Here is the syntax you can use to check your answers for Section 8.1. (Forward and Backward)

Say you want to know the \(Pr(\bar{X} < 5)\) and \(\bar{X} \sim \mathcal{N}(6,1.5)\)

pnorm(5, mean = 6, sd = 1.5 )
## [1] 0.2524925

Here is the syntax you can use to check if a “Backward” calculation is correct.

Say you know the probability to the left of \(\bar{x}\) = .04 and you want to know what the appropriate \(\bar{x}\) is. You also know that \(\bar{X} \sim \mathcal{N}(6,1.5)\)

qnorm(.04, mean = 6, sd = 1.5)
## [1] 3.373971

15.

  1. Sample mean is normal; mean = 80; and sample standard deviation = 2.
  2. 0.0668; We conclude that taking 100 simple random samples of n = 49 from a population with mean = 80 and sample standard deviation = 14 results in about 7 of the samples in a mean greater than 83.
  3. 0.0179; Taking 100 simple random samples of n = 49 from a population with mean = 80 and sample standard deviation = 14 results in about 2 of the samples in a mean less than or equal to 75.8.
  4. 0.7969; Taking 100 simple random samples of n = 49 from a population with mean = 80 and sample standard deviation = 14 results in about 80 of the samples in a mean between 78.3 and 85.1.

17.

  1. Population needs to be normally distributed in order to compute probabilities with the sample mean. The sampling distribution of the sample mean, with normal distribution, is normally distributed with the the mean = 64 and the sample standard deviation equals approximately 4.907.
  2. 0.7486; Taking 100 simple random samples of size n = 12 from a population normally distributed with mean = 64 and sample standard deviation = 17 results in about 75 of the samples in a mean that is less than 67.3.
  3. 0.4052; Taking 100 simple random samples of n = 12 from a population normally distributed with a mean = 64 and a sample standard deviation = 17 results in about 41 of the samples in a mean greater than or equal to 65.2.

19.

  1. 0.3520; If we randomly select 100 human pregnancies, 35 of pregnancies will last less then 260 days.
  2. The sampling distribution of the sample mean is normal with the mean = 266 and the sample standard deviation approximately 3.578.
  3. 0.0465; If we have 100 simple random samples with size n = 20 human pregnancies, about 5 of the samples will last 260 days or less.
  4. 0.0040; If we have 1000 simple random samples with size n = 50 human pregnancies, about 4 of the samples will result in mean gestation periods 260 days or less.
  5. Unusual, because a small proportion of 4 out 50 means it is highly likely that the mean gestation period for this population is less than 266 days.
  6. 0.9844; If we take 100 simple random samples of size n = 15 human pregnancies, 98 of samples result in a mean gestation period between 256 and 276 days inclusive.

21.

  1. 0.3085; If we select a simple random sample of n = 100 second grade students with a mean of 90 and a sample standard deviation of 10, 31 of the students will be able to read more than 95 words per minute.
  2. 0.0418; If we take 100 simple random samples of size n = 12 second grade students, about 4 of the samples will have a mean reading rate more than 95 words per minute.
  3. 0.0071; If we take 1000 simple random samples of size n = 24 second grade students, about 7 of the samples will have a mean reading rate more than 95 words per minute.
  4. The sample standard deviation decreases as the sample of size increases, so increasing the sample size will decrease the probability that a randomly selected student will be able to read more than 95 words a minute.
  5. We conclude that a mean reading rate of 92.8 words per minute is not unusual because the probability of a sample mean being greater than or equal to 92.8 is 0.1056. We can infer that the new reading program is effective but not effective enough to make a drastic difference in a relatively short time span.
  6. 5% chance that the mean reading speed of a random sample of 20 second grade students will exceed 93.9 words per minute.

23.

  1. 0.5675; If we select a simple random sample of n = 100 months, 57 of the months will have positive rates of return.
  2. 0.7291; If we have 100 simple random sampls of size n = 12 months, 73 of the samples will have a mean monthly rate of return that is positive.
  3. 0.8051; If we take 100 simple random samples of n = 24 months, 81 of the samples will have a mean monthly rate of return that is positive.
  4. 0.8531; If we have 100 simple random samples of n = 36 months, 85 of the samples will have a mean monthly rate of return that is positive.
  5. As the investment time horizon increases, the liklihood of earning a positive rate of return will also increase.

