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 Post subject: Binomial Distribution
PostPosted: Sat, 21 Apr 2012 14:19:30 UTC 
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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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PostPosted: Sat, 21 Apr 2012 15:23:07 UTC 
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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
\displaystyle
\sum_{n=55}^{275}\binom{275}{n}(0.2412)^n(1-0.2412)^{275-n}=0.9546542\dots
If you use a normal approximation, \mathrm{Binomial}(275,0.2412)\stackrel{\mathcal{D}}{\approx} N(66.33,50.331204), so you are looking at z_c=\dfrac{54.5-66.33}{\sqrt{50.331204}}=-1.6675, and \Phi(1.6675)=0.9523.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Sat, 21 Apr 2012 23:48:25 UTC 
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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
\displaystyle
\sum_{n=55}^{275}\binom{275}{n}(0.2412)^n(1-0.2412)^{275-n}=0.9546542\dots
If you use a normal approximation, \mathrm{Binomial}(275,0.2412)\stackrel{\mathcal{D}}{\approx} N(66.33,50.331204), so you are looking at z_c=\dfrac{54.5-66.33}{\sqrt{50.331204}}=-1.6675, and \Phi(1.6675)=0.9523.

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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PostPosted: Sun, 22 Apr 2012 00:43:44 UTC 
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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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PostPosted: Sun, 22 Apr 2012 05:16:03 UTC 
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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
\displaystyle
\sum_{n=55}^{275}\binom{275}{n}(0.2412)^n(1-0.2412)^{275-n}=0.9546542\dots
If you use a normal approximation, \mathrm{Binomial}(275,0.2412)\stackrel{\mathcal{D}}{\approx} N(66.33,50.331204), so you are looking at z_c=\dfrac{54.5-66.33}{\sqrt{50.331204}}=-1.6675, and \Phi(1.6675)=0.9523.

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: X\sim\mathrm{Binomial}(5,0.2412) giving \mathbb{P}(X\geq 1)=1-(1-0.2412)^5, repeat 55 days gives [1-(1-0.2412)^5]^{55}=1.199\times 10^{-7}.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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