Normal Distribution

[msdahlet]

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
be adjusted to? Why?
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
My Answer
1.012 original mean
1.00 adjusted 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

[JohnM]

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

[msdahlet]

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

1.22² x $80.00 = 119.07

1.02² x 80.00 = $83.23

7.76² x 80.00 =4,817.41

[JohnM]

(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...