Problem involving normal approximation to the binomial.

Question

I have the following problem, and I'm getting confused. I would appreciate any help:

Supposed that 20% of students don't have laptops. Let $X$ be the number of students without laptops in a random sample of size $n=50$. Use the normal approximation to the binomial to:

(a) Find the probability that between 6 and 16 (inclusive) of the selected students don't have laptops.

(b) Find the probability that the number of students without laptops is at least 30% in the next sample ($n=50$).

So (a) seems straightforward but I have a question about continuity correction:

We have $\mu=np=10$ and $\sigma^2=np(1-p)=8$. So:

$P(6 \leq X \leq 16) = P(\frac{6-10}{\sqrt{8}}\leq X \leq \frac{16-10}{\sqrt{8}}) = ...$

Do I need to use continuity correction? If so, does 6 become 5.5 and 16 become 16.5?

As for (b) I'm very much at a loss. Not even sure how to actually start.

Answer 1

For (a) looks good. Yes, use continuity correction. Subtract 0.5 from lower bound and add 0.5 to upper bound

for (b) you want the area of the gaussian above 0.3 * 50