Question

45. Too cool at the cabin? During the winter months, the temperatures at the Stameses' Colorado cabin can stay well below freezing $\left(32°\mathrm{F}\right.$or $\left.0°\mathrm{C}\right)$for weeks at a time. To prevent the pipes from freezing, Mrs. Stames sets the thermostat at $50°\mathrm{F}$. She also buys a digital thermometer that records the indoor temperature each night at midnight. Unfortunately, the thermometer is programmed to measure the temperature in degrees Celsius. Based on several years' worth of data, the temperature $T$in the cabin at midnight on a randomly selected night follows a Normal distribution with mean $8.5°\mathrm{C}$and standard deviation$2.25°\mathrm{C}$.
(a) Let $Y=$the temperature in the cabin at midnight on a randomly selected night in degrees Fahrenheit (recall that $\mathrm{F}=(9/5)\mathrm{C}+32)$. Find the mean and standard deviation of $Y$.

(b) Find the probability that the midnight temperature in the cabin is below $40°\mathrm{F}$. Show your work.

Short answer

(a) The mean and standard deviation of $Y$is $47.3^{\circ }F$ and $4.{05}^{\circ }F$.

(b) The probability that the midnight temperature in the cabin is below${40}^{\circ }F$is $0.0359$.

Step-by-step solution

Part (a) Step 1: Given information 

The temperature in the cabin at midnight on a randomly selected night in degrees Fahrenheit will be $Y$. Recall that $\mathrm{F}=(9/5)\mathrm{C}+32)$.

Part (a) Step 2: Explanation 

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

Then the standard deviation of $Y$is,

$$\begin{gathered} =\frac{9}{5}\times 2.25 \\ =4.05 \end{gathered}$$

Part (b) Step 1: Given information 

The probability that the midnight temperature in the cabin is $40°\mathrm{F}$.

Part (b) Step 2: Explanation 

The probability that the temperature is below$40°F$ as:
$$\begin{gathered} \begin{array}{c}P(Y<40)=P\left(Z<\frac{40-47.3}{4.05}\right)\end{array} \\ =P(Z<-1.80) \\ =0.0359 \end{gathered}$$