The distribution of the number of viewers for the American Idol television show

Question

The distribution of the number of viewers for the American Idol television show follows a normal distribution with a mean of 30 million with a standard deviation of 8 million.

What is the probability next week's show will:

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

Have at least 26 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

Exceed 49 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

The distribution of the number of viewers for the American Idol television show follows a normal distribution with a mean of 30 million with a standard deviation of 8 million.

What is the probability next week's show will:

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

Have at least 26 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

Exceed 49 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

Explanation / Answer

= 30, = 8

(a) z1 = (x1 - u)/ = (36 - 30)/8 = 0.75 and z2 = (x2 - )/ = (43 - 30)/8 = 1.63

P(36 < x < 43) = P(0.75 < z < 1.63) = 0.1751

(b) z = (x - )/ = (26 - 30)/8 = -0.50

P(x > 26) = P(z > -0.50) = 0.6915

(c) z = (x - )/ = (49 - 30)/8 = 2.38

P(x > 49) = P(z > 2.38) = 0.0087.

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