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 an σx̅=2.
  2. P(x̅>83)=0.0668. If 100 simple random samples are taken of n=49 size from a population of μ=80 and σ=14, then approximately 7 of the samples wil give a mean greater than 83.
  3. P(x̅<75.8)=0.0179. If 100 simple random samples are taken of n=49 size from a population with μ=80 and σ=14, the approximately 2 of the samples will give a mean less than or equal to 75.8.
  4. P(78.3<x̅<85.1)=0.7969. If 100 simple random samples are taken of n=49 size from a population with μ=80 and σ=14, then approximately 80 of samples will give a mean that is between 78.3 and 85.1.

17.

  1. In order to use the normal model to compute probabilities involving the sample mean, the distribution of the population must be normal. If so, then the sampling distribution of x̅ is aslo noraml with μx̅=64 and σx̅=4.907.
  2. P(x̅<67.3)=0.7486. If 100 simple random samples are taken of n=12 size from a population with a normal distribution with μ=64 and σ=17, then approximately 75 of the samples will give a mean less than 67.3 .
  3. P(x̅>65.2)=0.4052. If 100 simple random samples are taken of n=12 size from a population with a normal distribution with μ=64 and σ=17, then approximately 41 of the samples will give a mean greater than or equal to 65.2.

19.

  1. P(X<260)=0.3520. If 100 human preganancies are randomly selected, then approximately 35 of the pregnancies will last less than 260 days.
  2. x̅ has a normal sample distibution with μx̅=266 and σx̅=3.578.
  3. P(x̅<260)=0.0465. If 100 simple random samples are take of n=20 human pregnancies, then approximately 5 of the samples will give a mean gestation period of 260 days or less.
  4. P(x̅<260)=0.0040. If 1000 simple random samples are taken of n=50 human pregnancies, then approximately 4 of the samples will give a mean gestation period of 260 days or less.
  5. What I may conclude if a random sample of 50 pregnancies resulted in a mean gestation period of 260 days or less is this result is unusual. This sample likely came from a population whose mean gestation period is less than 266 days.
  6. P(256<x̅<276)=0.9844. If 100 simple random samples are taken of size n=15 human pregnancies , then approximately 98 of the samples will give a mean gestatin period between 256 and 276 days, inclusive.

21.

  1. P(X>95)=0.3085. If a simple random sample of n=100 second grade students is taken, then approximately 31 of the students will read more than 95 words per minute.
  2. P(x̅>95)=0.0418. If 100 simple random samples are taken of size n=12 second grade students, then approximately 4 of the samples will give a mean reading rate more than 95 words per minute.
  3. P(x̅>95)=0.0071. If 100 simple random samples are taken of size n=24 second graders, then approximately 7 of the samples will give a mean reading rate more than 95 words per minute.
  4. Increasing the sample size decreases P(x̅>95) because σx̅ decreases as n increases.
  5. What I would conclude based on this result is that a mean reading rate of 92.8 words per minute is not unusual due to P(x̅>92.8)=0.1056, which means that the new reading rate program is not much more effective than the old one.
  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 a simple random sampe of n=100 months is chosen, the approximately 57 months will have positive return rates.
  2. P(x̅>0)=0.7291. If 100 simple random samples are taken of size n=12 months, then approximately 73 of the samples wil give a mean positive monthly return rate .
  3. P(x̅>0)=0.8051. If 100 simple random samples are taken of size n=24 months , then appproximately 81 of the samples will give a mean positive monthly return rate.
  4. P(x̅>0)-0.8531. If 100 simple random samples are taken of size n=36 months, then approximately 85 of the samples will give a mean positive monthly return rate.
  5. The likelihood of earning a positive return rate increases as 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 normal where μp̂=0.8 and σp̂=0.046.
  2. P(p̂>0.84)=0.1922. Approximately 19 out of 100 random samples with size n=75 will give 63 or more individuals (84%+) with the characteristic.
  3. P(p̂<0.68)=0.0047. Approximately 5 out of 1000 random samples with size n=75 will give 51 or less individuals (68% or less) with the characteristic.

12.

  1. The sample distribution of p̂ is normal where μp̂=0.65 and σp̂=0.034.
  2. P(p̂>0.68)=0.1894. Approximately 19 out of 100 random samples with size n=200 will give 136 or more individuals with the characteristic.
  3. P(p̂<0.59)=0.0392 Approximately 39 out of 1000 random samples with size n=200 will give 118 or less individuals with the characteristic.

13.

  1. The sample distribution of p̂ is normal where μp̂=0.35 and σp̂=0.015.
  2. P(p̂>0.39)=0.0040. Approximately 4 out of 1000 random samples with size n=1000 will give 390 or more individuals (39%+) with the characteristic.
  3. P(p̂<0.32)=0.0233. Approximately 2 out of 100 random samples with size n=1000 will give 320 or less individuals (32% or less) with the characteristic.

14.

  1. The sample distribution of p̂ is normal where μp̂=0.42 and σp̂=0.013.
  2. P(p̂>0.45)=0.0104. Approximately 1 out of 100 random samples with size n=1460 will give 657 or more individuals with the characteristic.
  3. P(p̂<0.4)=0.0618. Approximately 2 out of 100 random samples with size n=1460 will give 584 or less individuals with the characteristic.

15.

  1. The sample distribution of p̂ is normal where μp̂=0.47 and σp̂=0.035.
  2. P(p̂>0.5)=0.1977. Approximately 20 out of 100 random samples with size n=200 will give more than 100 individuals (more than 50%) who can order a meal in a foreign language.
  3. P(p̂<0.4)=0.0239. Approximately 2 out of 100 random samples with size n=200 will give 80 or less individuals (40% or less) who can order a meal in a foreign language. This is an unusual result.

16.

  1. The sample distribution of p̂ is normal where μp̂=0.82 and σp̂=0.038.
  2. P(p̂>0.85)=0.2148. Approximately 21 out of 100 random samples with size n=100 will give more than 85 individuals who are satisfied with the way things are going in life.
  3. P(p̂<0.75)=0.0329. Approximately 3 out of 100 random samples with size n=100 will give 75 or less individuals who are satisfied with the way things are going in life. This is an unusual result.

17.

  1. The sample distribution of p̂ is normal where μp̂=0.39 and σp̂=0.022.
  2. P(p̂<0.38)=0.3228. Approximately 32 out of 100 random samples with size n=500 American adults will give less than 190 individuals (less than 38%) who believe marriage is obsolete.
  3. P(0.40<p̂<0.45)=0.3198. Approximately 32 out of 100 random samples with size n=500 American adults will give between 200 and 225 individuals (40%-45%) who believe marriage is obsolete.
  4. P(p̂>0.42)=0.0838. Approximately 8 out of 100 random samples with size n=500 American adults will give 210 or more individuals (42%+) who believe marriage is obsolete. This is not an unusual result.