Let $X_1,\dots , X_{16}$ be a random sample from a normal population with mean $\mu= 6$ and variance $\sigma^2 = 4$.
(a) What is the approximate distribution of X?
(b) Find $P( X< 4)$
(c) Find $P(S^2 < 4)$
Here is what I came up with:
(a) $\bar{X} \sim N(6, \frac{4}{16})$
(b) $P(\bar{X} <4) = P(Z < \frac{4-6}{\sqrt{\frac{4}{16}}})=P(Z<-4)=pnorm(-4)= 3.167124e-05$ Which I think is too small.
(c) Using theorem $\frac{(n-1)S^2}{\sigma^2}\sim X^2(n-1)$ I get
$P(S^2 < 4)= P(\frac{(16-1)S^2}{\sigma^2}< 4(16-1))= P(X<60) = pchisq(60, 15)= .999997$
For $b$ and $c$ the values are total opposites which make me think I'm not doing it correctly.
(a) and (b) are correct. By the way, the probability you got in (b) isn't actually too small. Think about it this way, $\mu$ is $6$ so the probability that the sample mean ($\bar{X}$) is less than $-4$ will be really small.
(c) You forgot to divide by $\sigma^2$ on the right hand side of the inequality:
$P(S^2<4)=P(X<\dfrac{4(16-1)}{4})=P(X<15)$, where $X\sim\chi_{15}^2$.