I am not sure how I am supposed to work this question out but I am given that the height of students from college A have a distribution written as:
$A$~$N(1.78,0.06^2)$
and the height of students from college B have a distribution written as:
$B$~$N(1.72,0.05^2)$
$A_1, A_2, A_3, A_4$ are the heights of $4$ students from college A. I have to find the distribution of their average height ${\bar A}=(A_1, A_2, A_3, A_4)/4$
How am I supposed to calculate this? I then have to do the same for college B although only taking the heights of $2$ students and from this, find the probability that the average height of the students from college B (just those 2) exceeds that of the students from college A (just those 4).
The average height $\bar{A}$ of the $4$ students from college A is $\frac{A_1+A_2+A_3+A_4}{4}$.
The sum of independent normals is normal, with mean the sum of the means, and variance the sum of the variances. If $X$ is normal, then $kX$ is normal with mean $k$ times the mean of $X$, and variance $k^2$ times the variance of $X$.
It follows that $\bar{A}$ is normal, mean $1.78$, and variance $\frac{1}{4}(0.06)^2$.
Similarly, thw average height $\bar{B}$ of the two students from college B is normal, mean $1.72$, variance $\frac{1}{2}(0.05)^2$.
Let $Y=\bar{B}-\bar{A}$. We want the probability that $Y\gt 0$.
The random variable $Y$ has normal distribution, mean $1.72-1.78=-0.06$.
The variance of $Y$ is $\frac{1}{2}(0.05)^2+\frac{1}{4}(0.06)^2$ (yes, plus).
Now you know the mean and variance of $Y$, so you can use software, or tables of the standard normal, to find $\Pr(Y\gt 0)$.