6. The distribution of the weight of a package of M&M plain candies produced by

Question

6. The distribution of the weight of a package of M&M plain candies produced by an assembly line is supposed to have population mean 16 oz with standard deviation 0.5 oz. Two quality control colleagues, Fisher and Pearson randomly take 64 and 100 packages from the line for weight inspection.

a. Fisher and Pearson calculate the average weight of their samples. In general, whose average would be closer to 16 oz by the law of large number?

Pearson’s average

b. By central limit theorem, what are the approximated sampling distributions of average weights of Fisher and Pearson’s samples if they randomly take 64 and 100 packages many times?
Fisher: N(16, 0.0625); Pearson: N(16, 0.05)

c. What are the probabilities that the average weights from Fisher and Pearson’s samples are less than 16 oz?
0.5, 0.5

d. If the regulation requires the average weight of package to be at least 16.125 oz, what are the probabilities that the average weights from Fisher and Pearson’s samples are below the regulation standard?

Fisher: 0.9772

Pearson: 0.9938

e. Could the average weights from Fisher and Pearson’s samples be above 16.125 oz? If yes, calculate both probabilities.
Yes. Fisher: 1- 0.9772 = 0.0228, Pearson: 1- 0.9938 = 0.0062

Please show work to get the correct answers in listed in BOLD. Thank you for the help!

Explanation / Answer

(a)

Pearson's package would be more closer to 16 oz because his sample size is larger, and hence the standard error of the mean would be lesser for his sampling distribution.

(b)

For Fisher:

Mean, m = Population mean = 16

Standard error, SE = S/n0.5 = 0.5/640.5 = 0.0625

For Pearson:

Mean, m = Population mean = 16

Standard error, SE = S/n0.5 = 0.5/1000.5 = 0.05

(c)

Both probabilities are equal to 0.5, because at X = 16, the corresponding z-score for both is:

z = (X-16)/SE = 0/SE = 0

At z = 0, p-value = 0.5

(d)

For Fisher:

z = (X-16)/SE = (16.125-16)/0.0625 = 2

At z = 2, p-value = 0.97725

For Pearson:

z = (X-16)/SE = (16.125-16)/0.05 = 2.5

At z = 2, p-value = 0.99379

(e)

This part is simple, just subtract the above calculated probabilities from 1 to get the corresponding anwers.

HOPE THIS HELPS !

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