Question

Cool pool? Coach Ferguson uses a thermometer to measure the temperature

(in degrees Fahrenheit) at $20$different locations in the school swimming pool. An analysis of the data yields a mean of $77°F$and a standard deviation of $3°F$(Recall that$°C=\frac{5}{9}°F-\frac{160}{9}$

a. Find the mean temperature reading in degrees Celsius.

b. Calculate the standard deviation of the temperature readings in degrees Celsius.

Short answer

Part (a) Mean temperature reading in degrees Celsius is $25°C$

Part (b) Standard deviation,${\sigma }_C\approx 1.6667°C$

Step-by-step solution

Part (a) Step 1: Given information

$°C=\frac{5}{9}°F-\frac{160}{9}$

Mean, ${\mu }_F=77°F$

Standard deviation,${\sigma }_F=3°F$

Part (a) Step 2: Calculation

Since

$°C=\frac{5}{9}°F-\frac{160}{9}$

In order to convert every data value from $°F$to $°C$, every data value needs to be multiplied by $\frac{5}{9}$and decreased by $\frac{160}{9}$

Because the mean represents the distribution's center.

When every data value is multiplied by $5/9$then the distribution's center is also multiplied by $\frac{5}{9}$

Similarly,

When every data value is decreased by $\frac{160}{9}$then the distribution's center is likewise reduced by $\frac{160}{9}$

Then

The new mean $°C$is the mean is$°F$multiplied by 5/9 and decreased by 160/9

New mean,

$\mu C=\frac{5}{9}{\mu }_F-\frac{160}{9}=\frac{5}{9}\left(77\right)-\frac{160}{9}=\frac{385}{9}-\frac{160}{9}=\frac{225}{9}=25°C$

Thus,

The mean temperature reading in degrees Celsius is $25°C$

Part (b) Step 1: Calculation

Since

$°C=\frac{5}{9}°F-\frac{160}{9}$

In order to convert every data value from $°F$to $°C$, every data value needs to be multiplied by $\frac{5}{9}$and decreased by $\frac{160}{9}$

Because the standard deviation is a measure of variance.

When every data value is multiplied by 5/9 , then the distribution's variability is likewise multiplied by $\frac{5}{9}$

And

When each data value is reduced by $\frac{160}{9}$ the variability of the distribution remains unchanged.

Then

The new standard deviation in $°C$is the standard deviation in $°F$multiplied by $\frac{5}{9}$but not decreased by $\frac{160}{9}$

New standard deviation,

Thus,

${\sigma }_C=\frac{5}{9}{\sigma }_F=\frac{5}{9}\left(3\right)=\frac{15}{9}=\frac{5}{3}\approx 1.6667°C$

The standard deviation of the temperature readings in degrees Celsius is $1.6667°C$