Chapter 2.1, Problem 23E

To determine

To find: mean and standard deviation of the temperature

Answer to Problem 23E

The mean temperature is u = 77°F and the standard deviation is $\sigma ={3.6}^{o}F$ .

Explanation of Solution

Given:

Mean = ${25}^{\circ }C$

Standard deviation = $2^{\circ }C$

Calculation:

We have $\mu ={25}^{o}C$ and $\sigma =2^{o}C$

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

Conversion of Celsius into Fahrenheit

The mean temperature of the swimming pool in Fahrenheit is calculated as follows:

  $\begin{array}{l}\mu {\text{=25}}^{o}C\\ =\left(\frac{9}{5}\right)\times 25+32\left(\text{using equation(1)}\right)\\ ={77}^{o}F\end{array}$

The standard deviation, we just multiply standard deviation in degree Celsius by $\frac{9}{5}$ . Since adding 32 just shifts the distribution and does not affect the spread, so we get:

  $\begin{array}{l}\sigma {\text{=2}}^{o}C\\ =\left(\frac{9}{5}\right)\times 2\\ ={3.6}^{o}F\end{array}$

Therefore the mean temperature of the swimming pool is obtained as u = 77°F and the standard deviation as $\sigma ={3.6}^{o}F$ .

Conclusion:

Therefore, the mean temperature is u = 77°F and the standard deviation is $\sigma ={3.6}^{o}F$ .