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 21 million and a standard deviation of 6 million What is the probability that next week's show will: Have between 27 and 30 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.) Have at least 18 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.) Exceed 36 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

Explanation / Answer

a) From information given, Xbar=21, s=6. Find z scores corresponding to Xi=27 and 30.

z=(Xi-Xbar)/s; z1=(27-21)/6=1 and z2=(30-21)/6=1.5

Find areas corresponding to z scores and subtract the smaller from the larger to get the required probbaility.

P(z2)=0.4332, P(z1)=0.3413, P(27<Xi<30): 0.4332-0.3413=0.0919

b) Find z score corresponding to Xi=18

z=(18-21)/6=-0.5

P(Xi>=18)=0.3085

c) Find z csore corresponding to Xi=36

z=(36-21)/6=2.5

P(Xi>36)=0.4938

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