Converting mean and std deviation of degrees from Fahrenheit to Celsius

Question

Given the conversion table of these data points of temperatures in Celsius and the respective value in Fahrenheit:

\begin{array}{|c|c|c|c|} \hline C & 10 & 20 & 30 \\ \hline F & 50 & 68 & 86 \\ \hline \end{array}

To calculate the mean in Celsius:

$\bar{x_{C}} = \dfrac{10 + 20 + 30}{3} = 20$

and standard deviation:

$\sigma_{C} = \sqrt{\dfrac{(10 - 20)^2 + (20-20)^2 + (30-20)^2}{3}} \approx 8.165$

Now, you could calculate mean and standard deviation for the values in Fahrenheit using the same process:

$\bar{x_{F}} = \dfrac{50 + 68 + 86}{3} = 68$

and standard deviation:

$\sigma_{F} = \sqrt{\dfrac{(50 -68)^2 + (68 - 68)^2 + (86 - 68)^2}{3}} \approx 14.7$

But, since we already have a formula for converting Celsius to Fahrenheit:

$ F = 1.8 \cdot C + 32 $

We can obtain the mean with:

$ \bar{x_{F}} = 1.8 \cdot \bar{x_C} + 32 $
$ \hspace{1.5em} = 1.8 \cdot 20 + 32 $
$ \hspace{1.5em} = 68 $

But if we try to obtain the standard deviation:

$ \sigma_{F} = 1.8 \cdot \sigma_{C} + 32 $
$ \hspace{1.5em} = 1.8 \cdot 8.165 + 32 $
$ \hspace{1.5em} \approx 46.7 $

And that:

$\sigma_{F} = 1.8 \cdot \sigma_{C}$
$\hspace{1.5em} = 1.8 \cdot 8.165$
$\hspace{1.5em} \approx 14.7$

Which is correct.

Trying it with other data values seems to show the same result.

It seems standard deviation isn't influenced by the displacement values, for example, adding 5 to each value in the data set would yield the same standard deviation, but multiplying by 2 would increase it. But, its still confusing. Can someone explain why the formula doesn't work? Thanks

Answer 1

Show that $$\frac{1}{n} \sum_{i=1}^n (ax_i + b) = a \left(\frac{1}{n} \sum_{i=1}^n x_i\right) + b$$ to see why you can obtain the Fahrenheit sample mean by applying the conversion to the Celsius sample mean.

Similarly, with $\overline{ax+b} := \frac{1}{n} \sum_{i=1}^n (ax_i + b)$, show that $$\sqrt{\frac{1}{n} \sum_{i=1}^n ((ax_i + b) - \overline{ax+b})^2} = a \sqrt{\frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2}$$ to see why the displacement $+b$ is irrelevant.