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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.

Confused Student
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1 Answers1

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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

  • I'm not sure I follow what you mean by "the area of the Gaussian above 0.3*50"? You mean that I ought to figure out the z for x=15, and p(z) is the result? – Confused Student Jun 28 '17 at 02:27
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    @ConfusedStudent yes, use the approximation to the binomial and find P(N>15) –  Jun 28 '17 at 02:29