How to find SD given distributions of SD's and means?

Question

A list of transactions contains 100 numbers: 60 gains and 40 losses. The gains are positive numbers and the losses are negative numbers. The units are thousands of dollars. For the 60 gains, the average is 18 and the SD is 7.5. For the 40 losses the average is – 20 and the SD is 9.2

a) Find the SD of the 100 transactions. (Ans: 20.35)

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Tried answering this for a friend as a math stats major (lol), but we can't seem to get the answer. My background is finding variance in terms of finding $E(X^2) - \mu^2$ and then taking square root of this to get SD (I'm aware were not given expectations in this prob) but she's a business major and tried subtracting the SD from the average and then using root mean squared. We're both getting answer close to 20.35, but not exactly 20.35.

Answer 1

We have 100 observation $x_1,\dots,x_100$ where the losses are $x_i$ with $i=1,\dots, 60$ and the gains $x_i$ with $i=61,\dots,100$.

The standard deviation is computed in the following way: $$\sigma^2=\frac{1}{100}\sum_{i=1}^{100}\left(x_i-\frac{1}{100}\sum_{k=1}^{100}x_i\right)^2.$$

We know the following quantities $$\mu_\mbox{gain}=\frac{1}{60}\sum_{i=1}^{60}x_i,\quad \mu_\mbox{loss}=\frac{1}{40}\sum_{i=61}^{100}x_i,\quad\sigma_\mbox{gain},\quad, \sigma_\mbox{loss}$$

The SD can also be computed in the following way \begin{eqnarray} \sigma^2&=&\frac{1}{100}\sum_{i=1}^{100}\left(x_i\right)^2+\left(\frac{1}{100}\sum_{k=1}^{100}x_i\right)^2 \\ &=&\frac{1}{100}\sum_{i=1}^{100}\left(x_i\right)^2+\frac{1}{100^2}\left(60\mu_\mbox{gain}+40\mu_\mbox{loss}\right)^2 \\ &=&\frac{1}{100}\sum_{i=1}^{100}\left(x_i\right)^2+\left(\frac{60\mu_\mbox{gain}}{100}\right)^2+\left(\frac{40\mu_\mbox{loss}}{100}\right)^2+\frac{2*60*40\mu_\mbox{gain}\mu_\mbox{loss}}{100^2} \end{eqnarray} But you also know the quadratic sum as $\sigma_\mbox{gain}^2=\frac{1}{60}\sum_{i=1}^{60}x_i^2+\mu_\mbox{gain}^2$ and similarly for $\sigma_\mbox{loss}^2$. Hence all quantities in the above equation are known.

Answer 2

I got answer 20.3511 following the approach spelled out below.

Let $X$ be the total sample, $X_1$ the sample consisting of gains, and $X_2$ the sample consisting of losses.

Then, $$ E[X^2] = Pr(x \in X_1)E[X^2|x \in X_1] + Pr(x \in X_2)E[X^2|x \in X_2] \\ = \frac{6}{10} E[X_1^2] + \frac{4}{10} E[X_2^2] $$ Similarly, $$E[X] = Pr(x \in X_1)E[X|x \in X_1] + Pr(x \in X_2)E[X|x \in X_2] \\ = \frac{6}{10} E[X_1] + \frac{4}{10} E[X_2] $$

You can compute $E[X]$ immediately. To find $E[X^2]$ you need to find $E[X_1^2]$ and $E[X_2^2]$. But you simply follow the formula $SD_i^2 = E[X_i^2] - \mu_i^2$, for i =1,2. Once you obtained these two objects you can obtain E[X^2]. And that point you have everything needed to compute the variance, and hence the standard deviation.

I find: $$ E[X_1^2] = 380.2500 \\ E[X_2^2] = 484.6400 \\ E[X^2] = 422.0060 \\ E[X^2] - E[X]^2 = 414.1660 $$