Question

1. (20 pts) Do this one by hand. Suppose we measured the height of 5,000 men and found that the data were normally distributed with a mean of 70.0 inches and a standard deviation of 4.0 inches. Answer the questions using Table A and show your work:

What proportion of men can be expected to have heights less than 66 inches? Less than 75 inches?

What proportion of men can be expected to have heights greater than 64 inches? Greater than 73 inches?

What proportion of men can be expected to have heights between 65 and 75 inches? Between 71 and 72 inches?

What height corresponds to the 43^rd percentile? What height corresponds to the 55^th percentile? What height corresponds to the 99^th percentile? (If necessary, round the numbers, but keep at least two decimal places.)

Answer 1

Part a)

X ~ N ( µ = 70 , σ = 4 )
P ( X < 66 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 66 - 70 ) / 4
Z = -1
P ( ( X - µ ) / σ ) < ( 66 - 70 ) / 4 )
P ( X < 66 ) = P ( Z < -1 )
P ( X < 66 ) = 0.1587

X ~ N ( µ = 70 , σ = 4 )
P ( X < 75 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 75 - 70 ) / 4
Z = 1.25
P ( ( X - µ ) / σ ) < ( 75 - 70 ) / 4 )
P ( X < 75 ) = P ( Z < 1.25 )
P ( X < 75 ) = 0.8944

Part b )

X ~ N ( µ = 70 , σ = 4 )
P ( X > 64 ) = 1 - P ( X < 64 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 64 - 70 ) / 4
Z = -1.5
P ( ( X - µ ) / σ ) > ( 64 - 70 ) / 4 )
P ( Z > -1.5 )
P ( X > 64 ) = 1 - P ( Z < -1.5 )
P ( X > 64 ) = 1 - 0.0668
P ( X > 64 ) = 0.9332

X ~ N ( µ = 70 , σ = 4 )
P ( X > 73 ) = 1 - P ( X < 73 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 73 - 70 ) / 4
Z = 0.75
P ( ( X - µ ) / σ ) > ( 73 - 70 ) / 4 )
P ( Z > 0.75 )
P ( X > 73 ) = 1 - P ( Z < 0.75 )
P ( X > 73 ) = 1 - 0.7734
P ( X > 73 ) = 0.2266

Part c)

X ~ N ( µ = 70 , σ = 4 )
P ( 65 < X < 75 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 65 - 70 ) / 4
Z = -1.25
Z = ( 75 - 70 ) / 4
Z = 1.25
P ( -1.25 < Z < 1.25 )
P ( 65 < X < 75 ) = P ( Z < 1.25 ) - P ( Z < -1.25 )
P ( 65 < X < 75 ) = 0.8944 - 0.1056
P ( 65 < X < 75 ) = 0.7887

X ~ N ( µ = 70 , σ = 4 )
P ( 71 < X < 72 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 71 - 70 ) / 4
Z = 0.25
Z = ( 72 - 70 ) / 4
Z = 0.5
P ( 0.25 < Z < 0.5 )
P ( 71 < X < 72 ) = P ( Z < 0.5 ) - P ( Z < 0.25 )
P ( 71 < X < 72 ) = 0.6915 - 0.5987
P ( 71 < X < 72 ) = 0.0928

Part d)

X ~ N ( µ = 70 , σ = 4 )
P ( X < ? ) = 43% = 0.43
Looking for the probability 0.43 in standard normal table to calculate critical value Z = -0.18
Z = ( X - µ ) / σ
-0.18 = ( X - 70 ) / 4
X = 69.28
P ( X < 69.28 ) = 0.43

X ~ N ( µ = 70 , σ = 4 )
P ( X < ? ) = 55% = 0.55
Looking for the probability 0.55 in standard normal table to calculate critical value Z = 0.13
Z = ( X - µ ) / σ
0.13 = ( X - 70 ) / 4
X = 70.52
P ( X < 70.52 ) = 0.55

Part e)

X ~ N ( µ = 70 , σ = 4 )
P ( X < ? ) = 99% = 0.99
Looking for the probability 0.99 in standard normal table to calculate critical value Z = 2.33
Z = ( X - µ ) / σ
2.33 = ( X - 70 ) / 4
X = 79.32
P ( X < 79.32 ) = 0.99