selecting without replacement from a small number of students, the observations are not independent. Number? 4 names are drawn. Success? The probability that a


student’s name has more than 6 letters does not change from one draw to the next. This is a binomial setting and Y would have a binomial distribution. 6.71 Binary? “Success” = person is left-handed. “Failure” = person is right-handed. Independent? Since students are selected randomly, their handedness is independent. Number? There is not a fixed number of trials for this chance process since you continue until you find a left-handed student. Since the number of trials is not fixed, this is not a binomial setting and V is not a binomial random variable. 6.72 Binary? “Success” = person is left-handed. “Failure” = person is right-handed. Independent? Even though we are selecting students without replacement, 15 students is less than 10% of the total student body of most schools so the observations can be considered to be independent. Number? 15 students are selected. Success? The probability of left-handedness remains constant from one student to the next, about 0.10. This is a binomial setting and W has a binomial distribution with 15n = and

0.10.p = 6.73 (a) A binomial distribution is not an appropriate choice for field goals made by the National Football League player, because given the different situations the kicker faces, his probability of success is likely to change from one attempt to another. (b) It would be reasonable to use a binomial distribution for free throws made by the NBA player because we have n = 150 attempts, presumably independent (or at least approximately so), with chance of success p = 0.8 each time. 6.74 (a) This is the binomial setting. We check the BINS. Binary? “Success” = reaching a live person. “Failure” = any other outcome. Independent? It is reasonable to believe that each call is independent of the others. Number? We have a fixed number of observations ( n = 15). Success? Each randomly-dialed number has chance p = 0.2 of reaching a live person. (b) This is not a binomial setting because there are not a fixed number of attempts.

6.75 ( ) ( ) ( )4 374 0.44 0.56 0.2304.

4P X

= = =

There is about a 23% chance that exactly 4 of the 7

people chosen have blood type O.

6.76 ( ) ( )( )9101 0.05 0.95 0.3151.

1P Y

= = =

There is about a 32% chance that exactly one of the 10

rhubarb plants will die before producing any rhubarb. 6.77 Using technology, ( ) ( )4 1 4 1 0.8598 0.1402.P X P X> = − ≤ = − = There is about a 14% chance that more than 4 people of the 7 chosen will have blood type O. This is not particularly surprising. 6.78 Using technology, ( ) ( )3 1 2 1 0.9885 0.0115.P Y P Y≥ = − ≤ = − = There is only about a 1% chance that 3 or more of the plants would die before producing rhubarb. This would be surprising if it occurred.

6.79 (a) ( ) ( ) ( )17 32017 0.85 0.15 0.2428.

17P X

= = =

(b) Using technology, ( )12 0.0059.P X ≤ = This

would indeed be surprising and should make Judy suspicious.


6.80 (a) ( ) ( ) ( )3 12153 0.10 0.90 0.1285.

3P W

= = =

(b) ( ) ( )4 1 3 1 0.9444 0.0556.P W P W≥ = − ≤ = − =

There is about a 6% chance of finding 4 or more lefties in a sample of 15. This would be moderately surprising, but not completely unexpected. 6.81 (a) ( )15 0.20 3.X npµ = = = You would expect to reach a live person in an average of 3 phone calls

when making 15 calls. (b) ( ) ( )( )1 15 0.20 0.80 1.55.X np pσ = − = = In actual practice, you would expect the number of live persons you reach to vary from 3 per 15 calls by 1.55 on average. 6.82 (a) ( )12 0.20 2.4.X npµ = = = You would expect to find an average of 2.4 people that the machine finds to be deceptive when testing 12 people actually telling the truth. (b)

( ) ( )( )1 12 0.20 0.80 1.39.X np pσ = − = = In actual practice, you would expect the number “deceivers” to vary from 2.4 by an average of 1.39. 6.83 (a) Y is also a binomial random variable. The only difference is that what we called a “failure” for X is a “success” for Y and vice versa. So the probability of success for Y is 0.80. Therefore

( )15 0.80 12.Y npµ = = = Notice that 3Xµ = and that 12 3 15.+ = In other words, if we reach an average of 3 live people in our 15 calls, we must not reach a live person in an average of 12 calls. (b) Notice that

( ) ( )( )1 15 0.80 0.20 1.55.Y np pσ = − = = This is the same thing as Xσ because we have just switched the definitions of p and 1 .p− 6.84 (a) Y is also a binomial random variable. The only difference is that what we called a “failure” for X is a “success” for Y and vice versa. So the probability of success for Y is 0.80.

