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. \(\bar{X} \sim \mathcal{N}(80,2)\)
  2. P(\(\bar{x}\) >83)=0.0668
  3. P(\(\bar{x}\) <=75.8)=0.0179
  4. P(78.3< \(\bar{x}\) <85.1)=0.7669

17. (a) \(\bar{X} \sim \mathcal{N}(64,4.907)\) (b) P(\(\bar{x}\) <83)=0.0668 (c) P(\(\bar{x}\) >=65.2)=0.4052

19.

  1. P(X <260)=0.3520
  2. \(\bar{X} \sim \mathcal{N}(266,3.578)\)
  3. P(\(\bar{x}\) <=260)=0.0465
  4. P(\(\bar{x}\) <260)=0.0040
  5. The result would be unusual. The sample probably was taken from a population whose mean gestation period was less than 266 days.
  6. P(256<= \(\bar{x}\) <=276)=0.9844

21.

  1. P(X >95)=0.3085
  2. P(\(\bar{x}\) >95)=0.0418
  3. P(\(\bar{x}\) >95)=0.0071
  4. Increasing the sample size will decrease P(\(\bar{x}\) >95). This is because standard deviation(sigma sub x-bar) decreases as n increases.
  5. A mean reading rate of 92.8 wpm is not unusual since P(\(\bar{x}\) >=92.8)= 0.1056. Which means that the new reading program is not way more effective than the old reading program.
  6. 93.9 wpm

23.

  1. P(X >0)=0.5675
  2. P(\(\bar{x}\) >0)= 0.7291
  3. P(\(\bar{x}\) >0)= 0.8051
  4. P(\(\bar{x}\) >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. \(\hat{P} \sim \mathcal{N}(0.8,0.046)\)
  2. P(\(\hat{p}\) >= 0.84)= 0.1922
  3. P(\(\hat{p}\) <= 0.32)= 0.0233

12.

  1. \(\hat{P} \sim \mathcal{N}(0.65,0.0337)\)
  2. P(\(\hat{p}\) >= 0.68)= 0.1867
  3. P(\(\hat{p}\) <= 0.59)= 0.0375

13.

  1. \(\hat{P} \sim \mathcal{N}(0.35,0.015)\)
  2. P(\(\hat{p}\) >= 0.39)= 0.0040
  3. P(\(\hat{p}\) <= 0.32)= 0.0233

14.

  1. \(\hat{P} \sim \mathcal{N}(0.42,0.0129)\)
  2. P(\(\hat{p}\) >= 0.45)= 0.0102
  3. P(\(\hat{p}\) <= 0.40)= 0.0606

15.

  1. \(\hat{P} \sim \mathcal{N}(0.47,0.035)\)
  2. P(\(\hat{p}\) > 0.5)= 0.1977
  3. P(\(\hat{p}\) <= 0.4)= 0.0239. The result is unusual.

16.

  1. \(\hat{P} \sim \mathcal{N}(.4,.07)\)
  2. P(\(\hat{p}\) >= 0.85)= 0.2177
  3. P(\(\hat{p}\) <= 0.75)= 0.0344 The result is unusual.

17.

  1. \(\hat{P} \sim \mathcal{N}(0.39,0.022)\)
  2. P(\(\hat{p}\) < 0.38)= 0.3228
  3. P(0.40 < \(\hat{p}\) < 0.45)= 0.3198
  4. P(\(\hat{p}\) >= 0.42)= 0.0838