Combine sample from different normal distributions 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 $[81,83]$
 A: It seems you have $n = 10$ random observations $X_1, X_2, \dots, X_{10}$ from
$\mathsf{Norm}(\mu=78,\sigma=7.21)$ and
$m = 10$ random observations $Y_1, Y_2, \dots, Y_{10}$ from
$\mathsf{Norm}(\mu=90,\sigma=12.96).$ And that you want
to know the probability that $\bar W = .5(\bar X + \bar Y)$ is in the interval $[81, 83].$
Then $\bar X  \sim \mathsf{Norm}(78, 7.21/\sqrt{10}),\;$ 
$\bar Y  \sim \mathsf{Norm}(80, 12.96/\sqrt{10}),$ 
and $$\bar W =.5(\bar X + \bar Y) \sim
\mathsf{Norm}\left(.5(78+80), \sqrt{.25(3.2+16.8)}\right).$$
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 $Var(\bar X) = \sigma^2/n.$
A: 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 $\bar X=\frac{B+G}2$.

*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
$$\int_{162}^{166}f_Z(z) dz$$
(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 $D=\{(b,g)\in\mathbb R^2 \colon 162<b+g<166\}$.
Both a) and b) will give the desired probability.

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