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

Question

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.)

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.

Explanation / Answer

(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)

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