2) Suppose that the inspector selects 200 random samples, each consisting of 5 sheets. The inspector examines each sheet and determines the number of flaws. The outcomes of this process can be recorded in a table consisting of 5 rows (sample size) and 200 columns (number of random samples).

Obtain the average number of flaws for each sample consisting of 5 sheets. Make sure that all 200 columns are included in the panel of the "Numerical Summaries" dialog box.

(a) Obtain a frequency histogram of the 200 averages with the bins starting at zero, ending at two and using a width of 0.2. (Note: R assumes that the right endpoint of each interval is included. Your histogram should include the left endpoints. In addition, you may determine bins in R by using breaks = seq(0,2,0.2) in R.) Paste the histogram into your report. The format of the histogram should be the same as the format of the histogram (labels at the axes, titles).

(b) Refer to the histogram obtained in part (a). Do the data appear to be normally distributed? Compare the distribution of the averages to the distribution of the individual observations of the center, spread, and degree of skewness.

(c) Obtain the mean, standard deviation, and standard error of the 200 means. Paste the summaries into your report. What does the standard deviation mean here? Compare the values with the mean and the standard deviation of the sampling distribution of the sample mean predicted by the theory of sampling distributions.

(d) Count the number of boxes of five sheets, with all five sheets in perfect condition. Compare the corresponding relative frequency with the percentage obtained in Question 1, part (c).

3) Now, suppose that the inspector selects 200 random samples, each consisting of 50 sheets this time. Again, the inspector examines each sheet and determines the number of flaws. The outcomes of this process can be recorded in a table consisting of 50 rows (sample size) and 200 columns (number of random samples).

Now obtain the average number of flaws for each sample consisting of 50 observations. Make sure all 200 columns are included in the panel of the "Numerical Summaries" dialog box.

(a) Obtain a frequency histogram of the 200 averages with the bins starting at zero, ending at one and using a width of 0.1. (Note: R assumes that the right endpoint of each interval is included. Your histogram should include the left endpoints. In addition, you may determine bins in R by using breaks = seq(0,1,0.1) in R.) Paste the histogram into your report. The format of the histogram should have proper tittle x and y axis labels.

(b) Repeat part (b) of question 2. Moreover, compare the histogram with the histogram obtained in Question 2, part (a).

(c) Repeat part (c) of question 2. Moreover, compare the values with both the predicted mean and standard deviation of a sampling distribution of a sample mean. What do you conclude?

(d) Recall that the number of flaws in a randomly selected glass sheet follows a Poisson distribution with a mean of = 0.5 and a standard deviation of = (0.5)1/2 0.7071. In addition, note that based on the Central Limit Theorem (CLT), the average number of flaws in a randomly selected box approximately follows a normal distribution if the sample size (n) is large enough (n 30). Use CLT to find the approximate probability that the average number of flaws in a randomly selected box of 50 sheets is greater than 0.70. You may select the Distributions option in the R Commander menu and then the "Normal distribution" among the "Continuous distributions" options to find the probability. Compare the probability with the observed percentage of the 50 boxes with the average number of flaws > 0.70 in the data table. Are the two values consistent?

Q2 and Q3 data will be provided below at the comment section.