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 mean and standard deviation of the sampling distribution of x bar is 2. Sample size ( n = 49 ≥ 30 ). Therefore, with using the Central Limit Theorem we conclude that the sampling distribution of x is approximately normal.
  2. P(Z>(83-80)/2)=P(Z>1.5)=1-P(Z<=1.5)=1-0.9332=0.0668 If we take 100 simple random samples, about 7 of the samples will result in a mean that is greater than 83.
  3. P(Z<=(75.8-80)/2)=P(Z<=-2.10)=0.0179 If we take 100 simple random samples, about 2 of the samples will result in a mean that is 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 we take 100 simple random samples, about 78 of the samples will result in a mean that is between 78.3 and 85.1.

17.

  1. The population is normally distributed. The mean and standard deviation of the sampling distribution are μx = μ = 64 and 4.907
  2. P(Z<(67.3-62)/4.907)=P(Z<0.67)=0.7486 If we take 100 simple random samples, about 75 of the samples will result in a mean that is less than 67.3.
  3. P(Z>=(65.2-64)/4.907)=P(Z<0.24)=1-0.5948=0.4052 If we take 100 simple random samples, about 41 of the samples will result in a mean that is greater than or equal to 65.2.

19.

  1. P(Z<(260-266)/16)=P(Z<-0.38)=0.3520 If we take 100 simple random samples, about 35 of the samples will last less than 260 days.
  2. Since the length of human pregnancies is normally distributed, the sampling distribution of x = μ x = 266, and σx = 16 ≈3.578.
  3. P(Z<(260-266)/3.578)=P(Z<-1.68)=0.0465 If we take 100 simple random samples of, about 5 of the samples will result in a mean gestation period of 260 days or less.
  4. P(Z<(260-266)/2.263)=P(Z<-2.65)=0.0040 If we take 1000 simple random sample about 4 of the samples will result in a mean gestation period of 260 days or less.
  5. The sample likely came from a population whose mean gestation period is less than 266 days.
  6. P(-2.42<=Z<=2.42)=0.9922−0.0078=0.9844 If we take 100 simple random samples, about 98 of the samples will have a mean gestation period within 10 days of the mean.

21.

  1. P(Z>(95-90)/10)=P(Z>0.5)=1-0.6915=0.3085 If we select a simple random sample, about 31 of the students would read more than 95 words per minute.
  2. P(Z>1.73)=1-P(Z<=1.73)=1-0.9582=0.0418 If we take 100 simple random samples, about 4 of the samples will result in a mean reading rate that is more than 95 words per minute.
  3. P(Z>2.45)=1-P(Z<=2.45)=1-0.9929=0.0071 If we take 1000 simple random samples, about 7 of the samples will result in a mean reading rate that is more than 95 words per minute.
  4. Increasing the sample size decreases the probability that x > 95 ,because σx decreases as n increases.
  5. The new reading program is not abundantly more effective than the old program. P(Z>1.25)=1-P(Z<=1.25)=1-0.8944=0.1056. If we take 100 simple random samples, about 11 of the samples will result in a mean reading rate that is above 92.8 words per minute.
  6. The area to the right of z is 0.05 if z=1.645, so 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(Z>-0.17)=1-P(Z<=-0.17)=1-0.4325=0.5675 If we select a simple random sample of,about 57 of the months would have positive rates of return.
  2. P (Z > −0.61)=1−P(Z ≤−0.61) =1−0.2709= 0.7291 If we take 100 simple random samples of size n = 12 months, then about 73 of the samples will result in a mean monthly rate that is positive.
  3. P(Z > −0.86)=1−P(Z ≤−0.86) =1−0.1949= 0.8051 If we take 100 simple random samples of size n = 24 months, then about 81 of the samples will result in a mean monthly rate that is positive.
  4. P(Z>-1.05)=1−P(Z ≤−1.05) =1−0.1469= 0.8531 If we take 100 simple random samples of size n = 36 months, then about 85 of the samples will result in a mean monthly rate that is positive.
  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. 10000(0.05) = 500; n = 75 , is less than 5% of the population size; np(1−p)=75(0.8)(0.2)=12≥10. So the distribution of pˆ is approximately normal, with mean μpˆ = p = 0.8 and standard deviation σ=0.046.
  2. P(Z>=0.87)=1-P(Z<0.87)=1-0.8078=0.1922 About 19 out of 100 random samples of size n = 75 will result in 63 or more individuals with the characteristic.
  3. P(Z ≤ −2.60)= 0.0047 About 5 out of 1000 random samples of size n = 75 will result in 51 or fewer individuals with the characteristic.

12.

  1. 25000(0.05) = 1250; n = 200, is less than 5% of the population size; np(1−p)=200(0.65)(0.35)=45.5≥10. So the distribution of pˆ is approximately normal, with mean μpˆ = p = 0.65 and standard deviation σpˆ≈0.034.
  2. P(Z ≥ 0.89)=1− P(Z < 0.89)=1−0.8133=0.1867 About 19 out of 100 random samples of size n = 200 will result in 136 or more individuals with the characteristic.
  3. P(Z<=-1.78)=0.0375 About 4 out of 100 random samples of size n = 200 will result in 118 or fewer individuals with the characteristic.

13.

  1. np(1-p)=1000(0.35)(0.65)=227.5>10, so the distribution of pˆ is approximately normal, with mean μpˆ = p = 0.35 and standard deviation
  2. P^=0.39, P(Z≥2.65) =1−P(Z <2.65)=1-0.996 = 0.004 About 4 out of 1000 random samples of size n = 1000 will result in 390 or more individuals with the characteristic.
  3. P^=0.32, P(Z ≤ −1.99)=0.0233 About 2 out of 100 random samples of size n = 1000 will result in 320 or fewer individuals with the characteristic.

14.

  1. 1500000(0.05)=75000; n = 1460 is less than 5% of the population size; np(1− p)=1460(0.42)(0.58)=355.656 ≥10. The distribution of pˆ is approximately normal, with mean μpˆ = p = 0.42 and standard deviation 0.013
  2. P^=0.45, P(Z ≥ 2.32)=1−P(Z<2.32)=1−0.9898=0.0102 About 1 out of 100 random samples will result in 657 or more individuals with the characteristic.
  3. P^=0.4, P(Z<=-1.55)=0.0606 About 6 out of 100 random samples of size will result in 584 or fewer individuals with the characteristic.

15.

  1. The sample size is less than 5% of the population size and np(1− p)= 200(0.47)(0.53) = 49.82 ≥10. The distribution of pˆ is approximately normal, with mean 0.47 and standard deviation 0.035.
  2. 0.1977 About 20 out of 100 random samples Americans will result in a sample.
  3. 0.0239 About 2 out of 100 random samples Americans will result in a sample.

16.

  1. The sample size is less than 5% of population and the distribution of P^ is approximately normal, with mean 0.82 and standard deviation 0.038
  2. 0.2177. About 22 out of 100 random samples will result in at least 85 who are satisfied with their lives.
  3. Yes, unusual. 0.0344. About 3 out of 100 random samples of size will result in fewer than 76 individuals who are satisfied with their lives.

17.

  1. np(1− p)=500(0.39)(0.61)=118.95≥10. The distribution of pˆ is approximately normal, with mean 0.39 and standard deviation 0.022.
  2. 0.3228. About 32 out of 100 random samples will result in less than 38% who believe marriage is obsolete.
  3. 0.3198. About 32 out of 100 random sample between 40% and 45% say that marriage is obsolete.
  4. 0.0838. It’s not unusual.