Cool pool? Coach Ferguson uses a thermometer to measure the temperature (in degrees Celsius) at $20$different locations in the school swimming pool.

An analysis of the data yields a mean of$25°C$and a standard deviation of $2°C$. Find the mean and standard deviation of the temperature readings in degrees Fahrenheit (recall that $°F=(9/5)°C+32)$

Mean =$25°C$

Standard deviation =$2°C$

The sample standard deviation is measured in the same units as the original data.

We have μ = $25°C$and $\sigma =2°C$

Also we know that $°F=(\frac{9}{5})°C+32......\left(1\right)$

Temperature conversion from Celsius to Fahrenheit

The following formula is used to compute the pool's average temperature in Fahrenheit:$$\begin{gathered} \mu =25°C \\ =(\frac{9}{5})\times 25+32(u\sin gequation(1\left)\right) \\ =77°F \end{gathered}$$

We just multiply the standard deviation in degrees Celsius by $\frac{9}{5}$ to get the standard deviation. Because the addition of$32$ only changes the distribution and has no effect on the spread, we get:$$\begin{gathered} \sigma =2°C \\ =\left(\frac{9}{5}\right)\times 2 \\ =3.6°F \end{gathered}$$

As a result, the swimming pool's mean temperature is $u=77°F$, and the standard deviation is $\sigma =3.6°F$

Therefore, the mean temperature is $u=77°F$ and the standard deviation is $\sigma =3.6°F$