The weight X of any particular banana is an independent random variable with PDF. E(X) = 5, VAR(X) = 16/12. Suppose we obtain 40 bananas and separate then into sets of “light bananas” (those that weigh less than 4.0 ounces) and “heavy bananas” (those that weigh more than 4.0 ounces). Determine the variance of L, our number of light bananas. How should this compare with the variance of H, the number of heavy bananas? Explain.
ans:the probability of the sets of 'light bananas' equals to the probability of the sets of 'heavy bananas'. the variance of L = 40*(3/4)*(1/4) = the variance of H.
above the question, I don't know why the variance of the variance of L equals to the variance of H. I also have no idea to how to compute the numerical value. Maybe it's a easy question, but I don't still understand the variance about this respect.
To explain that the variances of $L$ and $H$ are the same.
Let $p$ denote the probability that a banana is light and let $q$ denote the probability that it is heavy.
Then $p+q=1$.
We are dealing in both cases with binomial distributions. The one for $L$ has parameters $40$ and $p$. The one for $H$ has parameters $40$ and $q$.
In general $\text{Var}Y=nr(1-r)$ if $Y$ is binomially distributed with parameters $n$ and $r$.
So we find: $$\text{Var}L=40pq=40qp=\text{Var}H$$
More directly:$$\text{Var}H=\text{Var}\left(40-L\right)=\mathbb{E}\left(40-L\right)^{2}-\left[\mathbb{E}\left(40-L\right)\right]^{2}=\mathbb{E}L^{2}-\left[\mathbb{E}L\right]^{2}=\text{Var}\left(L\right)$$ This shows that the fact that $H+L$ is constant is allready enough for the conclusion. We only need the existence (one) of the variance(s).
I think it is not possible to find values for $p$ (and $q$) on base of the data, because these data do not determine the distribution of $X$. There might be a way to estimate them.