Question

The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid is $1000 and the standard deviation is $70.

7908609301000107011401210Distribution of Prices

What is the approximate percentage of buyers who paid more than $1210?
%

What is the approximate percentage of buyers who paid between $1000 and $1210?
%

What is the approximate percentage of buyers who paid more than $1140?
%

What is the approximate percentage of buyers who paid between $860 and $1000?
%

What is the approximate percentage of buyers who paid between $930 and $1070?
%

What is the approximate percentage of buyers who paid between $1000 and $1070?
%

Answer 1

Part a)

X ~ N ( µ = 1000 , σ = 70 )
P ( X > 1210 ) = 1 - P ( X < 1210 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 1210 - 1000 ) / 70
Z = 3
P ( ( X - µ ) / σ ) > ( 1210 - 1000 ) / 70 )
P ( Z > 3 )
P ( X > 1210 ) = 1 - P ( Z < 3 )
P ( X > 1210 ) = 1 - 0.9987
P ( X > 1210 ) = 0.0013

Percentage = 0.13%

Part b)

X ~ N ( µ = 1000 , σ = 70 )
P ( 1000 < X < 1210 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 1000 - 1000 ) / 70
Z = 0
Z = ( 1210 - 1000 ) / 70
Z = 3
P ( 0 < Z < 3 )
P ( 1000 < X < 1210 ) = P ( Z < 3 ) - P ( Z < 0 )
P ( 1000 < X < 1210 ) = 0.9987 - 0.5
P ( 1000 < X < 1210 ) = 0.4987

Percentage = 49.87%

Part c)

X ~ N ( µ = 1000 , σ = 70 )
P ( X > 1140 ) = 1 - P ( X < 1140 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 1140 - 1000 ) / 70
Z = 2
P ( ( X - µ ) / σ ) > ( 1140 - 1000 ) / 70 )
P ( Z > 2 )
P ( X > 1140 ) = 1 - P ( Z < 2 )
P ( X > 1140 ) = 1 - 0.9772
P ( X > 1140 ) = 0.0228

Percentage = 2.28%

Part d)

X ~ N ( µ = 1000 , σ = 70 )
P ( 860 < X < 1000 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 860 - 1000 ) / 70
Z = -2
Z = ( 1000 - 1000 ) / 70
Z = 0
P ( -2 < Z < 0 )
P ( 860 < X < 1000 ) = P ( Z < 0 ) - P ( Z < -2 )
P ( 860 < X < 1000 ) = 0.5 - 0.0228
P ( 860 < X < 1000 ) = 0.4772

Percentage = 47.72%

Part e)

X ~ N ( µ = 1000 , σ = 70 )
P ( 930 < X < 1070 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 930 - 1000 ) / 70
Z = -1
Z = ( 1070 - 1000 ) / 70
Z = 1
P ( -1 < Z < 1 )
P ( 930 < X < 1070 ) = P ( Z < 1 ) - P ( Z < -1 )
P ( 930 < X < 1070 ) = 0.8413 - 0.1587
P ( 930 < X < 1070 ) = 0.6827

Percentage = 68.27%

Part f)

X ~ N ( µ = 1000 , σ = 70 )
P ( 1000 < X < 1070 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 1000 - 1000 ) / 70
Z = 0
Z = ( 1070 - 1000 ) / 70
Z = 1
P ( 0 < Z < 1 )
P ( 1000 < X < 1070 ) = P ( Z < 1 ) - P ( Z < 0 )
P ( 1000 < X < 1070 ) = 0.8413 - 0.5
P ( 1000 < X < 1070 ) = 0.3413

Percentage = 34.13%