1

Hello can someone please help me to answer this question it, it a binomial distribution question:

An email message advertises the chance to win a prize if the reader follows a link to an online survey. The probability that a recipient of the email clicks on the link to the survey is 0.0016. How many emails, to the nearest hundred, need to be sent out in order to have a 99% probability that at least 1000 will be answered?

Nadun
  • 75
  • 1
    Is it a homework question? What did you try? – Clement C. Sep 07 '14 at 01:43
  • I know that the formula for binomial distribution is Pr(X = k) = {n\choose k}p^k(1-p)^{n-k} and in this case n has to be found and p=0.0016 and the condition is that pr(X is alteast 1000)=0.99 – Nadun Sep 07 '14 at 01:47
  • Suppose $X$ follows a $\mathrm{Binomial}(n,p)$ distribution. What is the probability (as a function of $n$) that $X \geq 1000$? And how can you use that to answer the question? – Clement C. Sep 07 '14 at 01:59
  • Dont you have to know like the prob when X=0 and then u can use the formula to work it out? – Nadun Sep 07 '14 at 02:02
  • If you know that $\mathbb{P}{X \geq 1000} = f(n)$ for some explicit expression $f(n)$, you then only have to find the smallest $n$ (rounded to hundreds) such that $f(n) \geq 0.99$. – Clement C. Sep 07 '14 at 02:11
  • I'm not sure if you can solve for $''n''$ analytically. Do they want you to use some sort of approximation like the Poisson distribution? – ClassicStyle Sep 07 '14 at 02:22

1 Answers1

2

I think that this may be a geometric distribution question. We can use a geometric distribution to find the number of trials (emails) before we get a success (someone follows the link). We can also find the number of trials needed to be 99% certain of getting 1 success. Then we multiply that by 1000. We have: X~Geo(0.0016) The number of trials needed to be 99% certain of one success is worked out as shown:

r= number of trials needed $$P(X\le{r})\ge{0.99}$$

Therefore: $$1-P(X\gt{r})\ge{0.99}$$

For a geometric distribution, $$P(X\gt{r})$$ is given by $$q^r$$ where q is (1-p)

Therefore: $$1-q^r\ge{0.99}$$

$$q^r\le{0.01}$$

$$(1-0.0016)^r\le{0.01}$$

$$0.9984^r\le{0.01}$$

These logs are to base 10:

$$r\log0.9984\le{\log0.01}$$

$$r\log0.9984\le{\log0.01}$$

$$r\ge{\frac{\log0.01}{log0.9984}}$$

$$r\ge{2875.928167}$$

As r must be an integer, $$r=2876$$ This finds r for 1 email being clicked on. Final answer=$$2876\times 1000$$ Final answer=2,876,000

That's quite a few e-mails.

Hope I helped!

B.Jenkins
  • 339