Vtyjkyuk

Distribution of marks out of 240 for girls and boys appearing for an examination is $N(78,32)$ and $N(80,168)$ respectively. $X$ is a sample consisting of 10 boys and 10 girls.

What is the probability that $bar X$ lies in $[81,83]$

It seems you have $n = 10$ random observations $X_1, X_2, dots, X_10$ from
$mathsfNorm(mu=78,sigma=7.21)$ and
$m = 10$ random observations $Y_1, Y_2, dots, Y_10$ from
$mathsfNorm(mu=90,sigma=12.96).$ And that you want
to know the probability that $bar W = .5(bar X + bar Y)$ is in the interval $[81, 83].$

Then $bar X sim mathsfNorm(78, 7.21/sqrt10),;$
$bar Y sim mathsfNorm(80, 12.96/sqrt10),$
and $$bar W =.5(bar X + bar Y) sim
mathsfNormleft(.5(78+80), sqrt.25(3.2+16.8)right).$$

From there you should be able to find $P(81 le bar W le 83) approx 0.149.$

A simulation of a million such average test scores
gives the histogram below. The density curve is for
the normal distribution of $bar W.$

set.seed(826); m = 10^6
a.w = replicate( m, .5*mean(rnorm(10,78,sqrt(32))) 
 + .5*mean(rnorm(10,80,sqrt(168))) )
mean(a.w >= 81 & a.w <= 83) 
[1] 0.148802 # aprx prob
diff(pnorm(c(81,83), 79, 2.236))
[1] 0.1487247 # exact prob

Note: The key formulas are $E(aX + bY) = aE(X) + bE(Y)$ and, provided $X$ and $Y$ are independent,
$$Var(aX + bY) = a^2Var(X) + b^2Var(Y).$$
These are used to prove that if $X_i$ are $n$ random observations from a population with mean $mu$ and variance $sigma^2,$ then the sample mean $bar X$ has
$E(bar X) = mu$ and $Var(bar X) = sigma^2/n.$

I don't know what you have tried or what ideas came to you already, but I'll suggest you a possible route that you could follow.

• Let's call the means of the marks of the 'boys' and 'girls' $B$ and $G$, respectively. What's their distribution? (*)




• Now prove that in this case $bar X=fracB+G2$.



• So you need to find
  $$P(162<B+G<166)$$. At least two paths are at hand:

a) You can try to find the distribution of $Y=B+G$ (you haven't mentioned it, but ---lacking any other information--- independence will have to be assumed) and then calculate
$$int_162^166f_Z(z) dz$$
(and by the way, there is a very well known property regarding sums of independent normal random variables...).

b) You can get the joint density of $B$ and $G$, which under independence is just the product of both marginal densities. Then, calculate the probability as
$$iint_D f_BG(b,g) dA,$$
where $D=(b,g)inmathbb R^2 colon 162<b+g<166$.

Both a) and b) will give the desired probability.

(*) They are both normals. Try to find their mean and variance.