Question

4. The tensile strength of a metal part is normally distributed with mean 35 pounds and standard deviation 5 pounds. Suppose 40,000 parts are produced and specifications on the part have been established as 35.0 ± 4.75 pounds. a) Find the percentage of parts that will fail to meet specifications. b) Find the number of parts that will fail to meet specifications   c) Find the tensile strength at which 10% of the parts exceed the upper specification limit

Answer 1

P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = 35 pounds

Standard deviation = 5 pounds

a) P(a part will fail to meet specifications) = P(X < 35-4.75) + P(X > 35+4.75)

= P(Z < -4.75/5) - P(Z < 4.75/5)

= P(Z < -0.95) + P(Z > 0.95)

= P(Z < -0.95) + 1 - P(Z < 0.95)

= 0.1711 + 1 - 0.8289

= 0.3422

= 34.22%

b) Number of parts that will fail to meet specifications = 40,000 x 0.3422

= 13,688

c) 17.11% of the parts exceeds upper specification limit

Let the upper specification limit be U

At U, 17.11-10 = 7.11% parts exceeds U

P(X > U) = 0.0711

P(X < U) = 0.9289

P(Z < (U - 35)/5) = 0.9289

Take Z value corresponding to U from 0.9289

(U - 35)/5 = 1.47

U = 42.35 pounds