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )10 2 11 1 12 012 12 1210 10 11 12 0.8 0.2 0.8 0.2 0.8 0.2

10 11 12P Y P Y P Y P Y

≥ = = + = + = = + +

0.2835 0.2062 0.0687 0.5584.= + + = Notice that ( )2 0.5584P X ≤ = as well. If we find 2 or fewer

lying, by definition we are saying that 10 or more are telling the truth. (b) ( )12 0.8 9.6.Y npµ = = = Notice that ( )12 0.20 2.4Xµ = = which is 12 .Yµ− Both Yσ and Xσ are the same.

( )( )12 0.8 0.2 1.39.Y Xσ σ= = = The amount of variability around the number “failures” should be the same as the amount around the number of “successes.” 6.85 (a) Check the BINS: Binary? “Success” = operates for an hour without failure. “Failure” = does not operate for an hour without failure. Independent? Whether one engine operates for an hour or not should not affect whether any other engine does or not. Number? We are looking at 350 engines. Success? Each engine has the same probability of success: 0.999. This is a binomial setting so, since X is counting the number of successes, it has a binomial distribution. (b) ( )350 0.999 349.65.X npµ = = = If we were to test 350 engines over and over again, we would expect that, on average, 349.65 of them would operate for an hour without failure. ( ) ( )( )1 350 0.999 0.001 0.591.X np pσ = − = = In individual tests we would expect to find the number of engines that operate for an hour without failure to vary from 349.65 by an average of 0.591. (c) ( ) ( ) ( ) ( )348 1 349 1 349 350P X P X P X P X≤ = − ≥ = − = − =


( ) ( ) ( ) ( )349 1 350 0350 3501 0.999 0.001 0.999 0.001 1 0.2468 0.7046 0.0486.

349 350

= − − ≈ − − =

There is about a

4% chance that 2 or more engines will fail (or 348 engines or fewer will operate correctly). Since there is such a small chance to 2 or more engines failing, we are convinced that they are less reliable than it is supposed to be. 6.86 (a) Check the BINS: Binary? “Success” = win a prize. “Failure” = don’t win a prize. Independent? Assuming that the company puts the caps on in a random fashion, knowing whether one bottle wins or not should not tell us anything about any other bottles. Number? We are sampling 7 bottles. Success? Assuming the company is correct, the probability of success is the same for all bottles.

It is 1 .6

This is a binomial setting. Since X is counting the number of successes, it has a binomial

distribution and therefore is a binomial random variable. (b) 17 1.167.6X npµ = = =

When buying 7

bottles, we would expect to 1.167 of them to win, on average. 1 57 0.986.6 6Xσ = =

In individual

samples of size 7, we would expect the number of winning bottles to vary from the mean (1.167) by 0.986, on average. (c) ( ) ( )3 1 2 1 0.9042 0.0958.P X P X≥ = − ≤ = − = There is about a 10% chance that if 7 friends buy a bottle each, that at least 3 of them would win. This is somewhat surprising, but not incredibly rare.

