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. The mean of the sample mean of a sample of size 49 with a mean of 80 and standard deviation 14 is also 80. The standard deviation of the sample mean is 2.

  2. .0668.

  3. .0179.

  4. 0.7969.

17

  1. The distribution of the population must be normal in order to use the normal model to compare probabilities involving the sample mean. Assuming this condition is true, the mean of the sample means is 64, with a standard deviation of 4.91.

  2. 0.7486.

  3. 0.4052.

19

  1. 0.3520.

  2. The distribution of the sample mean length of human pregnancies in a sample size of 20 is normal. The mean of the sample means is 266, with a standard deviation of 3.58.

  3. 0.0465.

  4. .0040.

  5. This result is highly unusual.

  6. .9844.

21

  1. .3085.

  2. .0418.

  3. .0071.

  4. Increasing the sample size decreases the probability that the mean of the sample is more than 95 words per minute. This makes sense since it becomes increasingly unlikely the more people u have that ur mean is more than the population mean.

  5. There was a .3944 probability of this happening.

  6. 93.7.

23

  1. .5675.

  2. .7291.

  3. .8051.

  4. .8531.

  5. As your time horizon increases, the probability of earning a positive return increases..

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

8.2

11

  1. Since the sample size is less than 5% of the population size (.05N = 500) and since np(1-p) is greater than or equal to 10 (12), the distribution of p-hat is normal with a mean of 0.8 and standard deviation of .0462.

  2. .1922.

  3. .0047.

12

  1. Since the sample size is less than 5% of the population size (.05N = 1250) and since np(1-p) is greater than or equal to 10 (45.5), the distribution of p-hat is normal with a mean of .65 and standard deviation of .0337.

  2. .1867.

  3. .0375.

13

  1. Since the sample size is less than 5% of the population size (.05N = 50000) and since np(1-p) is greater than or equal to 10 (227.5), the distribution of p-hat is normal with a mean of .35 and standard deviation of .0151.

  2. .004.

  3. .0233.

14

  1. Since the sample size is less than 5% of the population size (.05N = 75000) and since np(1-p) is greater than or equal to 10 (355.656), the distribution of p-hat is normal with a mean of .42 and standard deviation of .0129.

  2. .0099.

  3. .0606.

15

  1. The response to this question is qualitative, since it is a yes or a no question.

  2. p-hat is a random variable because it can change based on the sample of the population that is questioned. Different samples can lead to different results for p-hat.

  3. The distribution of p-hat is normal with a mean of .47 and standard deviation of .0353. A sample of 200 Americans is definitely less than 5% of the entire American population, and np(1-p) is greater than or equal to 10.

  4. .1977.

  5. It would be unusual in a sample of 200 Americans in which 80 or fewer individuals could order a meal in a foreign language because the probability of this happening is very low: .0239.

16

  1. The response to this question is qualitative, since the answer requires either a yes or a no.

  2. p-hat is a random variable because it can change based on the sample that is questioned. One sample of a 100 Americans might generate a different value of p-hat than another sample of 100 Americans.

  3. The distribution of p-hat is normal with a mean of .82 and a standard deviation of .0384. A sample of 100 Americans is definitely less than 5% of the entire American population and np(1-p) is greater than or equal to 10.

  4. .2177.

  5. It would be unusual for there to be 75 or fewer individuals in a sample of 100 Americans that are satisfied with the way things are going in their lives because the probability of this happening is very low: .0344.

17

  1. The distribution of p-hat is normal with a mean of .39 and a standard deviation of .0218. The sample size of 500 is definitely less than 5% of the American population and np(1-p) is greater than or equal to 10.

  2. .3228.

  3. .3198.

  4. Yes this would be unusual, since the probability of this happening is very low: .0838.