My Math Forum Central limit theorem question

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 December 11th, 2017, 08:38 AM #1 Member   Joined: Jan 2016 From: Blackpool Posts: 76 Thanks: 2 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.
 December 11th, 2017, 09:29 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,691 Thanks: 859 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 $n = 121$
 December 11th, 2017, 05:24 PM #3 Member   Joined: Jan 2016 From: Blackpool Posts: 76 Thanks: 2 Hi Romsek, where does the number 200 come from?
December 11th, 2017, 07:49 PM   #4
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Quote:
 Originally Posted by Jaket1 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.

 December 12th, 2017, 04:01 AM #5 Member   Joined: Jan 2016 From: Blackpool Posts: 76 Thanks: 2 haha thank you!
 December 12th, 2017, 12:23 PM #6 Senior Member     Joined: Sep 2015 From: USA Posts: 1,691 Thanks: 859 $n = \left \lfloor \dfrac{240 \mu +p^2 \sigma ^2-\sqrt{p^4 \sigma ^4+480 \mu p^2 \sigma ^2}}{2 \mu ^2} \right \rfloor$ $n = 70$

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