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 of x bar is approximately normal (n = 49 which is greater than 30). The mean is 80 and the standard deviation is 14/(49^0.5) = 2
  2. Change x bar to z: P(z > 83-80/2) = P(z > 1.5) = 1 - 0.9332 = 0.0668. About 7 in every 100 random samples with n = 49 from a population with mu = 80 and sigma = 14 will have a mean greater than 83
  3. P(z<75.8-80/2) = P(z<-2.1) = 0.0179. If 100 random samples of size n = 49 from a population where mu = 80 and sigma = 14 were taken, about two of the samples will have a mean less than or equal to 75.8
  4. P(78.3 < x bar < 85.1) = P(78.3-80/2 < z < 85.1-80/2) = P(-0.85 < z < 2.55) = 0.9946 - 0.1977 = 0.7969. If 100 random samples of size n = 49 from a population where my - 80 and sigma = 14 were taken, about 80 of the samples will have a mean that is between 78.3 and 85.1

17.

  1. The distribution of the population must be normally distributed (bell-shaped) in order to use the normal model to compute probabilities involving the sample mean. Assuming this condition is true, the sampling distribution of x bar is mu(x bar) = 64 and sigma(x bar) = 17/(12^0.5) = 4.907
  2. P(z < 67.3 - 64 / 4.907) = 0.7486
  3. P(z > to 65.2 - 64 / 4.907) = 1-0.5948 = 0.4052

19.

  1. P(z< 260-266/16) = 0.3520
  2. The sampling distribution of the sample mean would be mu = 266 days and sigma = 16/(SQRT(20)) = 3.578
  3. P(z<260-266/3.578) = 0.0465
  4. P(z<260-266/(16/SQRT(50))) = P(-2.65) = 0.0040
  5. I would conclude that the random sample is very unusual because the probability of selecting a random sample of 50 pregnancies resulting in a mean gestation period of 260 days or less is 0.0040
  6. P(-10/(16/SQRT(15))<z<10/(16/SQRT(15))) = P(-2.42<z<2.42) = 0.9922-0.0078 = 0.9844

21.

  1. P(z>95-90/10) = P(z>0.5) = 1-0.6915 = 0.3085
  2. 0.0418
  3. 0.0071
  4. Increasing the sample size decreases the probability because the standard deviation gets smaller as n increases (sigma / square root of n = as n goes up, standard deviation goes down)
  5. P(z> 92.8-90 / 10/SQRT(20)) = 0.1056. The new reading program is not significantly more effective than the old reading program
  6. 0.95 has a Z value around 1.65, and from back solving to find x bar = zsigma/SQRT(n) + mu = 1.65 (2.236) + 90 = 93.7 wpm. There is a 5% chance that the mean reading speed of a random sample of 20 second grade students will exceeed 93.7 wpm

23.

  1. P(x>0) = P(z>-0.007233/0.04135) = 1 - 0.4325 = 0.5675.
  2. P(x bar > 0) = P(z > -0.007233/(0.04135/SQRT(12))) = 1-0.2709 = 0.7291
  3. 0.8051
  4. 0.8531
  5. As the investment time horizon increases, you are more and more likely to earn a positive rate of return on stocks.

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 of p hat is approximately normal with mu of p hat = 0.8 and sigma of p hat = SQRT(0.8*0.2/75) = 0.046
  2. Convert p hat into z by 0.84-0.8/0.046 = 1-0.8078 = 0.1922. About 19 of 100 random sample size n=75 will have 63 or more individuals with the characteristic.
  3. P(p hat < or equal to 0.68) = 0.0047. Almost 5 out of 1000 random samples of size n=75 will have 51 or fewer individuals with the characteristic.

12.

  1. The sampling distribution of p hat is approximately normal, with mu of p hat = 0.65 and sigma of p hat = SQRT(0.65*0.35/200) = 0.0337
  2. P(z>0.89) = 1 - 0.8133 = 0.1867
  3. P(z<-1.78) = 0.0375

13.

  1. The sampling distribution of p hat is approximately normal (np*(1-p)=227.5 > 10), with mu of p hat = 0.35 and sigma of p hat = 0.015
  2. P(z>0.39-0.35/0.015) = P(z>2.67) = 1-0.9962 = 0.004
  3. P(z<0.32-0.35/0.015) = P(z<-2) = 0.023

14.

  1. The sampling distribution of p hat is approximately normal with mu of p hat = 0.42 and sigma of p hat = 0.0129
  2. P(z>0.45-0.42/0.0129) = P(z>2.33) = 1-0.9901 = 0.0099
  3. P(z<0.4-0.42/0.0129) = P(z<-1.55) = 0.0606

15.

  1. 200*0.47(1-0.47) = 49.8 > 10. The sampling distribution of p hat is approximately normal, with mu of p hat being 0.47 and the sigma of p hat being SQRT(0.47(0.53)/200) = 0.0353
  2. 0.1977. About 20 out of 100 random samples of size n = 200 will result in more than 50% of individuals being able to order a meal in a foreign language.
  3. 0.0239. Yes it would be unusual because about 2 in 100 random samples of size n = 200 Americans will result in 80 or fewer individuals who can order a meal in a foriegn language.

16.

  1. np(1-p) > 10 is satisfied: 100*0.82(1-0.82)=14.76. The sampling distribution of p hat is approximately normal, with mu of p hat = 0.82 and sigma of p hat = SQRT(0.82(1-0.82)/100) = 0.0384
  2. P(z>0.85-0.82/0.0384) = P(z>0.78) = P(z<-0.78) = 0.2177.
  3. P(z<0.75-0.82/0.0384) = P(z<-1.82) = 0.0344. Yes it would be unusual because about 3 in 100 random samples of n = 100 Americans will reveal that 75 individuals or fewer are satisfied with the way things are going in their life.

17.

  1. The sampling distribution of p hat is approximately normal with mu of p hat = 0.39 and sigma of p hat = SQRT(0.39(1-.39)/500) = 0.0218
  2. 0.3228
  3. 0.3198. About 32 out of 100 random samples of size n = 500 will have between 40% and 45% of adults believing that marriage is obsolete.
  4. 0.0838. It is not unusual because about 8 out of 100 random sample sizes of size n = 500 adult Americans will result in 210 or more individuals believing that marriage is obsolete. (Less than 0.05 is my reasoning)