Question

women's heights are normally distributed with mean 63.5 in and standard deviation of 2.5 in. A social organization for tall people has a requirement that women must be at least 70 in tall. what percentage of women meet that requirement?

Answer 1

Solution :

Given that ,

mean = = 63.5

standard deviation = = 2.5

P(x 70) = 1 - P(x   70)

= 1 - P[(x - ) / (70 - 63.5) /2.5]

= 1 -  P(z 2.6)  

= 1 - 0.9953

= 0.0047

0.47% percentage of women meet that requirement.