A shooter hits a target with probability $0.4$ and shoots at target $150$ times. Find at least one interval in which, with probability $0.8$, will be the number of shooter's hits of target.
A random variable $S_{150}: B(150;0.4)$ represents the number of hits. We are looking for an interval $(a,b)$.
Also, $ES_{150}=150*0.4=60$ and $DS_{150}=150*0.4*0.6=36$.
Using central limit theorem, I set the probability to be equal to $0.8$:
$P(a<S_{150}<b)=P(\frac{a-60}{6}<\frac{S_{150}-60}{6}<\frac{b-60}{6})=0.8$
I am just lost after this step, how do I find a and b?
You have $\sigma = \sqrt{0.6 \cdot 0.4}$. Using CLT, just plug in $n, \sigma, n, p$ that you are given: $$ P(\frac{a-np}{\sqrt{n p (1-p)}}\leq \frac{S_n - np}{\sqrt{n p(1-p)}} \leq \frac{b-np}{\sqrt{n p (1-p)}}) \approx \Phi(\beta) - \Phi(\alpha) = 0.8 $$ Since $Z \sim N(0,1)$ is symmetric around $0$, you can easily find for which $\alpha, \beta$ the area under its $\varphi$ is equal to $0.8$. Once you've got them, plug in the equations to get $a,b$.