Here is some code that will allow you to check your answers. For example if \(\bar{X} \sim \mathcal{N}(5,2)\) and you want to know what the \(Pr(\bar{X} < 3)\) is, you can use the following code. Meaning just change the mean and sd to fit the problem you are #working on.

pnorm(2, mean = 5, sd = 2)
## [1] 0.0668072

8.1

15

  1. Assume that sampling distribution is normal since the sample size is larger than 30 (it is 49). C.L.T applies.

  2. 100 random samples from n=49, from a population with a mean of 80 and SD of 14. 7 of the samples will have a mean greater than 83.

  3. 100 random samples from n=49, from a population with a mean of 80 and SD of 14, roughly 2 of the samples will have a mean that is less than or equal to 75.8.

  4. 100 random samples from n=49, from the population with a mean of 80 and SD of 14, then roughly 78 of the samples will have a mean between 78.3 and 85.1.

17

  1. Population must be normally distributed. If so, the sampling distribution of x(bar) is exactly normal.

  2. 100 random samples of n=12 from a population that is normally distributed with a mean of 64 and SD of 17, then roughly 75 of the samples will have a mean of less than 67.3.

  3. 100 random samples of n=12 from a population that is normally distributed witha mean of 64 and SD of 17, then roughly 41 of the samples will have a mean that is greater than or equal to 65.2.

19

  1. SRS of n=100 human pregnancies, roughly 35 of the pregnancies would last less than 260 days.

  2. Length of human pregnancies being normally distributed, sampling distribution of x(bar) is normal with mean of x(bar) being 266 and SD of x(bar) roughly 3.58.

  3. 100 random SRS of n=20 of human pregnancies, then about 5 of the samples will have a mean gestation period of 260 days or less.

  4. 1000 SRS if n=50 human pregnancies, then roughly 4 of the samples will have a gestation period of 260 days or less.

  5. Would be unusual, but with a relatively small sample size, not impossible. Would increase sample size.

  6. 100 SRS of n=15, then about 98 samples would have a mean gestation period between 256 and 276 days.

21

  1. SRS of n=100 2nd graders, then about 31 of those second graders would read more than 95 Words Per Minute.

  2. 100 SRS of n=12 2nd graders, then roughly 4 samples would have a ean reading rade that is more than 95 W.P.M.

  3. 1000 ranodm samples of n=24 2nd graders, then roughly 7 of the samples will have a mean more than 95 W.P.M

  4. Increasing the sample size decrease the probability that x(bar)>95. This is because SD of x(bar) decreases as n increases.

  5. No, this result would not be unusual. The results mean that the new reading program is not vastly more effective or efficient in teaching them than the old program.

  6. 93.7 W.P.M

23

  1. SRS of n=100 months, then about 57 of the months would have positive rates of return.

  2. 100 SRS of n=12 months, then roughly 73 of the samples will have a mean monthly rate that is positive.

  3. 100 SRS of n=24 months, then roughly 81 of the samples will have a mean monthly rate that is positive.

  4. 100 SRS of n=36 months, then roughly 85 of the samples will have a mean monthly rate that is positive.

  5. Based on “B” and “D” the chance of earning a positive rate of return increases as the investment time horizon increases.

Similarily you can use the above code to determine the \(Pr(\hat{P} < \hat{p})\)

8.2

11

  1. Sample size is less than 5% of the population size and np (1-p) is 12>10. Distribution of p(hat) is approximately normal, p=.08 and an SD of .046.

  2. About 19/100 SRS n=75 will have 63 or more individuals with the characteristic.

  3. About 5/1000 random samples of n=75 will result in 51 or fewer individuals with the characterstic.

12

  1. Sample size is less than 5% of the population size and np(1-p) is 45.5>10. Distribution of p(hat) is approximately normal at p=.65 and an SD of .0337.

  2. About 18/100 SRS n=200 will have 136 or more individuals with the characteristic.

  3. About 3/1000 random samples will have 118 or fewer individuals with that characteristic.

13

  1. Sample size is less than 5% of the population size and np(1-p) is 227.5>10. Distribution is approximately normal for p(hat). p=.35 and SD=.015.

  2. 4/1000 random samples of size n=1000 will result in 390 or more individuals with the charactersitic.

  3. 2/100 random samples n=1000 will result in 320 or fewer individuals with the characteristic.

14

  1. Sample size is less than 5% of the population size and np(1-p)=355.66>10. Distribution is approximately normal for p(hat). p=.42 and SD=.013.

  2. 1/100 random samples of size n=100 will result in 657 or more individuals with the characteristic.

  3. 6/100 random samples of size n=100 will result in 584 or fewer individuals with that statistic.

15

  1. Qualitative with 2 possible outcomes, people either can or can not order a meal in a foreign language.

  2. Sample proportion is a random variable because it varies by sample. The source of the variability is the individuals in the sample and their ability to order a meal in a foreign language.

  3. Sample size of n=200 is less than 5% of the population size and np(1-p)=49.82>10. Distribution is approximately normal. p=.47 and SD=.035.

  4. 20/100 random samples of n=200 Americans will have a sample where more than half of respondants can order food in a foreign language.

  5. 2/100 random samples of n=200 Americans will have a sample where 80 or fewer respondants can order a meal in a foreign language, which would be unusual.

16

  1. Qualitative since people are either happy with their lives or not happy with their lives.

  2. Sample proportion is a random variable because it varies by sample. The source of the variability is the individuals in the sample and their satisfaction with life, either yes or no.

  3. Sample size of n=100 is less than 5% of the population size and np(1-p)=14.76>10. Distribution is approximately normal. p=.82 and SD=.038.

  4. 22/100 random samples n=100 will have a sample where more than 85% respondants will be satisfied with their lives.

  5. Yes, it would be very unusual to get a z-score of .0329 if the survey have only 75/100 respondants say they were satisfied with life.

17

  1. n=500 is less than 5% of the population size and np(1-p)=118.95>10. Distribution of p(hat) is approximately normal. p=.39 and SD=.022.

  2. 32/100 random samples of n=500 adults will have a result less than 38% who believe marriage is obsolete.

  3. 32/100 random samples of n=500 will have between 40-45% of the respondants who say that marriage is obsolete.

  4. 8/100 random samples of n=500 will have at least 210 respondants say that marriage is obsolete.