6.87 We cannot use the binomial distribution here because the sample size (10) is more than 10% of the population (76). This means that the probability of success changes too much as we take each observation out of the population and do not replace it. 6.88 We can use the binomial distribution in this case because the sample size (7) is less than 10% of the population (100). Even though the make-up of the population will change somewhat with each tile drawn, it will not change enough to make a big difference to the binomial distribution. The probabilities computed using the binomial distribution will be approximately correct. 6.89 If the sample size is more than 10% of the population, the amount of change to the make-up of the population is too much. The probability of success changes too much to be considered constant. 6.90 When we look at the histogram for a binomial distribution, it is symmetric when p is 0.5. As p moves farther from 0.5 the distribution, for a fixed n, becomes more skewed. However, if we fix a p (not 0.5) and increase n, the distribution becomes less skewed. So we need a criteria that takes both n and p into consideration. When 10np ≥ and ( )1 10n p− ≥ then we know that the combination of n and p is such that the distribution is close enough to Normal to use the Normal approximation. 6.91 (a) Check the BINS: Binary? “Success” = visit an auction site at least once a month. “Failure” = don’t visit an auction site at least once a month. Independent? The sample was random, so one person’s actions shouldn’t affect anyone else’s and even though we are sampling without replacement, the sample size (500) is far less than 10% of all males aged 18 to 34. Number? Sampled 500 men. Success? The probability of success for any one individual is 0.50. This is a binomial setting so, since X measures the number of successes, X is approximately a binomial random variable. (b) In this case 500n = and

0.5,p = so ( )500 0.5 250np = = and ( ) ( )1 500 0.5 250n p− = = are both greater than 10. (c) First we need the mean and standard deviation for the binomial distribution. ( )500 0.5 250.X npµ = = =


(1 ) 500(0.5)(0.5) 11.18.x np pσ = − = = So, using the Normal approximation,

235 250( 235) ( 1.34) 0.9099.11.18

P X P Z P Z− ≥ = ≥ = ≥ − =

6.92 (a) Check the BINS: Binary? “Success” = Person identify themselves as black. “Failure” = person does not identify themselves as black. Independent? The sample was random, so one person’s identification shouldn’t affect anyone else’s and even though we are sampling without replacement, the sample size (1500) is far less than 10% of all American adults. Number? 1500 people were sampled. Success? The probability of success for any one individual is 0.12. This is a binomial setting so, since X measures the number of successes, X is approximately a binomial random variable. (b) The Normal approximation is quite safe: 180np = and ( )1 1320n p− = are both more than 10. (c) First we need the mean and standard deviation for the binomial distribution. ( )1500 0.12 180.X npµ = = =

1500(0.12)(0.88) 12.5857.xσ = = So, using the Normal approximation,

165 180 195 180(165 195) ( 1.19 1.19) 0.7660.12.5857 12.5857

P X P Z P Z− − ≤ ≤ = ≤ ≤ = − ≤ ≤ =

6.93 Let X be the number of 1’s and 2’s; then X has a binomial distribution with n = 90 and p = 0.477 (in the absence of fraud). Using the calculator or software, we find ( 29)P X ≤ = binomcdf(90, 0.477, 29) 0.0021. Using the Normal approximation (the conditions are satisfied), we find a mean of 42.93 and standard deviation of ( )( )90 0.477 0.523 4.7384σ = = . Therefore,

( )29 42.93( 29) 2.94 0.00164.7384

P X P Z P Z− ≤ = ≤ = ≤ − =

. Either way, the probability is quite small, so

we have reason to be suspicious. 6.94 Let X be the number of hits out of 500 times at bat. Then X has a binomial distribution with

500n = and 0.260.p = Using the Normal approximation (the conditions are satisfied), we find a mean

of 130 and a standard deviation of ( )( )500 0.26 0.74 9.808.σ = = Therefore

( ) ( )150 130150 2.04 0.0207.9.808

P X P Z P Z− ≥ = ≥ = ≥ =

This could happen but we would only expect it

to happen for about 2% of typical baseball players. 6.95 (a) Check the BITS: Binary? “Success” = get a card you don’t already have. “Failure” = get a card you already have. Independent? Assuming the cards are put into the boxes randomly at the factory, what card you get in a particular box should be independent of what card you get in any other box. Trials? We are not counting trials until the first success. This is not a geometric setting. (b) Check the BITS: Binary? “Success” = Lola wins some money. “Failure” = Lola does not win money. Independent? The outcomes of the games are independent of each other. Trials? We are counting the number of games until she wins one. Success? The probability of success on any given game is 0.259. This is a geometric setting. Let Y be the number of games Lola plays up to and including her first win. This is a geometric random variable. 6.96 (a) Check the BITS: Binary? “Success” = get an ace. “Failure” = do not get an ace. Independent? The trials are not independent of one another because we are not replacing the previous cards drawn back


