Question

2. The heights of all adult American women are normally distributed with a mean of 63.8 inches and a standard deviation of 6 inches.  Give the standard (z) score and approximate percentile (from the tables) for women with each of the following heights:

1. 64 inches

2. 62 inches

3. 60.5 inches

Answer 1

This is a normal distribution question with

a) x = 64
P(x < 64.0)=?
The z-score at x = 64.0 is,

z = 0.0769
This implies that
P(x < 64.0) = P(z < 0.0769) = 0.5306 = 53.06 percentile
b) x = 62
P(x < 62.0)=?
The z-score at x = 62.0 is,

z = -0.6923
This implies that
P(x < 62.0) = P(z < -0.6923) = 0.2444 = 24.44 percentile
c) x = 60.5
P(x < 60.5)=?
The z-score at x = 60.5 is,

z = -1.2692
This implies that
P(x < 60.5) = P(z < -1.2692) = 0.1022 = 10.22 percentile
PS: you have to refer z score table to find the final probabilities.
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