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 sampling distribution of x bar is cosidered normal. This is because the sample size is above 30, it is 49 and so due to the central limit theorem, the datas distribution will be fairly normal as well. The sample distribution of x bar, being the standard deviation over the square root of the sample size, in this case is 2.

  2. The probablilty that x bar is greater than 83 is 0.0668.

  3. The probablilty that x bar is less than or equal to 75.8 is 0.0179.

  4. The probability that x bar is less than 85.1 but greater than 78.3 is 0.7969

17

  1. In order to use the normal model to compare distributions involving the sample mean, the distribution must be normal. You need a sample of at least 30 for the shape of the data to be considered to have a normal distribution. In this case the data only has a sample size of 12 people. The sample distribution of x bar, being the standard deviation over the square root of the sample size, in this case is 4.907.

  2. The probability that x bar is less than 67.3 is 0.7486.

  3. The probability that x bar is greater than or equal to 65.2 is 0.4052.

19

  1. The probability that a pregnancy lasts less than 260 days is 0.3520.

  2. The sample is n=20. That is below 30. The sample distribution can be calculated as the standard deviation divided by the square root of the sample size, and in this case is 3.578.

  3. The probability that a random sample of 20 lasts less than or equal to 260 days is 0.0465.

  4. The probability that a random sample of 50 lasts less than or equal to 260 days is 0.0040.

  5. It would be very unlikey for the random sample of 50 to have a mean gestational period of less than or equal to 260 days because as found in part d, there is a very small percent chance, 0.4% of that happening. You can conclude that the sample of 50 most likley came from popualtion where the mean gestation period was less than 266 days.

  6. The probability that a random sample size of 15 will have a mean gestational period within 10 days of the mean is 0.9844.

21

  1. The probability that a student will read more than 95 words per minute is 0.3085.

  2. The probabilty that a radnom sample of 12 second grade students will have a mean reading rate of greater than 95 wpm is 0.0418.

  3. The probability that a random sample of 24 second grade students will have a mean reading rate of greater than 95 wpm is 0.0071.

  4. As you increase sample size, the probability that that sample will have a mean reading of greater than 95 wpm is decreased.

  5. This data concludes that the new program is not very effective. This is because the probability that the sample will have a reading level greater than 92.8 wpm is 0.1056. This means that only 10& of a 100 student sample would see improvment, thus does not say much about the new program being better than the old.

  6. There is a 5% chance that the mean reading speed of a random sample of 20 second grade students will exceed 93.7 wpm.

23

  1. The probability that there is a positive return rate is 0.5675.

  2. The probability that within the next 12 months there will be a positive mean monthly rate of return is 0.7291

  3. The probability that within the next 24 months there will be a positive mean monthly rate of return is 0.8051

  4. The probability that within the next 36 months there will be a positive mean monthly rate of return is 0.8531

  5. As the investment time horizon increases, the probability that there will be a positive mean monthly return rate increases as well.

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

8.2

11

  1. The sample size is greater than 30 so the is essentialy has a normal districution, with a spread of 0.046.

  2. The probabilty of obtaining more than or equal to 63 individuals with the charactersitics is 0.1922

  3. The probability of obtaining less than or equal to 51 individuals with the characteristic is 0.0047.

12

  1. The sample distribution is normal. With a spread of 0.034.

  2. The probability of obtaining 136 or more with the characteristic is 0.5478

  3. The probability of obtaining 118 or less with the charactersitic is 0.0392

13

  1. The districution is approximatley normal, with a spread of 0.015.

  2. The probability of obtaining 390 or more with the characteristic is 0.0040

  3. The probability of obtaining 320 or less with the characteristic is 0.0233.

14

  1. The distribution is normal. The spread is 0.0129

  2. The probability of obtaining 657 or more with the characteristic is 0.0099

  3. The probability of obtaining 584 or fewwer with the charactersitic is 0.0606

15

  1. This response is qualitative because it is based on the two options if they can or cannot order in a foreign language

  2. p hat sample proportion is a random variable because it will vary from sample to sample. The source of the variabiliy is the individuals in the sample ability to order in a foreign language.

  3. The distribution is fairly normal and the spread is 0.035

  4. The probability the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.1977

  5. It would be unusual because the probability of it happening would be 0.0239

16

  1. This is a qualitative data because they are either satisfied or not satisfied.

  2. p hat is random varaible because it varies from sample to sample if people are satisfied or not. The source of the random variable is the indivuduals answer as to if they are satisfied or not.

  3. The distrubution is normal. The spread is 0.0384

  4. The probability that the proportion who are satisfied with the way things are going exceeds 0.85 is 0.7823.

  5. It would be unusual because the probabiliy that 75 or fewwer are satisfied with their life is 0.0344, relativley low.

17

  1. The data is approximatley normal. The spread is 0.022.

  2. The probability that in a random sample of 500 peeople and less than 38% believe marriage is obsolete is 0.3228.

  3. The probability that a random sample of 500 people and between 40% and 45% believe marriage is obsolete is 0.3198.

  4. It would be unusual because the probability that 210 or more believe marriage is obsolete is 0.0838 and that is relativley small.