The distribution of the number of viewers for the American I

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

Solution

(a) P(36<X<43) = P((36-30)/8 <(X-mean)/s <(43-30)/8 )

=P(0.75<Z<1.63) = 0.1751 (from standard normal table)

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(b)P(X>26) = P(Z>(26-30)/8)

=P(Z>-0.5) = 0.6915 (from standard normal table)

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(c) P(X>49) = P(Z>(49-30)/8)

=P(Z>2.38) =0.0087 (from standard normal table)