into the deck. This is not a geometric setting. (b) Check the BITS: Binary? “Success” = get a bulls-eye. “Failure” = do not get a bulls-eye. Independent? Different shots should be independent of each other. Whether he makes one shot does not affect whether he makes any other shots. Trials? We are continuing until he gets his first bulls-eye. Success? The probability of success is 0.10 for all shots. This is a geometric setting. Let X be the number of shots Lawrence makes up to and including his first bulls-eye. This is a geometric random variable. 6.97 (a) ( ) ( ) ( )23 0.8 0.2 0.128.P X = = = (b) Using technology,

( ) ( )10 1 10 1 0.8926 0.1074.P X P X> = − ≤ = − =

6.98 (a) ( )45 15 0.0804.

6 6P X = = =

(b) Using technology, ( )8 0.7674.P X ≤ =

6.99 (a) The expected value is 1 1 38.138

X pµ = = = We would expect it to take 38 games to win if one

wins in 1 out of 38 games. (b) Using technology, ( )3 0.0769.P X ≤ = While this is not a usual occurrence, it would happen about 8% of the time, so it is not completely surprising.

6.100 (a) The expected value is 1 1 10.31.0.097X p

µ = = = We would expect to examine about 10.31

invoices in order to find the first 8 or 9. (b) Using technology, ( ) ( )40 1 39 1 0.9813 0.0187.P X P X≥ = − ≤ = − = The likelihood of this happening is reasonably rare.

We may begin to worry that the invoice amounts are a fraud. 6.101 b 6.102 c 6.103 b 6.104 c 6.105 c 6.106 This is an experiment; students were randomly (at the time they visited the site) assigned to a treatment. The explanatory variable is the login box (genuine or not), and the response variable is the student’s action (logging in or not). 6.107 (a) [INSERT S6_107] ( )smoke 0.086 0.0986 0.0874 0.272.P = + + = 27.2% of British men

smoke. (b) ( ) ( )( )

routine and manual and smoke 0.0874routine and manual | smoke 0.321.smoke 0.272

PP

P= = =

About 32% of the smokers are routine and manual laborers.


Chapter Review Exercises (page 408) R6.1 (a) ( ) ( ) ( ) ( ) ( )5 1 1 2 3 4 1 0.1 0.2 0.3 0.3 0.1.P X P X P X P X P X= = − = − = − = − = = − − − − = (b) For any individual, the probability that they report a 1 or a 2 is ( ) ( ) ( )1 or 2 1 2 0.1 0.2 0.3.P X X P X P X= = = = + = = + = Since the two people are chosen randomly,

their pain reports are independent. So ( ) ( ) ( ) ( )first says 1 or 2 second says 1 or 2 first says 1 or 2 second says 1 or 2 0.3 0.3 0.09.P P P∩ = = =

(c) ( ) ( ) ( ) ( ) ( )1 0.1 2 0.2 3 0.3 4 0.3 5 0.1 3.1.Xµ = + + + + =

( ) ( ) ( ) ( )2 22 1 3.1 0.1 ... 5 3.1 0.1 1.29.Xσ = − + + − = 1.29 1.136.Xσ = = R6.2 (a) The mean number of degrees off target is 550 − 550 = 0°C, and the standard deviation stays the same, 5.7°C, because subtracting a constant does not change the variability. (b) In degrees Fahrenheit,

the mean is 9 32 10225Y Xµ µ= + = °F and the standard deviation is 9

10.265Y xσ σ= =

°F.

