A has normal distribution of scores of students with $X \sim \mathcal N(625, 100)$
B has normal distribution of scores of students with $X \sim \mathcal N(600, 150)$
Now I have to calculate probablity of 2 A students's average marks to be greater than average of 3 B students,if 2 students from A and 3 from B appear in test.
Here what I supppose is i should calculate difference of averages from which i will get rrquired proabability.
$E(X-Y)=625-600=25$
$Var(X-Y)=Var(X)+Var(Y)-CoVar(X,Y)$
Now here i have $Var(x)$ and $Var(Y)$ but i don't have $CoVar(X,Y).
So now lets suppose somehow i calculated $Var(X-Y)
I will say $Z=(X-Y+25)/\sqrt{Var(X,Y)}$
I will say $Pr(X-Y>0)=Pr(Z>{some number i need to calculate after i have Var(X,Y) }=some probability
So here i got simple probability of distribution of A minus B how will I calculatethe probability of 2 students of A average greater than 3 students of B average.
Let the first group have mean $a$, variance $\sigma^2$, and let the second group have mean $b$, variance $\tau^2$.
We pick $m$ students from the first group, and $n$ from the second group. Let $\bar{X}$ be the average of the $m$ grades from the first group, and $\bar{Y}$ the average of the $n$ grades from the second group. Then $\bar{X}$ has mean $a$, variance $\frac{\sigma^2}{m}$. A similar result holds fpr $\bar{Y}$.
If $\bar{X}$ and $\bar{Y}$ are independent, then $W=\bar{X}-\bar{Y}$ has normal distribution, mean $a-b$, and variance $\frac{\sigma^2}{m}+\frac{\tau^2}{n}$. We want the probability that $W\gt 0$.
Put in the appropriate numbers, and calculate.