Question

The distribution of the number of viewers for the American Idol television show follows a normal distribution with a mean of 26 million with a standard deviation of 8 million. What is the probability next week's show will: a. Have between 30 and 37 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.) b. Have at least 21 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.) c. Exceed 49 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

Answer 1

Solution :

Given that ,

mean = = 26

standard deviation = = 8

(a)

P(30 < x < 37) = P((30 - 26 / 8) < (x - ) / < (37 - 26) / 8) )

P(30 < x < 37) = P(0.5 < z < 1.375)

P(30 < x < 37) = P(z < 1.38) - P(z < 0.5)

P(30 < x < 37) = 0.9162 - 0.6915 = 0.2247

Probability = 0.2247

(b)

P(x 21) = 1 - P(x 21)

= 1 - P((x - ) / (21 - 26) / 8)

= 1 -  P(z -0.63)  

= 1 - 0.2643   

= 0.7357

P(x 21) = 0.7357

Probability = 0.7357

(c)

P(x > 49) = 1 - P(x < 49)

= 1 - P((x - ) / < (49 - 26) / 8)

= 1 - P(z < 2.88)

= 1 - 0.9980   

= 0.0020

P(x > ) = 0.0020

Probability = 0.0020