Let $X_1,\ldots,X_{144}$ be a random sample from a population with mean $\mu = 20$ and variance $\sigma^2 = 64$.
(a) What is the approximate distribution of $\bar X$?
(b) Find $P( \bar{X} < 19)$.
(c) Find a constant $c$ such that $P(|\bar{X}-20| < c) = 0.95$
My attempt
(a) $\bar{X} \overset{\text{approx}}{\sim} N\left(20, \frac{64}{144}\right)$
(b) $P(Z<19)=P\left(Z<\dfrac{19-20}{2/3}\right) = P(Z<-\dfrac{3}{2})=\operatorname{pnorm}\left(-\dfrac{3}{2}\right)= 0.0668$
(c) To find $c$ in the past I was given a function which I don't have in this case. Any ideas what I need to use to solve for $c$. Is it the Central Limit Theorem formula?
You have $\displaystyle \bar X \sim N\left( 20, \left(\frac 2 3 \right)^2 \right)$ approximately. (Here you may have been slightly misled. Is this a reasonably close approximation? An answer to that would have to rely on information beyond what is specified in the problem. For some highly skewed distributions, $144$ would be nowhere near a big enough sample to warrant that conclusion; for more commonplace distributions $10$ is big enough for practical purposes.) You need $$ \Pr\left( \left|\frac{\bar X - 20}{2/3}\right| < \frac c {2/3} \right) = 0.95. $$ $$ \Pr\left( |Z| < \frac 3 2 c \right) = 0.95\quad\text{where } Z \sim N(0,1). $$ $$ \frac 3 2 c \approx 1.96. $$ $$c\approx 1.31$$