3 The distribution of the number of viewers for the American Idol television sho

Question

3 The distribution of the number of viewers for the American Idol television show follows a normal distribution with a mean of 31 million with a standard deviation of 10 million. What is the probability next week's show will a. Have between 35 and 42 million viewers? (Round your z-score computation to 2 decimal places and final answer to 4 decimal places.) Probability- b. Have at least 23 million viewers? (Round yourz-score computation to 2 decimal places and final answer to 4 decimal places Probability c. Exceed 50 million viewers? (Round your z-score computation to 2 decimal places and final answer to 4 decimal places.) Probability

Explanation / Answer

= 31 = 10

Z = (X - ) /

a. P(35 < X < 42)

= P(Z < ((42 - 31) / 10)) - P(Z < ((35 - 31) / 10))

= P(Z < 1.1) - P(Z < 0.4)

= 0.8643 - 0.6554

= 0.2089.

b. P(X >= 23)

= P(Z >= ((23 - 31) / 10))

= P(Z > = -0.8)

= 0.7881.

c. P(X >= 50)

= P(Z >= ((50 - 31) / 10))

= P(Z > = 1.9)

= 0.0287.

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