a problem with central limit theorem Student scores on exams given by a certain instructor
have mean 74 and standard deviation 14. This
instructor is about to give two exams, one to a class
of size 25 and the other to a class of size 64.
(a) Approximate the probability that the average
test score in the class of size 25 exceeds 80.
(b) Repeat part (a) for the class of size 64.
(c) Approximate the probability that the average
test score in the larger class exceeds that of
the other class by over 2.2 points.
(d) Approximate the probability that the average
test score in the smaller class exceeds that of
the other class by over 2.2 points.
My attempt is:
(a)Let Y be the average test score in the class of size 25
 $$P(Y>80)=P(\frac{Y-E[Y]}{ \sigma(Y) }>\frac{80-74}{ \frac{14}{5} })=1-\Phi (2,14)=0,0162 $$
(b)Let Z be the average test score in the class of size 64
 $$P(Z>80)=P(\frac{Z-E[Z]}{ \sigma(Z) }>\frac{80-74}{ \frac{14}{8} })=1-\Phi (3,43)=0,0003 $$
(c)
$$P(Z-Y>2,2)=P(\frac{(Z-Y)-E[Z-Y]}{ \sigma(Z-Y) }>\frac{2,2-0}{3,30})=1-\Phi (0,667)=1-0,7486=0,2514 $$
where
$$Var(Z-Y)=Var(Z)+Var(Y)=14^2/64+14^2/25$$
$$\sigma(Z-Y)= \sqrt{14^2/64+14^2/25}\approx3,30$$
(d)
$$P(Y-Z>2,2)=P(\frac{(Y-Z)-E[Y-Z]}{ \sigma(Y-Z) }>\frac{2,2-0}{ 3,30})=1-\Phi (0,667)=1-0,7486=0,2514 $$
where
$$Var(Y-Z)=Var(Y)+Var(Z)=14^2/25+14^2/64$$
$$\sigma(Y-Z)= \sqrt{14^2/25+14^2/64}\approx3,30$$
(I've correct the exercise thank to the help Graham Kemp)
While the solution for points (c) and (d) should be 0,2516.
What's wrong with my solution?
 A: The central limit theorem says that if $X_1,\dots X_n$ are independent and identically distributed with variance $\sigma ^2$ and mean $\mu$, then
$$\sum_{i=1}^nX_i=\sqrt{n\sigma ^2}Z+n\mu$$
Where $Z\sim \mathcal N(0,1)$. Dividing through by $n$ we get the average, i.e.
$$\frac 1n \sum_{i=1}^nX_i=\frac{\sigma}{\sqrt{n}}Z+\mu$$
Now let $X_i$ for $i=1,\dots, 25$ be the score of student $i$ in the first class and define $Y_j$ for $j=1,...,64$ analogously, then
$$\frac {1}{25} \sum_{i=1}^{25}X_i=\frac{14}{5}Z_1+74$$
$$\frac {1}{64} \sum_{j=1}^{64}Y_j=\frac{14}{8}Z_2+74$$
Now, for $a$ we would just have
$$\mathbb{P}\left(\frac{14}{5}Z_2+74>80\right)=\mathbb{P}\left(Z_1>\frac{80-74}{\frac{14}{5}}\right)$$
Similarly for $b$.
For $c$ we want
$$\mathbb{P}\left(\frac{14}{8}Z_2+74>2.2+\frac{14}{5}Z_1+74\right)=\mathbb{P}\left(\frac{14}{8}Z_2+74-\left(\frac{14}{5}Z_1+74\right)>2.2\right)=\mathbb{P}\left(\frac{14}{8}Z_2-\frac{14}{5}Z_1>2.2\right)$$
To calculate this last probability, remember that the sum of two normal r.v's is normally distributed with the sum of their means and variances, so $\frac{14}{8}Z_2-\frac{14}{5}Z_1\sim \mathcal N\left(0,\left(\frac{14}{5}\right)^2+\left(\frac{14}{8}\right)^2\right)$
Or $\frac{14}{8}Z_2-\frac{14}{5}Z_1=\sqrt{\left(\frac{14}{5}\right)^2+\left(\frac{14}{8}\right)^2} Z$ where $Z$ is standard normal. Now just find 
$$\mathbb{P}\left(\sqrt{\left(\frac{14}{5}\right)^2+\left(\frac{14}{8}\right)^2} Z>2.2\right)$$
Similarly for d.
A: 
$$Var(Y-Z)=Var(Y)+Var(Z)=14^2/5+14^2/8$$
  $$\sigma(Y-Z)= \sqrt{14^2/5+14^2/8}\approx 7,98$$

However, $~\mathsf{Var}(Y) = 14^2/25~$ and $~\mathsf{Var}(Z)=14^2/64~$.
