What is the logic behind converting these units of standard deviation?

Question

I was solving a question from high school, and it was asking for a conversion of units of one given standard deviation (σ). But I really didn't get why it is logical just converting this way without know the other variables. These are the data:

1 hectare= $10000m^2$.

1 plot= $30000m^2$.

The measured σ of the coffee production is 90kg/PLOT

The question is:

What is the variance (V) of the production in $(Sacks/Hectare)^2$ if 1 sack is equal to 60 Kg?

The solution given is this:

σ=$\sqrt V$

Coverting V to sacks/hectare:

90KG=1,5 Sack

1 Plot = 3 hectare

σ=$\sqrt 0.5$

$V=σ^2$

$V=0,5^2$

$V=0,25(sack/hectare)^2$

But I do not understand the reason it makes sense to just convert from plot to hectare without knowing how many hectares exist and defining them (and the amount of coffee in each). For example, I could have just 3 hectares, each one producing 2 sacks of coffee, or I could have one of them producing 6 and the others 0. That would change the variance.

Answer 1

If $X$ is production in kg/plot, and $Y$ is production in sacks/hectare, then

$$Y = \frac{X}{180}$$

You demonstrated dividing the SD $\sigma_X$ by $180$ to obtain an SD $\sigma_Y$ in sacks/hectare, then squaring that to obtain the variance $\sigma^2_Y$.

Alternatively, imagine converting $n$ samples from kg/plot to sacks/hectare first, then computing the variance. The original $\sigma_X$ would be

$$\sigma_X = \sqrt{\frac{(x_1-\mu_X)^2+\cdots + (x_n-\mu_X)^2}{n}} = 90 \text{ kg/plot}$$

The required variance in (sacks/hectare)$^2$ would be

$$\begin{align*} \sigma_Y^2 &= \frac{(y_1-\mu_Y)^2+\cdots +(y_n-\mu_Y)^2}{n}\\ &= \frac{\left(\frac{x_1}{180}-\frac{\mu_X}{180}\right)^2+\cdots +\left(\frac{x_n}{180}-\frac{\mu_X}{180}\right)^2}{n}\\ &= \frac{(x_1-\mu_X)^2+\cdots +(x_n-\mu_X)^2}{180^2n}\\ &= \frac{\sigma^2_X}{180^2}\\ &= \left(\frac{\sigma_X}{180}\right)^2\\ &= 0.5^2 \text{ (sacks/hectare)}^2 \end{align*}$$

So even if you choose to measure the samples in another unit, the conversion outcome would be the same.

Answer 2

This is just a change of scale, with squaring to get the variance.

• If a random variable $X$ has mean $\mu$, standard deviation $\sigma$ and variance $\sigma^2$

• then $kX$ has has mean $k\mu$, standard deviation $|k| \sigma$ and variance $k^2 \sigma^2$.

• So with $PLOT = 3\text{ hectares}$ and $1 \text{ sack} = 60 \, kg$

• you have $1 \, kg/PLOT = \frac{1}{3 \times 60} \text{ sacks} / \text{hectare}$ and $k= \frac1{180}$

• so a standard deviation of $90\, kg/PLOT$ is equivalent to a standard deviation of $\frac12 \text{ sack} / \text{hectare}$,

• which is equivalent to a variance of $\frac14 (\text{sack} / \text{hectare})^2$.