Question
The heights of American women between the ages of 18 and 24 are approximately normally distributed. The mean is 64.1 inches, and the standard deviation is 2.5 inches.
What percentage of such women are over 5 feet 8 inches tall?
z = (68 - 64.1)/2.5 = + 1.56
= P(z > 1.56) (100)
= 0.0594 (100)
= 5.9%
But 5.9% is marked as wrong still. Any ideas?
What percentage of such women are over 5 feet 8 inches tall?
z = (68 - 64.1)/2.5 = + 1.56
= P(z > 1.56) (100)
= 0.0594 (100)
= 5.9%
But 5.9% is marked as wrong still. Any ideas?
Answers
Steve
I get 5.94%, but you are correct. Bogus answer key.
Anonymous
bam
Shena
wrong answer. I still don't know how to solve but per my key answer it is 5.5%