Question

A survey found that​ women's heights are normally distributed with mean 62.2 in. and standard deviation 3.2 in. The survey also found that​ men's heights are normally distributed with mean 67.2 in. and standard deviation 3.5 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 55 in. and a maximum of 62 in. Complete parts​ (a) and​ (b) below  

what is The percentage of men who meet the height requirement?

If the height requirements are changed to exclude only the tallest​ 50% of men and the shortest​ 5% of​ men, what are the new height​ requirements?

Answer 1

This is a normal distribution question with

Mean(u) = 67.2 Standard Deviation(o) = 3.5 Since we know that score =-

a) P(55.0 < x < 62.0)=?

The 2 – score at r = 55.0is, 55.0 – 67.2 2= 3.5 z1 = -3.4857 The 2 – score at r = 55.0is, 62.0 – 67.2 3.5 22 = -1.4857

This implies that
P(55.0 < x < 62.0) = P(-3.4857 < z < -1.4857) = P(Z < -1.4857) - P(Z < -3.4857)
P(55.0 < x < 62.0) = 0.06867924532273449 - 0.000245425572017854
P(55.0 < x < 62.0) = 0.0684
b) Given in the question
P(X > x) = 0.5
This implies that
P(Z > 0.0) = 0.5
With the help of formula for z, we can say that

Τ = μ + εσ + 1 = 67.2 + (0.0)3.5

x = 67.2
Given in the question
P(X < x) = 0.05
This implies that
P(Z < -1.645) = 0.05
With the help of formula for z, we can say that

Τ = μ + εσ + 1 = 67.2 + (-1.645)3.5

x = 61.443

New minimum height requirement is 61.443in and maximaum height requirement is 67.2
PS: you have to refer z score table to find the final probabilities.
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