Question

A normal distribution has a mean of u = 60 and a

standard deviation of o = 16. For each of the following

scores, indicate whether the body is to the right or

left of the score and find the proportion of the distribution

located in the body.

a. X = 64

b. X = 80

c. X = 52

d. X = 28

Answer 1

Part a)

X = 64 lies above the mean, it is at right side of the mean

X ~ N ( µ = 60 , σ = 16 )
P ( X > 64 ) = 1 - P ( X < 64 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 64 - 60 ) / 16
Z = 0.25
P ( ( X - µ ) / σ ) > ( 64 - 60 ) / 16 )
P ( Z > 0.25 )
P ( X > 64 ) = 1 - P ( Z < 0.25 )
P ( X > 64 ) = 1 - 0.5987
P ( X > 64 ) = 0.4013

Part b)

X = 80 lies above the mean, it is at right side of the mean

X ~ N ( µ = 60 , σ = 16 )
P ( X > 80 ) = 1 - P ( X < 80 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 80 - 60 ) / 16
Z = 1.25
P ( ( X - µ ) / σ ) > ( 80 - 60 ) / 16 )
P ( Z > 1.25 )
P ( X > 80 ) = 1 - P ( Z < 1.25 )
P ( X > 80 ) = 1 - 0.8944
P ( X > 80 ) = 0.1056

Part c)

X = 52 lies below the mean, it is at left side of the mean

X ~ N ( µ = 60 , σ = 16 )
P ( X < 52 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 52 - 60 ) / 16
Z = -0.5
P ( ( X - µ ) / σ ) < ( 52 - 60 ) / 16 )
P ( X < 52 ) = P ( Z < -0.5 )
P ( X < 52 ) = 0.3085

Part d)

X = 28 lies below the mean, it is at left side of the mean

X ~ N ( µ = 60 , σ = 16 )
P ( X < 28 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 28 - 60 ) / 16
Z = -2
P ( ( X - µ ) / σ ) < ( 28 - 60 ) / 16 )
P ( X < 28 ) = P ( Z < -2 )
P ( X < 28 ) = 0.0228