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 29 million and a standard deviation of 10 million.

Have between 32 and 40 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 54 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 29 million and a standard deviation of 10 million.

Explanation / Answer

a) probability that next week's show will 32 and 40 million viewers=P(32<X<40)=P((32-29)/10<Z<(40-29)/10)

=P(0.3<Z<1.1) =0.8643-0.6179 =0.2464

b) P(X>26)=1-P(X<26)=1-P(Z<(40-26)/10)=1-P(Z<-0.3) =1-0.3821=0.6179

c)P(X>54)=1-P(X<54)=1-P(Z<(54-29)/10)=1-P(Z<2.5)=1-0.9938 =0.0062

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