Assuming that the heights of college women are normally distributed with mean 60
ID: 3180160 • Letter: A
Question
Assuming that the heights of college women are normally distributed with mean 60 inches and standard deviation 3.2inches, answer the following questions. (Hint: Use the figure below with mean and standard deviation .)
(a) What percentage of women are taller than 60 inches?
(b) What percentage of women are shorter than 60 inches?
(c) What percentage of women are between 56.8 inches and 63.2 inches?
(d) What percentage of women are between 53.6 and 66.4 inches?
Explanation / Answer
mean = 60 , sd = 3.2
a) x = 60
By normal distribution formula,
z = ( x - mean) / sd
= ( 60 - 60) / 3.2
= 0
Now, we need to find P(Z> 0)
P(X > 60 ) = p(z >0) = 0.5 = 50%
b)
x = 60
By normal distribution formula,
z = ( x - mean) / sd
= ( 60 - 60) / 3.2
= 0
Now, we need to find P(z< 0)
P(X < 60 ) = p(z <0) = 0.5 = 50%
c)
x1 = 56.8 , x2 = 63.2
By normal distribution formula,
z = (( x1 - mean) / sd < z < (x2 - mean)/sd)
= ( (56.8 - 60) / 3.2 < z < ( 63.2 - 60)/3.2)
= ( -1 <z < 1)
Now, we need to find P( -1 < z < 1)
P(56.8 < X < 63.2 ) = p( -1 < z < 1) = 0.6827 = 68.27%
d)
x1 = 53.6 , x2 = 66.4
By normal distribution formula,
z = (( x1 - mean) / sd < z < (x2 - mean)/sd)
= ( (53.6 - 60) / 3.2 < z < ( 66.4 - 60)/3.2)
= ( -2 <z < 2)
Now, we need to find P( -2 < z < 2)
P(53.6 < X < 66.4 ) = p( -2 < z < 2) = 0.9545= 95.45%