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 Post subject: Binomial DistributionPosted: Sat, 21 Apr 2012 14:19:30 UTC
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Joined: Wed, 15 Feb 2012 07:19:33 UTC
Posts: 31
The probability that a candidate is offered a scholarship is 0.2412. A panel interviews 5 candidates per day for a period of 55 days. Find the probability that, on average, at least 1 candidate is offered the scholarship per day out of this period of 55 days.

The answer is 0.235 (3 s.f.) but I don't know how to get there. You definitely can't use Poisson here right? Someone please help! Also, if the question seems weird, it's because it's the last part of a really long question. I've already solved the other parts of the question, 0.2412 being one of the solutions.

Thanks!

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 Post subject: Re: Binomial DistributionPosted: Sat, 21 Apr 2012 15:23:07 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6836
Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
thesocialnetwork wrote:
The probability that a candidate is offered a scholarship is 0.2412. A panel interviews 5 candidates per day for a period of 55 days. Find the probability that, on average, at least 1 candidate is offered the scholarship per day out of this period of 55 days.

The answer is 0.235 (3 s.f.) but I don't know how to get there. You definitely can't use Poisson here right? Someone please help! Also, if the question seems weird, it's because it's the last part of a really long question. I've already solved the other parts of the question, 0.2412 being one of the solutions.

Thanks!

I have no idea how they get 0.235.

5 candidates per day, for 55 days, total 275 candidates. Assuming the candidates' results are independent, and you want at least 55 successes out of them (to make the average at least one per day). So the exact probability is

If you use a normal approximation, , so you are looking at , and .

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 Post subject: Re: Binomial DistributionPosted: Sat, 21 Apr 2012 23:48:25 UTC
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Joined: Fri, 1 Jul 2011 01:17:26 UTC
Posts: 468
outermeasure wrote:
thesocialnetwork wrote:
The probability that a candidate is offered a scholarship is 0.2412. A panel interviews 5 candidates per day for a period of 55 days. Find the probability that, on average, at least 1 candidate is offered the scholarship per day out of this period of 55 days.

The answer is 0.235 (3 s.f.) but I don't know how to get there. You definitely can't use Poisson here right? Someone please help! Also, if the question seems weird, it's because it's the last part of a really long question. I've already solved the other parts of the question, 0.2412 being one of the solutions.

Thanks!

I have no idea how they get 0.235.

5 candidates per day, for 55 days, total 275 candidates. Assuming the candidates' results are independent, and you want at least 55 successes out of them (to make the average at least one per day). So the exact probability is

If you use a normal approximation, , so you are looking at , and .

Since your answer is so different from the answer posited in the original question, might there be a problem with the wording. You answered assuming the question was at least 55 scholarships be offered. Is it possible that the intent was "What is the probability of at least one scholarship per day?"

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 Post subject: Re: Binomial DistributionPosted: Sun, 22 Apr 2012 00:43:44 UTC
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Joined: Wed, 15 Feb 2012 07:19:33 UTC
Posts: 31
Yep, I agree. Even with an average of 55 results, there could be two offered on one day and none offered on another, which doesn't fulfil what the question asks for. Guess it's still open for discussion then. Thanks guys!

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 Post subject: Re: Binomial DistributionPosted: Sun, 22 Apr 2012 05:16:03 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6836
Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
mathematic wrote:
outermeasure wrote:
thesocialnetwork wrote:
The probability that a candidate is offered a scholarship is 0.2412. A panel interviews 5 candidates per day for a period of 55 days. Find the probability that, on average, at least 1 candidate is offered the scholarship per day out of this period of 55 days.

The answer is 0.235 (3 s.f.) but I don't know how to get there. You definitely can't use Poisson here right? Someone please help! Also, if the question seems weird, it's because it's the last part of a really long question. I've already solved the other parts of the question, 0.2412 being one of the solutions.

Thanks!

I have no idea how they get 0.235.

5 candidates per day, for 55 days, total 275 candidates. Assuming the candidates' results are independent, and you want at least 55 successes out of them (to make the average at least one per day). So the exact probability is

If you use a normal approximation, , so you are looking at , and .

Since your answer is so different from the answer posited in the original question, might there be a problem with the wording. You answered assuming the question was at least 55 scholarships be offered. Is it possible that the intent was "What is the probability of at least one scholarship per day?"

That would be tiny: giving , repeat 55 days gives .

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