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. x̅ is approximately normal with μx̅= 80 and σ x̅= 2
  2. P (x̅>83) = 0.0668.
  3. P(x̅ less than or equal to 75.8) = 0.0179.
  4. P(78.3 < x̅ < 85.1) = 0.7969

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 x̅ is also normally distributed with μx̅= 64 and σ x̅= 17/sqrt(12) = 4.907.
  2. P (x̅ < 67.3) = 0.7286.
  3. P(x̅ is more than or equal to 65.2) = 0.4052.

19.

  1. P(X<260) = 0.3520. If we randomly select 100 human pregnancies, then about 35 of the pregnancies will last less than 260 days
  2. The sampling distribution of x̅ is normal with μx̅=266 and σ x̅= 16/sqrt(20) = 3.578.
  3. P(x̅ less than or equal to 260) = 0.0465.
  4. P(x̅ less than or equal to 260) = 0.0040.=
  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 less than or equal to x̅ less than or equal to 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.
  2. P(x̅ > 95) = 0.0418.
  3. P(x̅ > 95) = 0.0071
  4. Increasing the sample size decreases P(x̅ > 95). This happens because σ x̅ decreases as n increases.
  5. A mean reading rate of 92.8 wpm is not usually P(x̅ greater than or equal to 95) = 0.1056. This mean that the new reading program is 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.
  2. P(x̅ > 0) = 0.7291.
  3. P(x̅ > 0) = 0.8051.
  4. P(x̅ > 0) = 0.8531.
  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 sample distribution of p̂ is approximately normal with μp̂ = 0.8 and σp̂= sqrt((0.8(0.2))/75) = 0.046
  2. P (p̂ is greater than or equal to 0.84) = 0.1922.
  3. P (p̂ is less than or equal to 0.68) = 0.0047.

12.

  1. The sample distribution of p̂ is approximately normal with μp̂ = 0.65 and σp̂= sqrt((0.35(0.65))/200) = 0.033
  2. P (p̂ is greater than or equal to 0.68) = 0.8159
  3. P (p̂ is greater than or equal to .59) = 0.0351

13.

  1. The sample distribution of p̂ is approximately normal with μp̂ = 0.35 and σp̂= sqrt((0.35(0.65))/1000) = 0.015
  2. P (p̂ is greater than or equal to 0.39) = 0.0040.
  3. P (p̂ is less than or equal to 0.32) = 0.0233.

14.

  1. The sample distribution of p̂ is approximately normal with μp̂ = 0..42 and σp̂= sqrt((0.42(0.58))/1460) = 0.013
  2. P (p̂ is greater than or equal to 0.45) = 0.0104
  3. P (p̂ is less than or equal to 0.40) = 0.0618

15.

  1. The sample distribution of p̂ is approximately normal with μp̂ = 0.47 and σp̂= sqrt((0.47(0.53))/200) = 0.035
  2. P (p̂ > 0.5) = 0.1977.
  3. P (p̂ is less than or equal to 0.4) = 0.0230. about 2 out of 100 random samples of size n=200 will result in 80 or fewer individuals (that is, 40% or less) who can order a meal in a foreign language. The result is unusual.

16.

  1. The sample distribution of p̂ is approximately normal with μp̂ = 0.82 and σp̂= sqrt((0.82(0.18))/100) = 0.038
  2. P (p̂ is greater than 0.85) = 0.214
  3. P (p̂ is less than 0.75) = 0.0329. Yes, this would be unusual.

17.

  1. The sample distribution of p̂ is approximately normal with μp̂ = 0.39 and σp̂= sqrt((0.39(0.61))/500) = 0.022
  2. P (p̂ < 0.38) = 0.3228.
  3. P (0.40<p̂ <0.45) = 0.3198.
  4. P (p̂ is greater than or equal to 0.42) = 0.0838. about 8 out of 100 random samples of size n=500 adult Americans will result in 210 or more individuals (that is, 42% or more) who believe that marriage is obsolete. This result is not unsual