A particular manufacturing design requires a shaft with a diameter between 17.90

Question

A particular manufacturing design requires a shaft with a diameter between 17.90 mm and 18.010 mm. The manufacturing process yields shafts with diameters normally distributed, with a mean of 18.003 mm and a standard deviation of 0.004 mm.Complete parts (a) through (c). a. For this process what is the proportion of shafts with a diameter between 17.90 mm and 18.00 mm ? The proportion of shafts with diameter between 17.90 mm and 18.00 mm is 0.2266 (Round to four decimal places as needed.) b. For this process what is the probability that a shaft is acceptable?

Explanation / Answer

Mean = 18.003

Standard deviation = 0.004

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

a) Proportion of shafts with a diameter between 17.90 mm and 18.00 mm = P(X < 18.00) - P(X < 17.90)

= P(Z < (18.00 - 18.003)/0.004) - P(Z < (17.90 - 18.003)/0.004)

= P(Z < -0.75) - P(Z < -25.75)

= 0.2266 - 0

= 0.2266

b) P(A shaft is acceptable)

= P(X < 18.01) - P(X < 17.90)

= P(Z < (18.01 - 18.003)/0.004) - P(Z < (17.90 - 18.003)/0.004)

= P(Z < 1.75) - P(Z < -25.75)

= 0.9599

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