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. normal shape; mean= 80; std dev= 2
  2. z-score = 1.50; P=1-0.9332; P(xbar>83)= 0.0668
  3. P(xbar<75.8)= 0.0179
  4. P(78.3<xbar<85.1)= 0.7969

17.

  1. the distribution of the population must be normal in order to use a normal model to compute probabilities involving the sample mean. Assuming this is true, the sample distribution of xbar is normal with mean=64 and standard deviation=4.91
  2. P(xbar<67.3)= 0.7486
  3. P(xbar>65.2)= 0.4052

19.

  1. z-score=(260-266)/16=-0.38; P(X<260)= 0.3520
  2. the sampling distribution of a random sample of 20 pregnancies is normal with mean=266 and standard deviation=3.58
  3. n=20; P(xbar<260)= 0.0465
  4. n=50; P(xbar<260)= 0.0040
  5. if a random sample of 50 pregnancies resulted in a mean gestation period of 260 days or less, this would be unusual due to the low probability of this happening. So, the sample may be from a population whose mean is closer to 260 days.
  6. n=15; P(256<xbar<276)= 0.9844

21.

  1. z-score= 5/10=0.50; P(X>95)=1-0.6915= 0.3085
  2. n=12; mean=90; standard dev=2.89; P(xbar>95)= 0.0418
  3. n=24; standard dev= 2.04; P(xbar>95)= 0.0071
  4. as the sample size increases, the probability that a sample of second graders with a mean reading rate of more than 95 words per minute decreases. This is because as the sample size increases, the standard deviation decreases.
  5. the probability that the sample of 20 students has a reading rate of more than 92.8 is 0.1056 (because P(xbar<92.8)=0.8944). so, it is not unusual that the sample’s mean is 92.8 but the reading program has not increased the mean by much/is not that much more effective in increasing children’s reading rates.
  6. P=0.95; 1.65=(x-90)/2.24; x= 93.7; there is a 5% chance that the mean reading speed of a random sample of 20 second grade students will exceed 93.7 words per minute

23.

  1. P(X>0)= 0.5675
  2. n=12; standard deviation= 0.00345; P(xbar>0)= 0.7291
  3. n=24; P(xbar>0)= 0.8051
  4. n=36; P(xbar>0)= 0.8531
  5. as the investment time horizon increases, the likelihood of earning a positive rate of return on stocks 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

Note: I am unable to use the greater/less than or equal signs when answering for probability, so I only use the < or > sign, but I am aware that they should be underlined.

11.

  1. The sampling distribution of phat is approximately normal (because (75)(0.8)(0.2)=12, which is greater than 10) with mean= 0.8 and standard deviation=0.046
  2. z-score= 0.87; P(phat>0.84)= 1-0.8078= 0.1922
  3. z-score= -2.61; P(phat<0.68)= 0.0045

12.

  1. approximately normal (200)(0.65)(0.35)>10; mean=0.65; standard deviation= 0.034
  2. z-score= 0.88; P(phat>0.68)= 1-0.8106= 0.1894
  3. z-score= -1.76; P(phat<0.59)= 0.0392

13.

  1. Approximately normal (1000)(0.35)(0.65)>10; mean=0.35; standard deviation=0.015
  2. z-score= 2.67; P(phat>0.39)= 1-0.9962= 0.0038
  3. z-score= -2.00; P(phat<0.32)= 0.0228

14.

  1. approximately normal (1460)(0.42)(0.58)>10; mean=0.42; standard deviation=0.013
  2. z-score= 2.31; P(phat>0.45)=1-0.9896= 0.0104
  3. z-score= -1.54; P(phat<0.40)= 0.0618

15.

  1. approximately normal because (200)(0.47)(0.53)>10; mean=0.47; standard deviation= 0.035
  2. z-score= 0.86; P(phat>0.5)= 1-0.8051= 0.1949
  3. 80/200=0.4; P(phat<0.4)= 0.0228; so the probability that 80 or less americans out of a sample of 200 can order in a foreign langues is only about 2%, and about 2 out of 100 random samples of 200 americans will have 80 or less people who are able to order in a foreign language, which is an unusual result.

16.

  1. approximately normal because (100)(0.82)(0.18)>10 14.76>10; mean=0.82; standard deviation=0.038
  2. z-score= 0.79; P(phat>0.85)=1-0.7852= 0.2148
  3. z-score= -1.84; P(phat<0.75)= 0.0329; so the probability that 75 or less americans are satisfied with their lives is about 3% and about 3 out of 100 surveys of random samples of 100 americans will have 75 or less people say that they are happy with their lives. this is unusual

17.

  1. approximately normal because (500)(0.39)(0.61)>10; mean=0.39; standard deviation=0.022
  2. z-score= -0.45; P(phat<0.38)= 0.3264
  3. z-score1= 0.45; z-score2= 2.73; P(0.40<phat<0.45)=0.9968-0.6736= 0.3232
  4. z-score= 1.36; P(phat>0.42)= 1-0.9131= 0.0869; this result is not unusual