1. ## Normal Distribution

I am stuck again on Question 7, it requires a look at question 2, therefore, I have included that portion with the answers. I have also submitted my attempt at answering Question #6.

2. If the lathe can be adjusted to have the mean of the lengths to any desired value, what should it
We determined it should be adjusted to 1.0

Problem 7 is what I am having trouble with.
7. Now assume that the mean has been adjusted to the best value found in part 2 at a cost of \$80. Calculate the reduction in standard deviation necessary to have 90%, 95% and 99% of the parts
1.012 original mean

P(.98 -1.00)
.98 -1.00 z equals 1.645 for a 90% confidence interval.
1.012 - 1.00 = 0.012 = 7.29
1.645 1.645

7.29² x \$80.00 = 4,251.53

.98 -1.00 z equals 1.96 for a 95% confidence interval.
1.012 - 1.00 = 0.012 = 6.122
1.96 1.96

6.122² x 80.00 = \$2,998.31
1.96 1.96

.98 -1.00 z equals 2.57 for a 99% confidence interval.
1.012 - 1.00 = 0.012 =4.669
2.57 2.57
4.669² x 80.00 = 1,743.96

2. From your original post, the mean of the pin cutting process was adjusted to be centered at 1.00, and it has a standard deviation of .018. The customer will only purchase pins within .02.

So we need to find the std dev necessary to get 90%, 95%, and 99% of the pins within 1.0 +/- .02.

The z scores for 90%, 95%, and 99% are 1.645, 1.96, and 2.576 respectively.

z = (x - mu)/s and x = 1.02, mu = 1.00

plug in the z score for 90%, then solve for s --> that will give you the necessary standard deviation to get 90% of the pins within 1.0 +/- .02

1.645 = (1.02-1.00)/s
s = (1.02-1.00)/1.645 = .0122

for 95% and 99%, repeat the process --> you should get smaller answers for s

3. For 90% which is 1.645 = (1.02-1.00)/s
s = (1.02-1.00)/1.645 = .0122

For 95% or 1.96
1.02 - 1.00/1.96 = 0.0102

For 99% or 2.567
1.02 - 1.00/2.576 = 7.76

Then do I perform the following operation to determine the cost at \$80.00

1.22&#178; x \$80.00 = 119.07

1.02&#178; x 80.00 = \$83.23

7.76&#178; x 80.00 =4,817.41

4. (1.02-1.00)/2.576 = 0.00776

I'm not sure I follow how you're determining (mathematically) the increased cost, but I follow the general idea - the tighter the variation, the more costly it is...

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