During the winter months, the temperatures at the Starneses’ Colorado cabin can stay well below freezing $(32°For0°C)$for weeks at a time. To prevent the pipes from freezing, Mrs. Starnes sets the thermostat at $50°F.$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 can be modeled by a Normal distribution with mean $8.5°C$and standard deviation $2.25°C$. Let $Y=$the temperature in the cabin at midnight on a randomly selected night in degrees Fahrenheit (recall that$F=(9/5)C+32)$.

a. Find the mean of $Y$.

b. Calculate and interpret the standard deviation of $Y.$

c. Find the probability that the midnight temperature in the cabin is less than $40°F$.

Given :

Mrs. Starnes sets the thermostat at : 50°F.

The temperature is modeled by a Normal distribution with mean: 8.5°C Standard deviation : 2.25°C.

Y = the temperature in the cabin at midnight

We are aware that the degrees in Fahrenheit are calculated by multiplying the degrees in Celsius by $Y(°F)=\frac{9}{5}T(°C)+32.$

Then it was increased by the same amount.

The same constant constant $32$is multiplied by every data value in the $T$distribution.

When the same constant is applied to each data value, the center of the distribution is also enlarged by the same constant.

Furthermore, if every data value is multiplied by the same constant, the distribution's center is multiplied by the same constant.

The mean is the center's measurement, as we all know.

As a result, the mean is :

$$\begin{gathered} {\mu }_Y=\frac{9}{5}{\mu }_T+32 \\ =\frac{9}{5}(8.5)+32 \\ =15.3+32 \\ =\mathbf{47}\mathbf{.}\mathbf{3}\mathbf{°}\mathit{F} \end{gathered}$$

Given :

Mrs. Starnes sets the thermostat at : $50°F$.

The temperature is modeled by a Normal distribution with mean: $8.5°C$ Standard deviation : $2.25°C$.

$Y=$the temperature in the cabin at midnight

We know, $Y(°F)=\frac{9}{5}T(°C)+32$

Then, in the $T$distribution, every data value is multiplied by the same constant $32$and raised by the same constant.

The spread of the distribution is unaltered if every data value is multiplied by the same constant.

Furthermore, if every data value is multiplied by the same constant, the distribution's spread is also multiplied by the same constant.

The standard deviation is a measure of the spread, as we all know.

Thus, the standard deviation is :

$$\begin{gathered} {\sigma }_Y=\frac{9}{5}{\sigma }_T \\ =\frac{9}{5}(2.25) \\ =\mathbf{4}\mathbf{.}\mathbf{05}\mathbf{°}\mathit{F} \end{gathered}$$

Given :

Mrs. Starnes sets the thermostat at : $50°F.$

The temperature is modeled by a Normal distribution with mean: $8.5°C$ Standard deviation : $2.25°C$.

$Y=$the temperature in the cabin at midnight

Calculate the $z-$score using the formula :

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{40-47.3}{4.05} \\ =\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{80} \end{gathered}$$

To find the equivalent probability, use the normal probability table in the appendix.

For $P(z<-1.80),$look at the row that starts with $-1.8$and the column that starts with $.00$of the ordinary normal probability table.

$P(x<40)=P(z<-1.80)=\mathbf{}\mathbf{0}\mathbf{.}\mathbf{0359}\mathbf{}$