R6.3 (a) The average payout on a $1 bet is $0.70 and the amount that individual payouts vary from this is $6.58, on average. (b) Let Y be the amount of Jerry’s payout. Then 5 .Y X= Therefore

( )5 5 0.70 $3.50Y Xµ µ= = = and ( )5 5 6.58 $32.90.Y Xσ σ= = = (c) Let W be the amount of Marla’s payout. Since she plays 5 separate $1 games, 1 2 3 4 5W X X X X X= + + + + where all of the iX have the same distribution as X and are independent. This means that

( )1 2 3 4 5

5 5 0.70 $3.50.W X X X X X Xµ µ µ µ µ µ µ= + + + + = = =

( )1 2 3 4 5

22 2 2 2 2 2 25 5 6.58 216.482.W X X X X X Xσ σ σ σ σ σ σ= + + + + = = = So 216.482 $14.71.Wσ = = (d) The casino would probably prefer Marla since there is less variability in her strategy. They are less likely to get great amounts from her, but also less likely to have to pay great amounts to her. R6.4 (a) The machine that makes the caps and the machine that applies the torque are not the same. (b) Let T denote the torque applied to a randomly selected cap and C denote the cap strength. T is N(7, 0.9) and C is N(10, 1.2), so C − T is Normal with mean 10 – 7 = 3 inch-pounds and standard deviation

2 20.9 1.2 1.5+ = inch-pounds. (c) ( ) ( ) ( )0 30 2 0.0228.1.5

P T C P C T P Z P Z− > = − < = < = < − =

R6.5 (a) Check the BINS: Binary? “Success” = candy is orange. “Failure” = candy is not orange. Independent? Assuming the candies are well mixed, the color of one candy chosen should not tell us anything about the color of any other candy chosen. Number? We are choosing a fixed sample of 8 candies. Success? Since the sample is far smaller than 10% of the population, even though we are sampling without replacement, the probability of success remains approximately constant at 0.20. This is a binomial setting so X has a binomial distribution. (b) ( )8 0.2 1.6.X npµ = = = We expect to find 1.6

orange candies in samples of size 8, on average. (c) ( ) ( )( )1 8 0.2 0.8 1.13.X np pσ = − = = Individual samples of size 8 will have the number of orange candies varying from 1.6 by 1.13 on average.

R6.6 (a) ( ) ( ) ( )0 880 0.2 0.8 0.1678.

0P X

= = =

Since the probability is about 17%, it would not be

surprising to get no orange M&M’s. (b) Using technology we


find ( ) ( )5 1 4 1 0.9896 0.0104.P X P X≥ = − ≤ = − = Since the probability is about 1%, it would be somewhat surprising to find 5 or more orange M&M’s. R6.7 Let Y be the number of spins to get a “wasabi bomb.” Y is a geometric random variable with

3 0.25.12

p = = Using technology we find ( ) ( )3 1 2 1 0.4375 0.5625.P Y P Y≥ = − ≤ = − =

R6.8 (a) Let X be the number of heads in 10,000 tosses. Assuming the coin is balanced, X has a binomial distribution with 10,000n = and 0.5.p = Therefore ( )10,000 0.5 5,000X npµ = = = and

(1 ) 10,000(0.5)(0.5) 50.x np pσ = − = = (b) The conditions are met to approximate the binomial

distribution by the Normal distribution: both 5,000np = and ( )1 5000n p− = are far larger than 10. (c)

( ) ( ) ( ) 4933 5000 5067 50004933 or 5067 4933 506750 50

P X X P X P X P Z P Z− − ≤ ≥ = ≤ + ≥ = ≤ + ≥

( ) ( )1.34 1.34 0.0901 0.0901 0.1802.P Z P Z= ≤ − + ≥ = + = If the coin were balanced, finding 5067 heads or more, or 4933 heads or less, would happen 18% of the time in 10,000 tosses. This is not particularly surprising so we do not have strong evidence that the coin was not balanced.

AP Statistics Practice Test (page 459) T6.1 b. Add the probabilities for the events that X is 3 and 4. T6.2 d. The mean for one person is 2.3 and the mean of the sum for 3 people is the sum of the means. T6.3 e. All other pairs of variables would likely change together (e.g. those who are taller, likely also weigh more). T6.4 d. 1.5Y X= , so to get the mean and standard deviation of Y, multiply the mean and standard deviation of X by 1.5. T6.5 d. The mean of the sum of random variables is the sum of the means of the individual random variables. T6.6 d. To find the standard deviation of the sum of random variables, add their variances and take the square root. T6.7 c. In part (a) we are looking for 2 successes, not 1, in part (b) the trials are not independent (not putting the cards back after dealing), in part (d) we have a fixed number of trials and are counting the number of successes (binomial random variable) and in part (e) we have a fixed number of trials and are counting the number of successes (binomial random variable). T6.8 b. This is a binomial setting so the number of cases that hospital has to deal with is a binomial random variable with 17n = and 0.4.p = The question is looking for ( ) ( )10 1 10 .P X P X> = − ≤


