91Ó°ÊÓ

$°C=\frac{5}{9}°F−\frac{160}{9}$

Mean, ${\mathrm{μ}}_F=77°F$

Standard deviation,${\mathrm{σ}}_F=3°F$

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,

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

Thus,

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

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,

${\mathrm{σ}}_C=\frac{5}{9}{\mathrm{σ}}_F=\frac{5}{9}\left(3\right)=\frac{15}{9}=\frac{5}{3}≈1.6667°C$

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