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December 11th, 2017, 08:38 AM   #1
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Central limit theorem question

Times spent on processing orders are independent random variables
with mean 1.5 minutes and standard deviation 1 minute. Let n be the number of orders an operator is scheduled to process in 2 hours. Use the CLT to find the largest value of n which give at least a 95% chance of completion in that time.
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December 11th, 2017, 09:29 AM   #2
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CLT says that if each task duration $T_k$ has mean $\mu$ and standard deviation $\sigma$ then

$S_n = \dfrac 1 n \sum \limits_{k=1}^n T_k \to N\left(\mu, \dfrac{\sigma}{\sqrt{n}}\right)$

so we need to find $n \ni P\left[S_n < \dfrac {200}{n}\right] > 0.95$

This in turn means that

$\Phi\left(\dfrac{\frac{200}{n}-\mu}{\frac{\sigma}{\sqrt{n}}}\right)> 0.95$

if we let $p=\Phi^{-1}(0.95)$ then we can solve for $n$ as

$n = \left \lfloor \dfrac{400 \mu +p^2 \sigma ^2-\sqrt{p^4 \sigma ^4+800 \mu p^2 \sigma ^2}}{2 \mu ^2}\right \rfloor$

using $\mu=1.5,~\sigma=1,~p=1.645$

we get

$n = 121$
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December 11th, 2017, 05:24 PM   #3
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Hi Romsek, where does the number 200 come from?
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December 11th, 2017, 07:49 PM   #4
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Quote:
Originally Posted by Jaket1 View Post
Hi Romsek, where does the number 200 come from?
jeeze.. from me multiplying 2x60 and getting 200 instead of 120.

Must have been before coffee.

Replace it with 120.
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December 12th, 2017, 04:01 AM   #5
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haha thank you!
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December 12th, 2017, 12:23 PM   #6
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$n = \left \lfloor \dfrac{240 \mu +p^2 \sigma ^2-\sqrt{p^4 \sigma ^4+480 \mu p^2 \sigma ^2}}{2 \mu ^2} \right \rfloor $

$n = 70$
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