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?

iescabroussexg

iescabroussexg

Answered question

2022-09-10

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?

Answer & Explanation

Everett Melton

Everett Melton

Beginner2022-09-11Added 12 answers

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 r ) 0.99
Therefore:
1 P ( X > r ) 0.99
For a geometric distribution,
P ( X > r )
is given by
q r
where q is (1-p)
Therefore:
1 q r 0.99
q r 0.01
( 1 0.0016 ) r 0.01
0.9984 r 0.01
These logs are to base 10:
r log 0.9984 log 0.01
r log 0.9984 log 0.01
r log 0.01 l o g 0.9984
r 2875.928167
As r must be an integer,
r=2876
This finds r for 1 email being clicked on. Final answer=
2876 × 1000
Final answer=2,876,000

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