Question

3. Suppose that X is normally distributed with mean 75 and standard deviation 10. Determine the following: a) P(X > 90) b) P(X> 60) e) P(65 <X < 75) d) Find the value for α that satisfies the equation P(75-α < X < 75+a) = 0.80128.

Answer 1

Part a)
X ~ N ( µ = 75 , σ = 10 )
P ( X > 90 ) = 1 - P ( X < 90 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 90 - 75 ) / 10
Z = 1.5
P ( ( X - µ ) / σ ) > ( 90 - 75 ) / 10 )
P ( Z > 1.5 )
P ( X > 90 ) = 1 - P ( Z < 1.5 )
P ( X > 90 ) = 1 - 0.9332
P ( X > 90 ) = 0.0668

Part b)
X ~ N ( µ = 75 , σ = 10 )
P ( X > 60 ) = 1 - P ( X < 60 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 60 - 75 ) / 10
Z = -1.5
P ( ( X - µ ) / σ ) > ( 60 - 75 ) / 10 )
P ( Z > -1.5 )
P ( X > 60 ) = 1 - P ( Z < -1.5 )
P ( X > 60 ) = 1 - 0.0668
P ( X > 60 ) = 0.9332

Part c)
X ~ N ( µ = 75 , σ = 10 )
P ( 65 < X < 75 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 65 - 75 ) / 10
Z = -1
Z = ( 75 - 75 ) / 10
Z = 0
P ( -1 < Z < 0 )
P ( 65 < X < 75 ) = P ( Z < 0 ) - P ( Z < -1 )
P ( 65 < X < 75 ) = 0.5 - 0.1587
P ( 65 < X < 75 ) = 0.3413

Part d )
X ~ N ( µ = 75 , σ = 10 )
P ( a < X < b ) = 0.80128
Dividing the area 0.80128 in two parts we get 0.80128/2 = 0.40064
since 0.5 area in normal curve is above and below the mean
Area below the mean is a = 0.5 - 0.40064
Area above the mean is b = 0.5 + 0.40064
Looking for the probability 0.09936 in standard normal table to calculate Z score = -1.2852
Looking for the probability 0.90064 in standard normal table to calculate Z score = 1.2852
Z = ( X - µ ) / σ
-1.2852 = ( X - 75 ) / 10
a = 62
1.2852 = ( X - 75 ) / 10
b = 88
P ( 62 < X < 88 ) = 0.80128

Lower limit = 75 - 62 = 13

Upper limit = 88 - 75 = 13

P ( 75 - a  < X < 75 + a ) = 0.80128

a = 13