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. The sampling distribution is approximately normal with a mean of 80 and a standard deviation of 2.
  2. .0668
  3. .0179
  4. .8969

17.

  1. Because the sample size is fairly low, the original distribution must be at least roughly normal for the sampling distribution to also be normal. Assuming that it is, that distribution would be approximately normal with a mean of 64 and a standard deviation of 4.91.
  2. .7496
  3. .4052

19.

  1. .3520
  2. The distribution would be approximately normal with a mean of 266 and a standard deviation of 3.58.
  3. .0465
  4. .0040
  5. That the original data was unreliable; the probability of getting such a result is extremely low. Or I guess you could assume that the sample was not random and it was taken in a wing of a hospital specializing in premature births.
  6. .9844

21.

  1. .3085
  2. .0418
  3. .0071
  4. Increasing the sample size reduces the probability of getting a sample mean further out in the tails, as the distribution becomes progressively more normal. The more normal it is, the more you would expect results to tend towards the mean.
  5. You would probably conclude that the program was (statatistically speaking anyway…2.8 words per minute improvement doesn’t sound great in another context) effective, because there is just barely a 10% chance of finding 20 students who can read at 92.8 wpm. The fact that 20 students in the program had improved that much suggests the success of the program.
  6. 93.696

23.

  1. .5714
  2. .7291
  3. .8023
  4. .8531
  5. Effectively, the longer you hold the stock without selling it, the higher the probability of a positive return.

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 is approximately normal with a mean of .8 and a standard deviation of .046.
  2. .1922
  3. .0045

12.

  1. The sampling distribution is approximately normal with a mean of .65 and a standard distribution of .034.
  2. .1894
  3. .0392

13.

  1. The sampling distribution is approximately normal with a mean of .35 and a standard deviation of .015.
  2. .0038
  3. .0228

14.

  1. The sampling distribution is approximately normally distributed with a mean of .42 and a standard deviation of .013.
  2. .0104
  3. .0618

15.

  1. The sampling distribution is approximately normal (passes both tests) with a mean of .47 and a standard deviation of .035.
  2. .1949
  3. Yes, it would be somewhat odd, as the probability of that is only .0228.

16.

  1. The sampling distribution is approximately normal (passes both tests) with a mean of .82 and a standard deviation of .038.
  2. .2148
  3. Yes, it would be weird because the probability of that is only .0329.

17.

  1. The sampling distribution is approximately normal (passes both tests) with a mean of .39 and a standard deviation of .022.
  2. .3264
  3. .6667
  4. Yes, it would a bit odd since the probability of that is only .0869.