Question

For questions 6 and 7, use the following information: The heights of women are normally distributed with a mean of 65.5 inches and a standard deviation of 2.5 inches.

A recent job posting on Disney's website specified that women applicants must be between 63 and 68 inches. What is the probability a woman meets this requirement? Write your answer as a decimal rounded to three places.

What height would put a woman at the 35^th percentile? Round your answer to one decimal place.

Answer 1

X ~ N ( µ = 65.5 , σ = 2.5 )
P ( 63 < X < 68 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 63 - 65.5 ) / 2.5
Z = -1
Z = ( 68 - 65.5 ) / 2.5
Z = 1
P ( -1 < Z < 1 )
P ( 63 < X < 68 ) = P ( Z < 1 ) - P ( Z < -1 )
P ( 63 < X < 68 ) = 0.8413 - 0.1587
P ( 63 < X < 68 ) = 0.683

X ~ N ( µ = 65.5 , σ = 2.5 )
P ( X < x ) = 35% = 0.35
To find the value of x
Looking for the probability 0.35 in standard normal table to calculate Z score = -0.3853
Z = ( X - µ ) / σ
-0.3853 = ( X - 65.5 ) / 2.5
X = 64.5
P ( X < 64.5 ) = 0.35