Chapter 6.2, Problem 45E

a.

To determine

To find: The mean and standard deviation of Y

Answer to Problem 45E

The mean and standard deviation are 47.3 and 4.05.

Explanation of Solution

Given:

Population mean $\left(\mu \right)=8.5$

Population standard deviation $\left(\sigma \right)=2.25$

  $Y=\frac{9}{5}X+32$

Calculation:

Here, $Y=\frac{9}{5}X+32$

The mean and standard deviation of Y are:

  $\begin{array}{c}{\mu }_Y=\frac{9}{5}{\mu }_X+32\\ =\frac{9}{5}\left(8.5\right)+32\\ =47.3\end{array}$

  $\begin{array}{c}{\sigma }_Y=\frac{9}{5}\times {\sigma }_X\\ =\frac{9}{5}\times 2.25\\ =4.05\end{array}$

Thus, the mean and standard deviation are ${47.3}^{°}F\text{ and }{4.05}^{°}F$

b.

To determine

To find: The probability that the temperature is below ${40}^{°}F$

Answer to Problem 45E

The required probability is 0.0359.

Explanation of Solution

Calculation:

The probability that the temperature is below ${40}^{°}F$ can be calculated as:

  $\begin{array}{c}P\left(Y<40\right)=P\left(Z<\frac{40-47.3}{4.05}\right)\\ =P\left(Z<-1.80\right)\\ =0.0359\end{array}$

Thus, the required probability is 0.0359.