Question

The distribution of the number of viewers for the American Idol television broadcasts follows the normal distribution with a mean of 29 million and a standard deviation of 4 million.

  

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

(a)

= 29

= 4

To find P(31<X<38):

Case1: For X from mid value to 31:

Z = (31- 29)/4 = 0.50

Table of Area Under Standard Normal Curve gives area = 0.1915

Case2: For X from mid value to 38:

Z = (38- 29)/4 = 2.25

Table of Area Under Standard Normal Curve gives area = 0.4878

So,

P(31<X<38) = 0.4878 - 0.1915 = 0.2963

So,

Answer is:

0.2963

Probability

0.2963

(b)

To find P(X20):

Z = (20 - 29)/4= - 2.25

Table of Area Under Standard Normal Curve gives area = 0.4878

So,

P(X20) = 0.5 + 0.4878 = 0.9878

So,

Answer is:

0.9878

Probability

0.9878

(c) To find P(X>40):
Z = (40 - 29)/4 = 2.75

Table of Area Under Standard Normal Curve gives area = 0.4970

So,
P(X>40) = 0.5 - 0.4970 = 0.0030

So,

Answer is:

0.0030

Probability

0.0030