Human heights are known to be normally distributed Men have

Human heights are known to be normally distributed. Men have a mean height of 70 inches and females a mean height of 64 inches. Both have a population standard deviation of 3 inches.

(a) Find the 1^st Quartile (Q1) of the female height distribution

(b) Find the height of a female in the 90^th percentile

(c) What is the probability that a randomly selected female is above 70 inches height?

Assume a random sample of 100 males is selected.

(d) What is the standard deviation of the sample mean (also known as the standard error of the mean)?

(e) What is the probability that the mean height of samples is less than 68 inches tall?
HINT: Since we know the population standard deviation, you will always use the Normal Distribution

if possible can you show work so i understand this? thanks

Solution

Female Data
Mean ( u ) =64
Standard Deviation ( sd )=3
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P ( Z < x ) = 0.25
Value of z to the cumulative probability of 0.25 from normal table is -0.674
P( x-u/s.d < x - 64/3 ) = 0.25
That is, ( x - 64/3 ) = -0.67
--> x = -0.67 * 3 + 64 = 61.978                  
b)
P ( Z < x ) = 0.9
Value of z to the cumulative probability of 0.9 from normal table is 1.282
P( x-u/s.d < x - 64/3 ) = 0.9
That is, ( x - 64/3 ) = 1.28
--> x = 1.28 * 3 + 64 = 67.846                  

c)
P(X > 70) = (70-64)/3
= 6/3 = 2
= P ( Z >2) From Standard Normal Table
= 0.0228                  
d)
Number ( n ) = 100
Standard Deviation ( sd )= (sd/Sqrt(n) = 3/ Sqrt ( 100 ) = 0.3

e)
P(X < 68) = (68-64)/3/ Sqrt ( 100 )
= 4/0.3= 13.3333
= P ( Z <13.3333) From Standard NOrmal Table
= 1