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. x bar is approximately normal with mu x-bar = 80, sigma x-bar = 2.
  2. P(x-bar > 83) = 0.0668. if we take 100 simple random samples of size n = 49 from a population with mu = 80 and sigma = 14, then about 7 of the samples will result in a mean that is greater than 83.
  3. P(x-bar <= 75.8) = 0.0179. if we take 100 simple random samples of size n = 49 from a population with mu = 80 and sigma = 14, then about 2 of the samples will result in a mean that is less than or equal to 75.8.
  4. P(78.3 < x-bar < 85.1) = 0.7969. if we take 100 simple random samples of size n = 49 from a population with mu = 80 and sigma = 14, then about 80 of the samples will result in a mean that is between 78.3 and 85.1.

17.

  1. The population must be normally distributed to compute probabilities involving the sample mean. If the population is normally distributed, then the sampling distribution of x-bar is also normally distributed with mu x-bar = 64 and sigma x-bar = 4.907.
  2. P(x-bar < 67.3) = 0.7486. if we take 100 simple random samples of size n = 12 from a population with mu = 64 and sigma = 17, then about 75 of the samples will result in a mean that is less than 67.3.
  3. P(x-bar >= 65.2) = 0.4052. if we take 100 simple random samples of size n = 12 from a population with mu = 64 and sigma = 17, then about 41 of the samples will result in a mean that is greater than or equal to 65.2.

19.

  1. P(X < 260) = 0.3520. If we take 100 simple random samples of size n = 20 human pregnancies, then about 35 of the samples will result in a mean gestation period of 260 days or less.
  2. The sampling distribution of x-bar is normal with mu x-bar = 266 and sigma x-bar = 3.578.
  3. P(x-bar <= 260) = 0.0465. If we take 100 simple random samples of size n = 20 human pregnancies, then about 5 of the samples will result in a mean gestation period of 260 days or less.
  4. P(x-bar <= 260) = 0.0040. If we take 1000 simple random samples of size n = 50 human pregnancies, then about 4 of the samples will result in a mean gestation period of 260 days or less.
  5. This result would be unusual, so the sample likely came from a population whose mean gestation period is less than 266 days.
  6. P(256 <= x-bar <= 276) = 0.9844.If we take 100 simple random samples of size n = 15 human pregnancies, then about 98 of the samples will result in a mean gestation period between 256 and 276 days, inclusive.

21.

  1. P(X > 95) = 0.3085. If we select a simple random sample of n = 100 second grade students, then about 31 of the students will read more than 95 words per minute.
  2. P(x-bar > 95) = 0.0418. If we take 100 simple random samples of size n = 12 second grade students, then about 4 of the samples will result in a mean reading rate that is more than 95 words per minute.
  3. P(x-bar > 95) = 0.0071.If we take 1000 simple random samples of size n = 24 second grade students, then about 7 of the samples will result in a mean reading rate that is more than 95 words per minute.
  4. Increasing the sample size decreases P(x-bar > 95). This happens because sigma x-bar decreases as n increases.
  5. A mean reading rate of 92.8 wpm is not unusual since P(x-bar >= 92.8) = 0.1056. This means that the new reading program is not abundantly more effective than the old program.
  6. There is a 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. P(X > 0) = 0.5675. If we select a simple random sample of n = 100 months, then about 57 of the months will have positive rates of return.
  2. P(x-bar > 0) = 0.7291. If we take 100 simple random samples of size n = 12 months, then about 73 of the samples will result in a mean monthly rate of return that is positive.
  3. P(x-bar > 0) = 0.8051.If we take 100 simple random samples of size n = 24 months, then about 81 of the samples will result in a mean monthly rate of return that is positive.
  4. P(x-bar > 0) = 0.8531. If we take 100 simple random samples of size n = 36 months, then about 85 of the samples will result in a mean monthly rate of return that is positive.
  5. The likelihood of earning a positive rate of return increases as the investment time horizon increases.

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 p-hat is approximately normal with mu p-hat = 0.8, sigma p-hat = 0.046.
  2. P(p-hat >= 0.84) = 0.1922 About 19 out of 100 random samples of size n = 75 will result in 63 or more individuals with the characteristic.
  3. P(p-hat <= 0.68) = 0.0045. About 5 out of 1000 random samples of size n = 75 will result in 51 or fewer individuals with the characteristic.

12.

  1. The sampling distribution of p-hat is approximately normal with mu p-hat = 0.65, sigma p-hat = 0.034.
  2. P(p-hat >= 0.68) = 0.1894 About 19 out of 100 random samples of size n = 200 will result in 136 or more individuals with the characteristic.
  3. P(p-hat <= 0.59) = 0.0392. About 4 out of 100 random samples of size n = 200 will result in 118 or fewer individuals with the characteristic.

13.

  1. The sampling distribution of p-hat is approximately normal with mu p-hat = 0.35, sigma p-hat = 0.015.
  2. P(p-hat >= 0.39) = 0.0038 About 4 out of 1000 random samples of size n = 1000 will result in 390 or more individuals with the characteristic.
  3. P(p-hat <= 0.32) = 0.0228. About 2 out of 100 random samples of size n = 1000 will result in 320 or fewer individuals with the characteristic.

14.

  1. The sampling distribution of p-hat is approximately normal with mu p-hat = 0.42, sigma p-hat = 0.0129.
  2. P(p-hat >= 0.45) = 0.0099 About 10 out of 1000 random samples of size n = 1460 will result in 657 or more individuals with the characteristic.
  3. P(p-hat <= 0.40) = 0.0606. About 6 out of 100 random samples of size n = 1460 will result in 584 or fewer individuals with the characteristic.

15.

  1. The sampling distribution of p-hat is approximately normal with mu p-hat = 0.47, sigma p-hat = 0.036.
  2. P(p-hat >= 0.5) = 0.2033 About 20 out of 100 random samples of size n = 200 will result in 100 or more individuals with the characteristic.
  3. P(p-hat <= 0.40) = 0.0262. About 3 out of 100 random samples of size n = 1460 will result in 80 or fewer individuals with the characteristic.

16.

  1. The sampling distribution of p-hat is approximately normal with mu p-hat = 0.82, sigma p-hat = 0.038.
  2. P(p-hat > 0.85) = 0.2177 About 22 out of 100 random samples of size n = 200 Americans will result in more than 100 individuals who can order a meal in a foreign language.
  3. P(p-hat <= 0.75) = 0.0329. About 3 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.This result is unusual.

17.

  1. The sampling distribution of p-hat is approximately normal with mu p-hat = 0.39, sigma p-hat = 0.022.
  2. P(p-hat < 0.38) = 0.3228. About 32 out of 100 random samples of size n = 500 adult Americans will result in fewer than 190 individuals who believe that marriage is obsolete.
  3. P(0.4 < p-hat < 0.45) = 0.3196. About 32 out of 100 random samples of size n = 500 adult Americans will result in between 200 and 225 individuals who believe that marriage is obsolete.
  4. P(p-hat >= 0.42) = 0.0869. About 9 out of 100 random samples of size n = 500 adult Americans will result in 210 or more individuals who believe that marriage is obsolete. This result is not unusual.