Question

Assuming that the heights of college women are normally distributed with mean 62 inches and standard deviation 2.6 inches, answer the following questions. (Hint: Use the figure below with mean and standard deviation o.) Area Under a Normal Curve 34% 34% 13.5% 2.35% (a) What percentage of women are taller than 62 inches? 50 % (b) What percentage of women are shorter than 62 inches? (c) What percentage of women are between 59.4 inches and 64.6 inches? (d) What percentage of women are between 56.8 and 67.2 inches?

Answer 1

Given that

$\mu$ = 62

$\sigma$ = 2.6

a)the percentage women are between 59.4 and 64.6

We know that z = x-$\mu$/$\sigma$

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.