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.

<em>Let X = number of viewers for the American Idol television show</em>

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

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

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

4 0

3 years ago

Use the table of values to find the function's values. If x = 0, then f(0) = . If f(x) = 27, then x =

Afina-wow [57]

Answer:

x = 0 =

f ( 0 ) = 27

x = 27

7 0

3 years ago

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