Here is some code that will allow you to check your answers. For example if \(\bar{X} \sim \mathcal{N}(5,2)\) and you want to know what the \(Pr(\bar{X} < 3)\) is, you can use the following code. Meaning just change the mean and sd to fit the problem you are #working on.

pnorm(2, mean = 5, sd = 2)
## [1] 0.0668072

8.1

15

  1. Since Mu=80 and Standard Deviation = 14, We are not told that the population is normally distributed, but we do have a large sample size (n+ 49_>30). Therefore, we can use the Central Limit Theorem to say that the sampling distrubtion of x bar is normal.

  2. If we take 100 simple randoms samples of size n=49 from a population with Mu = 80 and standard deviation = 14, then about 7 of the samples will results in a mean that is greater than 83.

  3. If we take 100 simple random samples of size n=49 from a population with Mu = 80 and standard deviation = 14, then about 2 of the samples will result in a mean that is less than or equal to 75.8

  4. If we take 100 simple random samples of size n=49 from a population with Mu=80 and standard deviation=14 then about 78 of the samples will result in a mean that is vetween 78.3 and 85.1.

17

  1. The population must be normally distributed. If this is the case, then the sampling distribution of x bar is exactly normal. the mean and standard deviation of the sampling distribution are Mu x bar = 64 and Standard deviation = 4.907

  2. If we take 100 simple random samples of size n = 12 from a population that is normally distributed with Mu=64 and standard deviation=17, then about 75 of the samples will results in a mean that is less than 67.3

  3. If we take 100 simple random samples of size n=12 from a population that is nomrally distributed with Mu=64 and Standard deviation=17 then about 41 of the samples will result in a mean that is greater than or equal to 65.2.

19

  1. If we select a simple random sample of n=100 human pregnancies, then about 35 of the pregnancies would last less than 260 days.

  2. Since the lenth of human pregnancies is normally distributed, the sampling distribution of x bar is normal with Mu x bar = 266 and Standard deviation x bar is approximately 3.578.

  3. If we take 100 simple random samples of size n = 20 human preganncies, then about 5 of the samples will result in a mean gestation period of 260 days or less.

  4. If we take 1000 simple random samples of size n=50 human pregnancies, then about 4 of the samples will result in a mean gestation period of 260 days or less.

  5. We can conclude that the sample likely came from a population whose mean gestation period is less than 266 days.

  6. If we take 100 simple random samples of size n=15 human prengnacies, then about 98 of the samples will result in a mean gestation period between 256 and 276 days inclusive.

21

  1. If we select a simple random sample of n =100 second grade students, then about 31 of the students would read more than 95 words per minute.

  2. If we take 100 simple random samples of size n=12 second grade students, then about 4 of the samples will result in a mean reading rate that is more than 95 words per minute.

  3. If we take 1000 simple random samples of size in n=24 second grade students, then 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 bar is greater than 95. This happens because stnadard deviation x bar decreases as n increases.

  5. No, this result would not be unusual because P(x bar > 92.8) = 0.1056. If we taje 100 simple random samples of size n=20 second grade students, then about 11 of the samples will result in a mean reading rate that is above 92.8 words per minute. This result does not qualify as unusual, it just means that the new reading program is not exponentially more effective than the old one.

  6. The area to the right of z is 0.05 if z = 1.645. Solving for c, we get 93.7 words per minute.

23

  1. If we select a simple random sample of n =100 months, then about 57 of the months would have positive rates of return.

  2. 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. If we take 100 simple random samples of n=24 months, then about 81 of the samples will result in a mean monthly rate that is positive.

  4. 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 positve.

  5. The likelihood of earning a positive rate of return increases as the investment time horizon increases.

Similarily you can use the above code to determine the \(Pr(\hat{P} < \hat{p})\)

8.2

11

  1. 25,000(0.05) = 1250; the sample size n=75 is less than 5% of the population size and np(1-p) = 75(0.8)(0.2) = 12 is greater than or equal to 10. The distribution of p hat is approximately normal with mean 0.8 and standard deviation 0.046.

  2. P = 0.1922. Although 19 out of 100 random samples of size n=75 will result in 63 or more individuals (84% or more) with the characteristic .

  3. About 5 out of 1000 random samples of size n=75 will result in 51 or fewer individuals (68% or less) with the characteristic.

12

  1. Type answer here.

  2. Type answer here.

  3. Type answer here.

13

  1. Standard deviation p hat is 0.015

  2. About 4 out of 1000 random samples of size n=1000 willr esults in 390 or more individuals with the characterisitc

  3. About 2 out of 100 random samples of size n=1000 will result in 320 or fewer individuals with the characteristic

14

  1. Type answer here.

  2. Type answer here.

  3. Type answer here.

15

  1. the response to the question is qualitative with two possible outcomes, can order a meal in a foreign language or not

  2. the smaple proportion is a random variable because it varies from sample to sample

  3. standard eviation p hat = 0.035

  4. About 20 out of 100 random samples of size n=200 Americans willr esults in a sample where more than half can order a meal in a foreign language

  5. about 2 out of 100 random samples of size n=200 Americans will result in a sample where 80 or fewer can order a meal in a foreign language. this result is unusual.

16

  1. Type answer here.

  2. Type answer here.

  3. Type answer here.

  4. Type answer here.

  5. Type answer here.

17

  1. standard deviation p hat is apporximately 0.022

  2. about 32 our of 100 random samples of size n=500 adults will result in less tha 38% who believe marriage is obselete.

  3. about 32 out of 100 random samples of size n=500 adults will have between 40% and 45% of the respondents who say marrriage is obselete

  4. about 8 out of 100 random samples of size n=500 adults will have at least 210 adults who say that marriage is obselete.