Question

The number of viewers of American Idol has a mean of 29 million with a standard deviation of 4 million. Assume this distribution follows a normal distribution.

  

What is the probability that next week's show will:

  

(a)	Have between 31 and 38 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

  

Probability

  

(b)	Have at least 20 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

  

Probability

  

(c)	Exceed 40 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

  

Probability

Answer 1

Part a)
X ~ N ( µ = 29 , σ = 4 )
P ( 31 < X < 38 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 31 - 29 ) / 4
Z = 0.5
Z = ( 38 - 29 ) / 4
Z = 2.25
P ( 0.5 < Z < 2.25 )
P ( 31 < X < 38 ) = P ( Z < 2.25 ) - P ( Z < 0.5 )
P ( 31 < X < 38 ) = 0.9878 - 0.6915
P ( 31 < X < 38 ) = 0.2963

Part b)
X ~ N ( µ = 29 , σ = 4 )
P ( X >= 20 ) = 1 - P ( X < 20 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 20 - 29 ) / 4
Z = -2.25
P ( ( X - µ ) / σ ) > ( 20 - 29 ) / 4 )
P ( Z > -2.25 )
P ( X >= 20 ) = 1 - P ( Z < -2.25 )
P ( X >= 20 ) = 1 - 0.0122
P ( X >= 20 ) = 0.9878

Part c)
X ~ N ( µ = 29 , σ = 4 )
P ( X > 40 ) = 1 - P ( X < 40 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 40 - 29 ) / 4
Z = 2.75
P ( ( X - µ ) / σ ) > ( 40 - 29 ) / 4 )
P ( Z > 2.75 )
P ( X > 40 ) = 1 - P ( Z < 2.75 )
P ( X > 40 ) = 1 - 0.997
P ( X > 40 ) = 0.0030