Question

A particular manufacturing design requires a shaft with a diameter of 22.000 ​mm, but shafts with diameters between 21.989 mm and 22.011 mm are acceptable. The manufacturing process yields shafts with diameters normally​ distributed, with a mean of 22.003 mm and a standard deviation of 0.005 mm. Complete parts​ (a) through​ (d) below.

a. For this​ process, what is the proportion of shafts with a diameter between 21.989 mm and 22.000 mm​?

The proportion of shafts with diameter between 21.989 mm and 22.000 mm is 0.2717. ​(Round to four decimal places as​ needed.) (solved)

b. For this​ process, what is the probability that a shaft is​ acceptable?

The probability that a shaft is acceptable is 0.9426. ​(Round to four decimal places as​ needed.) (solved)

c. For this​ process, what is the diameter that will be exceeded by only 5​% of the​ shafts?

Answer 1

Solution:-

a) The proportion of shafts with diameter between 21.989 mm and 22.000 mm is 0.2717.

x₁ = 21.989

x₂ = 22.000

By applying normal distruibution:-

z₁ = - 2.80

z₂ = - 0.60

P( -2.80 < z < - 0.60) = P(z > -2.80) - P(z > - 0.60)

P( -2.80 < z < - 0.60) = 0.9974 - 0.7257

P( -2.80 < z < - 0.60) = 0.2717

b) The probability that a shaft is​ acceptable is 0.9426.

x₁ = 21.989

x₂ = 22.011

By applying normal distruibution:-

z₁ = - 2.80

z₂ = 1.60

P( -2.80 < z < 1.60) = P(z > -2.80) - P(z > 1.60)

P( -2.80 < z < 1.60) = 0.9974 - 0.0548

P( -2.80 < z < 1.60) = 0.9426

c) The dimeter that will be exceeded by only 5% of the shafts is 22.01123.

p-value for the top 5% = 0.95

z-score for the p-value = 1.645

By applying normal distruibution:-

x = 22.01123