Question

The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid is $1200 and the standard deviation is $135.

What is the approximate percentage of buyers who paid between $1065 and $1335? ____%

What is the approximate percentage of buyers who paid between $930 and $1200? ___%

What is the approximate percentage of buyers who paid less than $795?____%

What is the approximate percentage of buyers who paid more than $1470? ____%

What is the approximate percentage of buyers who paid between $1065 and $1200? ____%

What is the approximate percentage of buyers who paid between $1200 and $1605? ____%

Answer 1

Part a)
X ~ N ( µ = 1200 , σ = 135 )
P ( 1065 < X < 1335 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 1065 - 1200 ) / 135
Z = -1
Z = ( 1335 - 1200 ) / 135
Z = 1
P ( -1 < Z < 1 )
P ( 1065 < X < 1335 ) = P ( Z < 1 ) - P ( Z < -1 )
P ( 1065 < X < 1335 ) = 0.8413 - 0.1587
P ( 1065 < X < 1335 ) = 0.6827
Percentage = 0.6827 * 100 = 68.27%

Part b)
X ~ N ( µ = 1200 , σ = 135 )
P ( 930 < X < 1200 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 930 - 1200 ) / 135
Z = -2
Z = ( 1200 - 1200 ) / 135
Z = 0
P ( -2 < Z < 0 )
P ( 930 < X < 1200 ) = P ( Z < 0 ) - P ( Z < -2 )
P ( 930 < X < 1200 ) = 0.5 - 0.0228
P ( 930 < X < 1200 ) = 0.4772
Percentage = 0.4772 * 100 = 47.72%

Part c)
X ~ N ( µ = 1200 , σ = 135 )
P ( X < 795 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 795 - 1200 ) / 135
Z = -3
P ( ( X - µ ) / σ ) < ( 795 - 1200 ) / 135 )
P ( X < 795 ) = P ( Z < -3 )
P ( X < 795 ) = 0.0013
Percentage = 0.0013 * 100 = 0.13%

Part d)
X ~ N ( µ = 1200 , σ = 135 )
P ( X > 1470 ) = 1 - P ( X < 1470 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 1470 - 1200 ) / 135
Z = 2
P ( ( X - µ ) / σ ) > ( 1470 - 1200 ) / 135 )
P ( Z > 2 )
P ( X > 1470 ) = 1 - P ( Z < 2 )
P ( X > 1470 ) = 1 - 0.9772
P ( X > 1470 ) = 0.0228
Percentage = 0.0228 * 100 = 2.28%

Part e)
X ~ N ( µ = 1200 , σ = 135 )
P ( 1065 < X < 1200 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 1065 - 1200 ) / 135
Z = -1
Z = ( 1200 - 1200 ) / 135
Z = 0
P ( -1 < Z < 0 )
P ( 1065 < X < 1200 ) = P ( Z < 0 ) - P ( Z < -1 )
P ( 1065 < X < 1200 ) = 0.5 - 0.1587
P ( 1065 < X < 1200 ) = 0.3413
Percentage = 0.3413 * 100 = 34.13%

Part f)
X ~ N ( µ = 1200 , σ = 135 )
P ( 1200 < X < 1605 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 1200 - 1200 ) / 135
Z = 0
Z = ( 1605 - 1200 ) / 135
Z = 3
P ( 0 < Z < 3 )
P ( 1200 < X < 1605 ) = P ( Z < 3 ) - P ( Z < 0 )
P ( 1200 < X < 1605 ) = 0.9987 - 0.5
P ( 1200 < X < 1605 ) = 0.4987
Percentage = 0.4987 * 100 = 49.87%