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?
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!