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

Answer:

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

Step-by-step 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 = $\frac{X-\mu}{\sigma}$ ~ 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( $\frac{X-\mu}{\sigma}$ < $\frac{37-26}{8}$ ) = P(Z < 1.38) = 0.91621

    P(X $\leq$ 30) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{30-26}{8}$ ) = 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.