Question

A particular manufacturing design requires a shaft with a diameter between 21.91 mm and 22.018 mm. The manufacturing process yields shafts with diameters normally​ distributed, with a mean of 22.003 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 a diameter between 21.91 mm and 22.00 mm question mark The proportion of shafts with diameter between 21.91 mm and 22.00 mm is nothing.

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

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

Answer 1

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

a) the proportion of shafts with a diameter between 21.91 mm and 22.00 mm is same as the probability that a randomly selected shaft has a diameter between 21.91 mm and 22.00 mm

ans:  the proportion of shafts with a diameter between 21.91 mm and 22.00 mm is 0.3085

b) the probability that a shaft is acceptable is same as the probability that a randomly selected shaft has a diameter between 21.91 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 the shafts.

That means the probability that a randomly selected shaft has a diameter 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, we get 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 exceed by only 1% of the shafts is 22.0170 mm