Here is the syntax you can use to check your answers for Section 8.2. (Forward and Backward)

Say you want to know the \(Pr (\hat{P} < .35)\) and \(\hat{P} \sim \mathcal{N}(.4,.07)\)

pnorm(.35, mean = .4, sd = .07 )
## [1] 0.2375253

Here is the syntax you can use to check if a “Backward” calcuation is corect.

Say you know the probability to the left of \(\hat{p}\) = .04 and you want to know what the appropriate \(\hat{p}\) is. You also know that \(\hat{P} \sim \mathcal{N}(.4,.07)\)

qnorm(.05, mean = 12, sd = 4)
## [1] 5.420585

Section 8.2

11.

  1. The sampling distribution of the sample proportion is normal with the mean of the sample proportion being 0.8 and the standard deviation of the sample proportion being approximately 0.046.
  2. 0.1922; About 19 out of 100 random samples of size n = 75 will result in 63 or more individuals.
  3. 0.0047; About 5 out of 1000 random samples of size n = 75 will result in 51 or fewer individuals.

12.

  1. The sampling distribution of the sample proportion is normal. The mean of the sample proportion is 0.65 and the standard deviation of the sample proportion is approximately 0.0337.
  2. 0.8106; About 81 out of 100 random samples of n = 200 will result in a population of 136 or more individuals.
  3. 0.0384; About 4 out of 100 random samples of size n = 200 will result in a population of 118 or fewer individuals.

13.

  1. The sampling distribution of the sample proportion is normal. The mean of the sample proportion is 0.35 and the standard deviation of the sample proportion is 0.015.
  2. 0.0040; About 4 out of 1000 random samples of size n = 1000 will result in 390 or more individuals.
  3. 0.0233; About 2 out of 1000 random samples of size n = 1000 will result in 320 or fewer individuals.

14.

  1. The sampling distribution of the sample proportion is normal. The mean of the sample proportion is 0.42 and the standard deviation of the sample proportion is 0.013.
  2. 0.9656; About 96 out of 1000 random samples of size n = 1460 will result in a population of 657 individuals or more.
  3. 0.8962;About 90 out of 1000 random samples of size n = 1460 will result in a population of 584 individuals or less.

15.

  1. The sampling distribution of the sample proportion is normal. The mean of the sample proportion is 0.47 and the standard deviation of the sample proportion is 0.035.
  2. 0.1977; About 20 out of 100 random samples of size = 200 Americans will result in more than 100 individuals who can order a meal in a foreign language.
  3. 0.0239; Result is unusual; about 2 out of 100 random samples of size n = 200 Americans will result in 80 or fewer individuals who can order a meal in a foreign language.

16.

  1. The sampling distribution of the sample proportion is normal. The mean of the sample proportion is 0.82 and the standard deviation of the sample proportion is 0.038.
  2. 0.2148; About 21 of the 100 random samples of size = 100 Americans will result in 21 or more individuals who are satisfied with the way things are going in their life.
  3. 0.0043; Result is unusual; Having a mere 75 or fewer Americans satisfied with the way things are going in their life means that about 4 out of 1000 random samples of size n = 100 Americans say that they are satisfied with the way things are going in their life.

17.

  1. The sampling distribution of the sample proportion is normal. The mean of the sample proportion is 0.39 and the standard deviation of the sample proportion is 0.022.
  2. 0.3228; This means about 32 out of 100 random samples of size n = 500 adult Americans will result in fewer than 190 individuals who believe marriage is obsolete.
  3. 0.3198; This means about 32 out of 100 random samples of size n = 500 adult Americans will result in between 200 and 225 individuals who believe marriage is obsolete.
  4. 0.0838; Result is not unusual. This means about 8 out of 100 random samples of size n = 500 adult Americans will result in between 210 or more individuals who believe that marriage is obsolete.