Probability and Statistics problem

a) What is the approximate distribution of the sample mean X(X hat)? Justify.

The Central Limit Theorem states that as $n\rightarrow \infty $ , the Sample Mean $\bar{X} $ will converge to the Normal Distribution with E( $\bar{X} $) = μ and Variance( $\bar{X} $) = $σ^{2}/n$. For most distributions, this approximation is considered valid if the sample size is over 30. Since n=100, it is reasonable to assume $\bar{X} $ will be approximately Normal with
E( $\bar{X} $) = μ = 1.5 and Variance( $\bar{X} $) = $σ^{2}/n$ = 9/100 = 0.09 and StdDev($\bar{X} $) = $\sqrt{0.09}=0.3 $ 

b) Find P(2 < X(X hat) < 3).

I used a Normal Calculator tool at this site: https://onlinestatbook.com/2/calculators/normal_dist.html. You can use whatever tool you have. 

E( $\bar{X} $) = μ = 1.5 and  StdDev($\bar{X} $) = $\sqrt{0.09}=0.3 $ 
P(2 < $\bar{X} $ < 3) = 0.0478 

c) Assume the sample is normal. Find P(9 < S < 10)

This is probably a typo and your instructor probably meant $S{^2}$.  I will show the answer both ways.

If X is normal, then $S{^2}(n-1)/σ^{2}$ has a chi-square distribution $χ^{2}$with n-1=99 degrees of freedom.
I will use a Chi-square cdf calculator at https://homepage.divms.uiowa.edu/~mbognar/applets/chisq.html


What your professor actual said:

P(9 < S < 10) = P(81 < $S{^2}$ < 100)
                         = P[81(99)/9 < $χ^{2}$ < 100(99)/9)
                         = P(891 < $χ^{2}$ < 1100) 
                         = P($χ^{2}$< 1000) - P($χ^{2}$<891) = 1-1 = 0 

What your professor probably meant:

P(9 < $S{^2}$ < 10) = P[9(99)/9 < $χ^{2}$ < 10(99)/9)
                               = P(99 < $χ^{2}$ < 110)
                               =  P($χ^{2}$< 110) - P($χ^{2}$<99) = 0.78858 - 0.51890 = 0.26968