Question

Problem 6. The net weights of Maxwell House coffee in cans produced by an assembly line follow a distribution with a mean of 11.5 ounces and a standard deviation of 0.25 ounce. Two product quality controllers, Walter Shewhart and Edwards Deming, randomly select 64 and 100 cans from the assembly line for inspection, respectively. a. (4 points) Shewhart and Deming calculate the average net weights in their samples. In general, whose average will be closer to 11.5 ounces by the law of large numbers? b. (4 points) By the central limit theorem, what are the approximate sampling distributions of the average net weights in Shewhart and Deming's samples, respectively, if they repeatedly take 64 and 100 cans randomly many times? c. (4 points) What are the probabilities that the total net weights in Shewhart and Deming's samples are less than 738 and 1154 ounces, respectively?

Answer 1

When a variable follows normal distribution then

X Nuo

z-score =

T

We therefore convert the variable value to z-score and then use the normal probability tables for the values

There are two samples :- Walter Shewhart n = 64 and Edwards Deming n = 100

a) The mean and SD given are for populations. When we choose sample we can't expect the same mean to same as population value but it can be closer. When the sample size is large, there is more accuracy that the difference between population and sample is less when compared to smaller samples. Therefore

According to law of large numbers , Deming's avg will be closer to 11.5 since it's 'n' is larger.

b)

Central limit theorem states that if the sample size is large ( n > 30) then the distribution of the means of similar sample size will approximately follow normal distribution.

X ~Nu, o/n)

$1592313286926_blob.png$ = 11.5  $1592313286935_blob.png$ = 0.25

Therefore for Shewhart

n = 64

$\bar{X}\sim N(11.5,0.03125)$

For Deming

n = 100

$\bar{Y}\sim N(11.5,0.025)$

c)

Just like for the sampling distribution of mean, we also have the sampling distribution for the sample total.

ΣΧο Ν(ημ, ση)

Therefore for Shewhart

n = 64

$\sum {X}\sim N(736,2)$

Probability that total weigt is less than 738

P( Sum X < 738) = P(Z < 1)

Ans: 0.84135

For Deming

n = 100

$\sum{Y}\sim N(1150,2.5)$

Probability that this is less than 1154

P(Sum Y < 1154) = P( Z < 1.6)

Ans: 0.9452

d)

To be within regulations the average weigth should not be less than 11.45 that is more than 11.45

We want to know that avergae is more than 11.45.

P( X > 11.45) =P( Z > -1.6)

=P( Z < 1.6)

For this we can directly calculate the probability since the mean for both is 11.5, the z-score will be negative.P(Z >-a)= P( Z < a)

		z-score	P(Z < z-score)
Shewhart	X=11.45	-1.6	0.9452
Deming	Y=11.45	-2	0.9772

e)

The above values for not violating the regulations so to get the probabilities of violating, we subtract above ones from '1'.

	P(Z < z-score)	Violating
Shewhart	0.9452	0.0548
Deming	0.9772	0.0228

Given the small sample size and higher variance for Shewhart, the probability of violating the regulations is more likely for Shewhart.