Question

1. Given a normal distribution of weekly incomes where the mean is $995 and the standard deviation is $90:

1. What is the area under the normal curve between $840 and $1200?
2. What is the percent of earnings $1245 or more?
3. What is the income level at which 15% of the remaining income levels are above it?
4. What is the income level at which 60% of the remaining income levels are above it?

Answer 1

Part a)

X ~ N ( µ = 995 , σ = 90 )
P ( 840 < X < 1200 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 840 - 995 ) / 90
Z = -1.7222
Z = ( 1200 - 995 ) / 90
Z = 2.2778
P ( -1.72 < Z < 2.28 )
P ( 840 < X < 1200 ) = P ( Z < 2.28 ) - P ( Z < -1.72 )
P ( 840 < X < 1200 ) = 0.9886 - 0.0425
P ( 840 < X < 1200 ) = 0.9461

Part b)

X ~ N ( µ = 995 , σ = 90 )
P ( X >= 1245 ) = 1 - P ( X < 1245 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 1245 - 995 ) / 90
Z = 2.7778
P ( ( X - µ ) / σ ) > ( 1245 - 995 ) / 90 )
P ( Z > 2.7778 )
P ( X >= 1245 ) = 1 - P ( Z < 2.7778 )
P ( X >= 1245 ) = 1 - 0.9973
P ( X >= 1245 ) = 0.0027

percentage = 0.0027 * 10 = 0.27%

Part c)

X ~ N ( µ = 995 , σ = 90 )
P ( X > x ) = 1 - P ( X < x ) = 1 - 0.15 = 0.85
To find the value of x
Looking for the probability 0.85 in standard normal table to calculate Z score = 1.0364
Z = ( X - µ ) / σ
1.0364 = ( X - 995 ) / 90
X = 1088.276
P ( X > 1088.276 ) = 0.15

Part d)

X ~ N ( µ = 995 , σ = 90 )
P ( X > x ) = 1 - P ( X < x ) = 1 - 0.6 = 0.4
To find the value of x
Looking for the probability 0.4 in standard normal table to calculate Z score = -0.2533
Z = ( X - µ ) / σ
-0.2533 = ( X - 995 ) / 90
X = 972.203
P ( X > 972.203 ) = 0.6