Question

The distribution of the number of viewers for the American Idol television show follows a normal distribution with a mean of 23 million and a standard deviation of 6 million.

a)What is the probability next week's show will: Have between 27 and 35 million viewers? (Round your z-score computation to 2 decimal places and final answer to 4 decimal places.)

b)Have at least 15 million viewers? (Round your z-score computation to 2 decimal places and final answer to 4 decimal places.)

c)Exceed 37 million viewers? (Round your z-score computation to 2 decimal places and final answer to 4 decimal places.)

Answer 1

Solution :

Given that ,

mean = = 23

standard deviation = = 6

a)

P(27 < x <35 ) = P((27-23)/ 6) < (x - ) /  < (35-23) / 6) )

= P(0.67 < z < 2.00)

= P(z <2.00 ) - P(z <0.67 )

= 0.9772 - 0.7486

= 0.2286

Probability = 0.2286

b)

P(x 15) = 1 - P(x   15)

= 1 - P((x - ) / (15-23) / 6)

= 1 -  P(z -1.33)  

= 1 - 0.0918

= 0.9082

Probability = 0.9082

c)

P(x > 37) = 1 - P(x < 37)

= 1 - P((x - ) / < (37-23) / 6)

= 1 - P(z <2.33 )

= 1 - 0.9901   

= 0.0099

Probability = 0.0099