Given that
= 62
= 2.6
a)the percentage women are between 59.4 and 64.6
We know that z = x-/
p(59.4 < x <64.6) = p(z > 59.4-62/2.6) + p(z > 64.6-62/2.6)
= p(z > -1.00) + p( z > 1.00)
= 0.3413 + 0.3413
p(59.4 < x < 64.6) = 0.6826
So the 68.26% women between 59.4 and 64.6
b) the percentage of women are between 56.8 and 67.2.
p(56.8< z < 67.2) =p(z > 56.8-62/2.6) + p(z > 67.2-62/2.6)
= p(z > - 2.00) + p(z > 2.00)
= 0.4772 + 0.4772
p(56.8 < z < 67.2) = 0.9544
Therefore 95.44% women are between 56.8 and 67.2.
Assuming that the heights of college women are normally distributed with mean 62 inches and standard deviation 2.6...
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