Central limit theorem problem verification 
Attempt:
a.$\mu=(1/6)(1)+(1/6)(2)+(1/6)(3)+(1/6)(4)+(1/6)(5)+(1/6)(6)=3.5$
$\sigma=\sqrt{EX^2-(EX)^2}=\sqrt{2.9167}$
b.$\mu=(1/216)((1)(3)+(3)(4)+(6)(5)+(10)(6)+(15)(7)+(21)(8)+(25)(9)+(27)(10)+(27)(11)+(25)(12)+(21)(13)+(15)(14)+(10)(15)+(6)(16)+(3)(17)+(1)(18))=10.5$
$\sigma=\sqrt{EX^2-(EX)^2}=\sqrt{8.75}$
c.By the law of large numbers, $\mu_{\overline{X}}=10.5,\sigma_{\overline{X}}=(\sqrt{8.75})/(\sqrt{25})$
d.$P\{\overline{X}>10\}=P\{Z>-.845\}=.8$
e.In this case $P\{\overline{X}>10\}=P\{Z>-1.195\}=.88$ so the value increases.
This is not a graded assignment, but an exercise I am practicing.
 A: Thanks for showing your work:
(a) $\mu = 3.5$ is correct. Checking $\sigma=1.7078,$
using R as a calculator. OK
x = 1:6; p = 1/6
vr = sum((x-3.5)^2 * p); vr
[1] 2.916667
sd = sqrt(vr); sd
[1] 1.707825

(b) Easier way to get mean and variance for $S = X_1 + X_2 + X_3,$ the sum on three dice.
$$\mu_S = E(S)=E(X_1+E(X_2)+E(X-3) = 3(3.5) = 10.5.$$
Because the three dice rolls are independent,
$$\sigma_S^2 = Var(S) = Var(X_1)+Var(X_2)+Var(X_3)\\ = 3(2.916667) = 8.75.$$
Notice that variances of independent random variables add, but standard deviations don't. $\sigma_S = \sqrt{8.75} =  2.95804.$
(c) Let $\bar S = \frac{1}{25}\sum_{i=1}^{25} S_i.$
$E(\bar S) = \mu = 10.5$ and $SD(\bar X) = \sigma/\sqrt{n}
= 2.95804/5 = 0.591608
 \approx 0.5916.$ As you say.
(d) $P(\bar S > 10) = 1 - P(\bar S \le 10),$ where
$\bar S \sim \mathsf{Norm}(10.5, \sigma = 0.5916).$
You have done this by standardizing and using printed
CDF tables of the standard normal distribution.
In R, pnorm is a normal CDF, and I checked your answer
using a direct computation in R to find $P(\bar S > 10) = 0.801:$ Your z-score corresponding to $10$ would be $\frac{10-10.5}{0.5916} \approx -0.8452,$ giving about the same answers from printed tables.
(10-10.5)/(0.5916)
[1] -0.8451657
1 - pnorm(-0.8452)
[1]  0.8010004

1 - pnorm(10, 10.5, 0.5915)
[1] 0.8010307

