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 a normal distribution.
  2. .0668
  3. .0179
  4. .7969

17.

  1. It must be normally distributed to solve the propabilities with the sample mean. If the population is normally distributed, X-Bar is also a normal distribution as well as the standard deviation.
  2. .7486
  3. .4052

19.

  1. .3520
  2. 3.578
  3. .0465
  4. .0040
  5. There would be an unusual result, so the sample most likely was from a population with a digestion period less than 266 days.
  6. .9844

21.

  1. .3085
  2. .0418
  3. .0071
  4. Increasing the sample size decreases
  5. .1056
  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. .5675
  2. .07291
  3. .8051
  4. .8531
  5. The likelihood of earning a positive rate of return increase 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 Phat is approximately normal
  2. .1922
  3. .0047

12.

  1. The sample distribution is normal.
  2. .8133
  3. .0375

13.

  1. Type answer here.
  2. .0040
  3. .0233

14.

  1. Type answer here.
  2. Type answer here.
  3. Type answer here.

15.

  1. The sampling distribution is normal because .35 does not stray to far from the standard deviation, which is .015
  2. .1977, which means that 2 out of 100 random samples when n=200 Americans will result in more than 50% of individuals who can order in another language.
  3. .0239, which means that about 2 out of 10 random sample sizes of n=200 Admericans will result in 8 or fewer people.

16.

  1. The sampling distribution is approximately normal, with the square root of (.82(1-.81)/100)= .0384
  2. .9032
  3. .9324

17.

  1. .0022
  2. .3228
  3. .3198
  4. .0838