The diameter of a pipe is normally distributed with a mean of 0.4 inch and a var

Question

The diameter of a pipe is normally distributed with a mean of 0.4 inch and a variance of 0.0004. What is the probability that the diameter of a randomly selected pipe will exceed 0.44 inch? Express your answer as a percent to two decimal places.

Explanation / Answer

P ( x > 0.44 ) convert this to z .... z = ( x - mean ) / standard devation z = ( 0.44 - 0.4 ) / v 0..0004 z = 0.04 / 0.02 z = 2 P ( x > 0.44 ) = P ( z > 2 ) = 1 - P ( z < 2 ) use standard normal table to find the value of z P (z < 2 ) = 0.9772 P ( x > 0.4 4) = 1 - 0.9772 = 0.0228 is the final answer

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