T6.9 b. This cannot be a geometric distribution because the bar above 1X = is not the tallest. Using a calculator you can verify that binompdf(8, 0.3, 7) = 0.001224. T6.10 c. This is a geometric random variable and we are looking for ( )5 .P X = T6.11 (a) If we want at least 10 unbroken eggs, that means no more than 2 broken eggs. So ( )2 0.78 0.11 0.07 0.96.P Y ≤ = + + = There is a 96% chance that at least 10 eggs are unbroken in a

randomly selected carton of “store brand” eggs. (b) ( ) ( ) ( ) ( ) ( )0 0.78 1 0.11 2 0.07 3 0.03 4 0.01 0.38.Yµ = + + + + = We expect to find 0.38 broken eggs on

average in a carton of a dozen eggs. (c) ( ) ( ) ( ) ( ) ( ) ( )2 2 22 0 0.38 0.78 1 0.38 0.11 ... 4 0.38 0.01 0.6756.Yσ = − + − + + − = So 0.6756 0.8219.Yσ = =

The number of broken eggs in individual cartons will vary from 0.38 broken eggs by 0.8219 broken eggs, on average. (d) Let X stand for the number of cartons inspected until (and including) the first carton with at least 2 broken eggs is found. X is a geometric random variable with 0.11.p = We are looking for

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )23 1 2 3 0.11 0.89 0.11 0.89 0.11 0.2950.P X P X P X P X≤ = = + = + = = + + = T6.12 (a) Check the BINS. Binary? “Success” = dog owner greets dog first. “Failure” = dog owner does not greet dog first. Independent? Dog owners are selected at random, so their behaviors should be independent. Number? We have a fixed number of trials (12). Success? The probability of success is constant for all trials (0.66). This is a binomial setting and since X counts the number of successes in the

12 trials it is a binomial random variable. (b) ( ) ( ) ( )4

12

0

124 0.66 0.34 0.0213.i i

iP X

i−

=

≤ = =

∑ If the

Ladies Home Journal is correct, we would only have about a 2% chance of getting a sample where 4 or fewer owners greeted their dogs first when coming home. This is reasonably unlikely, so we would suggest that the Ladies Home Journal may be wrong. T6.13 (a) Letting D A E= − means that 50 25 25D A Eµ µ µ= − = − = minutes. Assuming that the amount of time they spend on homework is independent of each other 2 2 2 100 25 125D A Eσ σ σ= + = + = so

125 11.18Dσ = = minutes. (b) If Ed spends more time on a particular assignment that means that

0.D < ( ) ( )0 250 2.24 0.0125.11.18

P D P Z P Z− < = < = < − =

T6.14 (a) Let X stand for the number of Hispanics in the sample. X has a binomial distribution with

1200n = and 0.13.p = Therefore ( )1200 0.13 156X npµ = = = and

( ) ( )( )1 1200 0.13 0.87 11.6499.X np pσ = − = = (b) If the sample contains 15% Hispanics, that means

that there were ( )1200 0.15 180= Hispanics in the sample. What is the probability of getting at least this many Hispanics in a sample if, in fact, only 13% of the population is Hispanic? In symbols, this is

( ) ( ) ( )1200

1200

180

1200180 0.13 0.87 .i i

iP X

i−

=

≥ =

∑ Using technology, this probability is 0.0235. If we use the

Normal approximation to the binomial distribution, we get

( ) ( )180 156180 2.06 0.0197.11.65

P X P Z P Z− ≥ = ≥ = ≥ =

chapter 6apstatsgrabowski.weebly.com/.../3/0/3/4/30343621/ch06_solutions.pdfthis means that about...

Documents