Assume that Acme tire life is normally distributed and that all both all weather

Question

Assume that Acme tire life is normally distributed and that all both all weather tires and high performance tires have a mean life of 50,000 miles. The standard deviation of tire life for all weather tires is 8192.42 miles and the standard deviation of tire life for high performance tires is 7986.46 miles.
What is the probability that a sample of 16 all weather tires will have an average tire life of more than 52,000 miles?

Explanation / Answer

a. z = (xbar - mu)/(sigma/vn) z = (52000 - 50000)/(6361.174452/v16) z = 2000 / 1590.2936 z = 1.26 p = 0.8957 BUT since we want "greater than," we need to subtract the probability from 1 p = 1 - 0.8957 = 0.1043 b. z = (xbar - mu)/(sigma/vn) z = (52000 - 50000)/(6361.174452/v64) z = 2000 /(NNN) NNN-NNNNbr/>z = 2.52 p = 0.9941 BUT since we want "greater than," we need to subtract the probability from 1 p = 1 - 0.9941 = 0.0059 c. z = (xbar - mu)/(sigma/vn) z = (40000 - 50000)/(7691.39991/v5) z = -10000 / 3439.6986 z = -2.91 p = 0.0018 d. z = (xbar - mu)/(sigma/vn) z = (40000 - 50000)/(7691.39991/v20) z = -10000 / 1719.8493 z = -5.81 p = approximately 0

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