5.1 The statistics are 81% of women and 43% of men. The populations are all women and all men. The parameters would be the percent of all men and all women who have experienced some form of sexual harassment.

5.3 Assuming he plays on other days of the week, and considering his future winnings, then his collection serves as an approximation to a sampling distribution. Without other assumptions, his records can be considered a population and would not represent a sampling distribution.

5.5 The centers would be the same. The spread for the SRS of 400 would be smaller than for the SRS of 100 people.

5.7 μx¯=25, σx¯=2.

5.9 x¯∼N(25, 2). About 95% of the time x¯ should be between 21 and 29.

5.11 Answers will vary.

5.13 P(x¯<8)=0.5.

5.15

1. 1240.
2. Answers will vary.
3. If X is Yes, then the count is 904.
4. p^=904/1240.

5.17 B(35, 0.5).

5.19

1. P(X=0)=0.0404; P(X≥6)=0.0253.
2. P(X=8)=0.1556; P(X≤2)=0.0043.
3. The number of failures in (a) has the distribution in (b).

5.21

1. μX=2.7, σX=1.37.
2. μX=6.3, σX=1.37.
3. The means add up to 9; the standard deviations are the same.

5.23

1. μX=81.84, σX=8.86.
2. P(X≤56)=0.0018.

5.25

1. P(X=7)=0.1200.
2. P(X≤5)=0.5461.

5.1

1. The population is all students from your college.
2. The parameter of interest is the proportion of students who are interested in requiring a statistics course.
3. The sample is 30 students from your college.
4. The statistic would be the number of Yes answers out of the 30 students.
5. Answers will vary.

5.3

1. The population is all people aged 18 to 25; the sample is 18,875 people aged 18 to 25.
2. The population is all restaurant workers; the sample is 250 restaurant workers.
3. The population is all milkweed plants in Yosemite Valley; the sample is 55 milkweed plants.

5.5

1. Drawings will vary. We would expect the distribution to be right-skewed, causing the mean to be larger than the median.

5.7

1. The shape of the sampling distribution should be roughly Normal, centered at 0.4.
2. The shape of the sampling distribution should be roughly Normal, centered at 3.75.

5.9

1. and
2. The mean should be “close” to 50.5.
3. The histogram theoretically should be a Normal distribution, centered at 50.5.

5.11 Using the applet, answers will vary.

2. The mean and standard deviation should be close to the true values.
3. The shape of the curve should be approximately normal.

5.13

1. σx¯=2010.

2. Larger sample sizes result in smaller standard deviations of x¯.

3. Only the standard deviation depends on the sample size.

4. The sample size does not depend on the population size.

5.15

1. μ=90.7.

2. The center of the histogram should theoretically be close to μ.

5.17

1. Larger.

2. For ME=0.17 need σx¯≤0.085.

3. We need n=387.

5.19 μx¯=250, σx¯=0.12.

5.21

2. P=0.7114.
3. P=0.4006.

5.23

1. x¯ is not systematically higher than or lower than μ.
2. With large samples, x¯ is more likely to be close to μ.

5.25 0.0681.

5.27

1. Separate flips are independent.
2. The coin is fair. The probabilities are still P(H)=P(T)=0.5.
3. The parameters for a binomial distribution are n and p.
4. This is best modeled with a Poisson distribution.

5.29

1. A B(200, p) distribution seems reasonable for this setting.
2. This setting is not binomial; there is no fixed value of n.
3. A B(500, 1/12) distribution seems appropriate for this setting.
4. This is not binomial because separate cards are not independent.

5.31 The probability that a digit is greater than 4 is 0.5, and the probability that the digit is not greater than 5 is 0.5.

1. 0.9688.
2. μ=20.

5.33

1. B(15, 0.25).
2. B(15, 0.75).
3. 0.0173.
4. 0.0173.

5.35

1. 3.75 and 11.25; they add up to 15.
2. 1.68.
3. The mean is 0.25; the standard deviation is 0.11.
4. 37.5 and 112.5; they add up to 150. The standard deviation is 5.3 for the number of undergraduates. The mean for p^ is 0.25, with a standard deviation of 0.035.

5.37

1. Yes, use binomial.
2. No, you should use the normal approximation to the binomial.

5.39

1. The mean is μ=0.69, and the standard deviation is σ=0.0008444.
2. 68.83% to 69.17%.
3. It is more reasonable to assume that the population proportion has changed over time.

5.41

1. 0.5355.
2. 0.0595.
3. 0.16798.
4. We can only calculate the probability the sum of the goals would be 2 but not the distribution of the goals between the teams.

5.43

1. μ=50.
2. σ=7.071. P(X>60)=0.0793. Software gives 0.0786.

5.45

1. x¯∼N(137,0.064).
2. 0.0294.

5.47

1. μx¯=0.43, σx¯=0.078.
2. 0.0618.
3. Yes, n=150 is a large enough sample size to be able to use the central limit theorem.

5.49 The probability that the first digit is 1, 2, or 3 is 0.602, so the answer is 0.0409.

5.51

1. μX=3.75.
2. P(X≥10)=0.000795.
3. P(X≥540)=0.0213 (0.0192 using the Normal approximation).

5.53

1. Approximately Normal, with mean 2.21 and standard deviation 0.161.
2. 0.0968.
3. Yes, because n=140 is large.

5.55

1. 0.6826.
2. 0.6826.
3. As the sample size grows, the probability stays the same, so the approximation is precise for large sample sizes.

5.57 P(Y≥200)=0.

5.59 Y has possible values 1, 2, 3, . . . P(Y=k)=(5/6)k−1(1/6).

5.61 The Poisson distribution is not appropriate because the rate is not constant and increases during the midnight and 6 a.m. period.

5.63

1. p=0.25.
2. P(X≥10)=0.0139.
3. μ=5 and σ=1.9365.
4. No. The trials would not be independent because the subject may alter his or her guessing strategy based on this information.

5.65

1. m=15.
2. μ=13.52 and σ=3.629.
3. Without the continuity correction, P(X≥15)=0.3409. With the continuity correction, we have P(X≥14.5)=0.3936. The continuity correction is much closer.

5.67

1. p^F is approximately N(0.82, 0.01921), and p^M is approximately N(0.88, 0.01625).
2. p^M−p^F is approximately N(0.06, 0.02516).
3. P(p^F>p^M)=0.0087 (software gives 0.0085).

5.69

1. 2.
2. 0.
3. Decrease because the standard deviation will decrease.

5.71

1. y¯ has a N(μY, σY/m) distribution, and x¯ has a N(μX, σX/n) distribution.
2. y¯−x¯ has a Normal distribution with mean μY−μX and standard deviation σy¯2+σx¯2.

5.73

1. p^=0.28.
2. p^ is approximately N(0.28, 0.0317) P(p^≥0.28)=0.5.

5.75

1. X has a B(900, 1/5) distribution, with mean μ=180 and σ=12 successes.
2. For p^, the mean is μp^=0.2 and σp^=0.01333.
3. P(p^>0.24)=0.0013.
4. 208 or more successes (correct guesses) in 900 attempts.