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Assume that z-scores are normally distributed with a mean of 0 and a standard deviation of...

Assume that z-scores are normally distributed with a mean of 0 and a standard deviation of 1. If P ( − b < z < b ) = 0.8632 P(-b

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Answer #1

Solution:

Given that,

P(-b Z b) = 0.8632

P(Z b) - P(Z -b) = 0.8632

2P(Z b) - 1 = 0.8632

2P(Z b) = 1 + 0.8632 = 1.8632

P(Z b) = 1.8632 / 2 = 0.9316

P(Z 1.49) = 0.9316

Z-score = 1.49

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