Using variance properties to find the standard deviation of a sample

Question

The question is: The standard deviation of the mean mass of a sample of 2 aubergines is 20 g smaller than the standard deviation in the mass of a single aubergine. Find the standard deviation of the mass of an aubergine.

So, I chose X as the mass of an aubergine, and set up an equation:

$\sqrt{Var(X_1 + X_2)+20}=\sqrt{Var(X)}$

$Var(X_1 + X_2)+20=Var(X)$

$2Var(X)+20=Var(X)$

But then I get a negative value of $Var(X)$. What did I do wrong?

Edit: I realized that since it's a sample, it should be $Var(\frac{X_1 + X_2}{2})$. So I did make an equation like that, assuming $X_1+X_2=2X$. But then again I don't have the answer for some reason.

$\frac{1}{4}(2)Var(X)+20=Var(X)$

$\frac{1}{2}Var(X)+20=Var(X)$

$\frac{1}{2}Var(X)=20$

$Var(X)=40$

And so the standard deviation would be $\sqrt{40}$. But apparently the answer is 68.3g. Anybody know what I did wrong?

Answer 1

Your problems are that the mean mass is $\frac{X_1+X_2}2$, not $X_1+X_2$ and that the equation should be with $+20$ outside the square root:

$$\sqrt{\text{Var}\left(\frac{X_1+X_2}2\right)} +20 = \sqrt{\text{Var}(X)}$$

Answer 2

The mean is $\frac{X_1+X_2}{2}$

And $Var\left(\frac{X_1+X_2}{2} \right)=\frac14\cdot (Var(X_1)+Var(X_2))$

Since $X_1$ and $X_2$ are iid, $\frac14\cdot (Var(X_1)+Var(X_2))=\frac{Var(X_1)}2$

Thus the equation $\frac{\sigma_1}{\sqrt 2}+20=\sigma_1$

Therefore $\sigma_1=\sigma_2=20(2+\sqrt 2)\approx 68.28$

Answer 3

\begin{align} \sqrt{Var(\bar{X})} + 20 &= \sqrt{Var(X)} \\ \sqrt{Var(X)/2} + 20 &= \sqrt{Var(X)} \\ \sqrt{Var(X)} &= \frac{20}{1-\sqrt{1/2}} = \frac{20\sqrt2}{\sqrt2-1} = 20\sqrt2 (\sqrt2 + 1) \end{align}