Approximate the probability that the average grade of class A exceeds the average grade of class B by more than $0.5$ points.
Hint: Consider what needs to happen with the sum of the grades for each course.
Solution:
By the Central Limit Theorem: Let $X_1 , X_2 , \ldots$ be a sequence of independent and identically distributed random variables, each having mean $7.4$ and variance $1.2$. Then,
$\frac{\sqrt{n} \ (\bar{X} - 7.4)}{\sqrt{1,2}} \rightarrow^{d} N(0,1)$
Let $X_1, X_2, \ldots$ be the students of class A.
Let $Y_1, Y_2, \ldots$ be the students of class B.
I think I need to calculate:
$P\big(\frac{\sum_{i=1}^{40} X_i}{40} - \frac{\sum_{i=1}^{40} Y_i}{40} > 0.5\big) $
but I don't know how to proceed. Any help? Thanks
I am assuming that you are given a population mean $\mu$ and variance $\sigma^2$, with each student's grade (across all classes) independently and identically distributed according to this population. Let's denote a generic student's grade from class $A$ or class $B$ by $X_i$ or $Y_i$ respectively. There are $n$ students in each class.
You want to compute $$P\left({1 \over n}\sum_i X_i -{1 \over n}\sum_i Y_i>0.5 \right)=P\left({1 \over n}\sum_i X_i - Y_i>0.5 \right).$$
Note $E[X_i-Y_i]=0,\text{Var}(X_i-Y_i)=2\sigma^2$ so CLT tells us that
$$\sqrt n\left({1 \over n}\sum_i X_i - Y_i\right)\rightarrow^{d}N(0,2\sigma^2),$$
so we obtain the following approximation:
$$\begin{align} P\left({1 \over n}\sum_i X_i - Y_i>0.5 \right)&=P\left(\sqrt{{n \over 2\sigma^2}}\left({1 \over n}\sum_i X_i - Y_i\right)>0.5\sqrt{{n \over 2\sigma^2}} \right)\\ &=1-P\left(\sqrt{{n \over 2\sigma^2}}\left({1 \over n}\sum_i X_i - Y_i\right)\leq 0.5\sqrt{{n \over 2\sigma^2}} \right)\\ &\approx 1-\Phi\left(0.5\sqrt{{n \over 2\sigma^2}}\right) \end{align}$$
where $\Phi$ is the CDF of a standard normal.