Question

A particular manufacturing design requires a shaft with a diameter between 20.89 mm and 21.015 mm. The manufacturing process yields shafts with diameters normally distributed with a mean of 21.004 mm and a standard deviation of 0.005 mm. Complete (a) through (c)
a. The proportion of shafts with a diameter between 20.891 mm and 21.00 mm is_____
(Round to four decimal places as needed)

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

Answer)

As the data is normally distributed we can use standard normal z table to estimate the answers

Z = (x-mean)/s.d

Given mean = 21.004

S.d = 0.005

A)

P(20.891<x<21) = P(x<21) - P(x<20.891)

P(x<21)

Z = (21 - 21.004)/0.005 = -0.8

From z table, P(z<-0.8) = 0.2119

P(x<20.891)

Z = -22.6

From z table, P(z<-22.6) = 0

Required proportion is 0.2119 - 0 = 0.2119

B)

P(20.89 < x < 21.015)

P(x<21.015)

Z = 2.2

From z table, P(z<2.2) = 0.9861

P(x<20.89)

Z = -22.8

From z table, P(z<-22.8) = 0

Required porportion is = 0.9861 - 0 = 0.9861

C)

From z table, P(z<-2.33) = 1%

-2.33 = (x - 21.004)/0.005

X = 20.99235