The weights of a population of mice fed a certain diet follow a normal distribution with mean $\mu=100$ grams and standard deviation $\sigma=20$ grams. A random sample of $8$ such mice is taken. Let $X_1$ denote the number of mice with weights between $80$ and $100$ grams. Let $X_2$ denote the number with weights above $100$ grams. I have to find $E(X_2|X_1=2)$ and Var$(X_2|X_1=2)$.
I know the formula for $E(X_2|X_1=2)$ = $\mu_2 + {p\sigma_2\over\sigma_1}(x_1-\mu_1)$. The only thing that gets me is that there is a sample of $8$ mice so I don't know how I would find $\mu_1, \mu_2, \sigma_1, and \sigma_2$. Would I do that $\mu_1={(80+100)\over 2}=90$, but I feel like that can't be right since it is a normal distribution. and little $x_1=2$, which would be the number of mice having nothing to do with the mean, so I'm a little confused. Any help would be great.
Thanks
I think that, conditionally on $X_{1}=2$, the random variable $X_{2}$ has a binomial distribution.
Condition $X_{1}=2$ can be interpreted as: we are looking at $6$ mice here and the weight of each of them is smaller than $80$ or larger than $100$. Indeed we started by taking $8$ mice, but $2$ of them where 'uninteresting' because their weight did not fall in one of these categories. This tells us that $X_{2}$ conditionally on $X_{1}=2$ has the distribution $\text{Bin}\left(6,p\right)$ for
$p=P\left(W>100\mid W<80\vee W>100\right)=\frac{P\left(W>100\right)}{P\left(W<80\right)+P\left(W>100\right)}$ where $W\sim N\left(100,400\right)$.
It is evident that $P\left(W>100\right)=\frac{1}{2}$ so $p=\frac{1}{2P\left(W<80\right)+1}$. This $p$ can be calculated.
Then $E\left(X_{2}\mid X_{1}=2\right)=6p$ and $\text{var }\left(X_{2}\mid X_{1}=2\right)=6p(1-p)$.