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 32 million with a standard deviation of 4 million.

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

Exceed 43 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 32 million with a standard deviation of 4 million.

Explanation / Answer

a)

P(36 <X< 43)

Z = (X - 32)/4

P(1 < Z< 2.75)

= 0.1557

b)

P(X > 29)

=P(Z> -0.75)

= 0.7734

c)

P(X > 43) =

P (Z>2.75)=0.003

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