Suppose an NFL team is average. They play 16 games(assume no ties). Two quants are trying to estimate the number of wins.
One quant assumes a probability of 1/2 of winning each game. Using this assumption, he calculates the standard deviation SD1 of the number of wins.
The second quant assumes a probability of 0.6 for home games and 0.4 for away games. Using this assumption she calculates the standard deviation SD2.
What is the ratio of SD1/SD2 to two decimals?
Well my experience and exposure with binomials is very limited so what i thought which i am aware it is wrong is that 0.5 * 16 would be 8 wins from quant 1 and then 0.6 * 8 = 4.8 but there is no such thing 4.8 wins so 4 and 0.4 * 8 = 3.2 so 3 wins for a total of 7 wins for quant 2 which would result in a 8/7 ratio.
Solution 2 i tried was SD1= sq.root of 16 * 1/2 * (1- 1/2) which is 1 and the same for SD2 which gives me 0.96 for a SD1/SD2 = 1/0.96
It seems you are mismatching your parameters. You have two binomial variables $X$ and $Y$ both with parameters $n,p$ and $n,r$. Here $n=16$, $p=0.5$ and $r=0.6$.
Then, recall the expected number of games is the number of trials multiplied by chance of success: $np=8$ and $nr=9.6$. It is ok to have a decimal there. While you can't win $9.6$ games exactly you will, on average, win $9.6$ games (think about the arithmetic average of $0$ and $1$ for an analogy: it is $0.5$).
Next, recall the variance is $npq$ where $q$ is the chance of failure, that is, $q=1-p$ and for your second RV, you have $nrs$ where $s=r-1$.
Can you finish from here? Comment for clarifications.
Edit: I have misread one aspect. Let me recap. Quant1 thinks the number of games won ought to be modeled as $X\sim B(n,p)$ with $n=16$ and $p=0.5$. Quant2 thinks it should be $W=Y+Z$ where $Y\sim B(a,r)$, $Z\sim B(h,s)$, and where $a,h$ are the number of away games and home games, so that $a+h=n$ and $r=0.4$ and $s=0.6$ are chances of winning away and home games, respectively. We assume $Y$ and $Z$ are independent. Then the variance of $W$ is the sum of the variances of $Y$ and $Z$, but we know these since they are binomial, as discussed above. Thus, in the general form, the SD of number of games won in Quant2's model is $\sqrt{ar+hs}$. So the general form for the desired ratio is $$\frac{\sqrt{np}}{\sqrt{ar+hs}},$$ Which will only give you a numerical answer if you know what $a$ and $h$ are (I assume, they want the case $a=h$).
Comment for any clarifications.
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