What is the 90th percentile of this distribution? A machine that produce ball be

Question

What is the 90th percentile of this distribution?

A machine that produce ball bearings has initially been set so that the target average diameter of the bearings it produces is 1.3 cm. A bearing is acceptable if its diameter is within 0.016 cm of this target diameter. Suppose, however, that the machine is misaligned, and the ball bearings it produces follow a normal distribution with a mean of 1.25 cm and standard deviation of 0.05 cm.

Explanation / Answer

Given X~Normal(mean=1.25, s=0.05) So the 90th percentile of this distribution is P(X P((X-mean)/s < (c-1.25)/0.05) =0.9 --> P(Z (c-1.25)/0.05 =1.28 (check standard normal table) --> c = 1.25 + 1.28*0.05 = 1.314

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