0
$\begingroup$

This question goes,

"According to a study, 4% of a population in a region have lung disease. Of those people without lung disease, 13% are smokers. Of those people with lung disease, 92% are smokers.

(a) If 20 people are randomly selected from the region, what is the probability that at least one of them has lung disease?

For (a) I tried to solve by using $\mu = np$ and $\sigma=\sqrt{npq}$, I got $\mu=0.8$ and $\sigma=0.4$. So to find at least 1 of them has lung cancer, I let $P(X \ge 1)=P(z \ge (\frac{1-0.8}{0.4/\sqrt{20}})=0.0125$. But from my lecture notes it says that to use $\mu=np$ it is only for approximating binomial distribution to normal if $n$ is large enough. However for this question, since they are asking "at least one of them has lung cancer", can I still use this? If not, another way I thought to solve this is to use sample proportion where $P(X \ge 1) = P(Z \ge \frac{0.05-0.04}{\sqrt{0.04 \times 0.96/20}} )$ where $0.05$ comes from $1/20$.

Improve this question
$\endgroup$
1

1 Answer 1

Reset to default
0
$\begingroup$

You've overthought this problem, haha.

$P(X \ge 1)$ is the same as $1-P(X=0)$. (If you're not greater or equal than 1, you have to be 0.)

Improve this answer
$\endgroup$
3
  • $\begingroup$ Have a look at our self-study question guidelines. We generally prefer to give hints, rather than a complete answer. Could you edit this answer to make it more of a hint than a full solution? $\endgroup$
    – Silverfish
    Commented Nov 18, 2015 at 17:41
  • $\begingroup$ Is this better, or should it be more vague? $\endgroup$
    – Kontorus
    Commented Nov 18, 2015 at 17:44
  • 1
    $\begingroup$ That's better, thanks. I'd be tempted to mention why the original poster (OP) has "overthought" the problem - since it's only necessary to find $P(X=0)$, there's no need to use a normal approximation. $\endgroup$
    – Silverfish
    Commented Nov 18, 2015 at 17:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.