Question

The distribution of the number of viewers for the American Idol television show follows a normal distribution with a mean of 26 million with a standard deviation of 8 million. What is the probability next week's show will: Have between 30 and 37 million viewers

Answer 1

Probability that next week's show will have between 30 and 37 million viewers is 0.2248.

Explanation:

We are given that the distribution of the number of viewers for the American Idol television show follows a normal distribution with a mean of 26 million with a standard deviation of 8 million.

Let X = number of viewers for the American Idol television show

So, X ~ N(
`\mu=26,\sigma^(2)=8^(2)`)

Now, the z score probability distribution is given by;

Z =
`(X-\mu)/(\sigma)` ~ N(0,1)

where,
`\mu` = population mean = 26 million

`\sigma` = standard deviation = 8 million

So, probability that next week's show will have between 30 and 37 million viewers is given by = P(30 < X < 37) = P(X < 37) - P(X
`\leq` 30)

P(X < 37) = P(
`(X-\mu)/(\sigma)` <
`(37-26)/(8)` ) = P(Z < 1.38) = 0.91621

P(X
`\leq` 30) = P(
`(X-\mu)/(\sigma)`
`\leq`
`(30-26)/(8)` ) = P(Z
`\leq` 0.50) = 0.69146

Therefore, P(30 < X < 37) = 0.91621 - 0.69146 = 0.2248

Hence, probability that next week's show will have between 30 and 37 million viewers is 0.2248.