Question

The distribution of the number of viewers for the American Idol television show follows a normal distribution with a mean of 32 million with a standard deviation of 4 million. What is the probability next week's show will:

Have between 36 and 43 million viewers? (Round your z-score computation to 2 decimal places and final answer to 4 decimal places.)

Have at least 29 million viewers? (Round your z-score computation to 2 decimal places and final answer to 4 decimal places.)

Exceed 43 million viewers? (Round your z-score computation to 2 decimal places and final answer to 4 decimal places.)

Answer 1

Solution :

Given that ,

mean = = 32

standard deviation = = 4

P(36 < x < 43) = P[(36 - 32) / 4) < (x - ) /  < (43 - 32) / 4 ) ]

= P(1.00 < z < 2.75)

= P(z < 2.75 ) - P(z < 1.00)

Using z table,

= 0.9970 - 0.8413

= 0.1557

P(x 29) = 1 - P(x   29 )

= 1 - P[(x - ) / (29 - 32) / 4]

= 1 -  P(z - 0.75)

= 1 - 0.2266

= 0.7734

P(x > 43) = 1 - p( x< 43)

=1- p P[(x - ) / < (43 - 32) / 4 ]

=1- P(z < 2.75)

= 1 - 0.9970

= 0.0030