Need Help, Sampling Distribution of Pins

Hey I have been toiling over this homework for a while now. I think I may have it correct but I am not certain. Here is the background info for the problem.


PECO Ltd. supplies pins in bulk to a customer. The company uses an automatic lathe to produce the pins. Factors such as vibration, temperature, wear and tear affect the pins, so that the lengths of the pins made by the machine are normally distributed with a mean of 1.008 inches and a standard deviation of 0.045 inch. The company supplies the pins in large batches to a customer. The customer will take a random sample of 50 pins from the batch and compute the sample mean. If the sample mean is within the interval
1.000 +- 0.010 inch, then the customer will buy the whole batch.

I am not certain I am doing question 2 or 3 right. If I can get those right then I can do the other questions.

Here are the questions with my answers.

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

- I answered 1.000 inch, since this is the customers desired mean. It is easier and overal cheaper to adjust the mean to this value.

3. Suppose the mean cannot be adjusted, but the standard deviation can be reduced. What maximum value of the standard deviation would make 90% of the parts acceptable to the customer? (Assume the mean to be 1.008.)

- I got the z value to equal 1.29 for 90%. So,
1.010-1.008/(unknown deviation/sq root of 50) = 1.29

1.29*sq root of 50= 1.010-1.008/Unknown deviation

7.07*unknown deviation= 0.002

Unknown deviation = 0.002/7.07

Max standard deviation = .000282

I don't think I am doing question 3 right, can you tell me what I may be doing wrong? Also does the answer to problem 2 look correct?

I really appreciate any help, thank you very much.

P.S. sorry if I posted this thread twice, my browser messed up so I resubmited the thread.


TS Contributor
Number 2 is correct - often they're just looking for the "common sense" answer.

On number 3, you used the Z score formula for means, not individual parts, and the question mentions 90% of the parts, not 90% of the batches.

So, it would be z = (x - mu)/s --> 1.282 = (1.01 - 1.008)/s
solve for s --> s = 0.00156

Technically, there could also be out-of-spec parts on the lower end of the distribution as well, but the z score at 0.99 is -11.5, so the area is 0 for practical purposes.