# A particular manufacturing des

A particular manufacturing design requires a shaft with adiameter between 21.91 mm and 22.018 mm. The manufacturing processyields shafts with diameters normally distributed, with a mean of22.003 mm and a standard deviation of 0.006 mm. Complete parts (a)through (c).

a. For this process what is the proportion of shafts with adiameter between 21.91 mm and 22.00 mm question mark The proportionof shafts with diameter between 21.91 mm and 22.00 mm isnothing.

b. For this process, the probability that a shaft is acceptableis _______ (Round to four decimal places as needed)

c. For this process the diameter that will exceed by only 1% ofthe shafts is ____mm (Round to four decimal places as needed

Answer:

Let X be the diameter of a randomly selected shaft. We know thatX is normally distributed with mean and standard deviation

a) the proportion of shafts with a diameter between 21.91 mm and22.00 mm is same as the probability that a randomly selected shafthas a diameter between 21.91 mm and 22.00 mm

ans: the proportion of shafts with a diameter between21.91 mm and 22.00 mm is 0.3085

b) the probability that a shaft is acceptable is same as theprobability that a randomly selected shaft has a diameter between21.91 mm and 22.018 mm

ans: the probability that a shaft is acceptable is 0.9938

c. Let q mm be the diameter that will exceed by only 1% of theshafts.

That means the probability that a randomly selected shaft has adiameter greater than q is 0.01

We will get the z value such that

Using the standard normal tables, we can get that for z=2.33, weget P(Z<2.33)=0.99

That is

P(Z>2.33) = 0.01

Now, we can equate the z value to the z score of q

That is, P(X>22.0170) = 0.01

ans: For this process the diameter that will exceedby only 1% of the shafts is 22.0170 mm