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December 11th, 2017, 07:38 AM  #1 
Member Joined: Jan 2016 From: Blackpool Posts: 95 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, 08:29 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,977 Thanks: 1026 
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, 04:24 PM  #3 
Member Joined: Jan 2016 From: Blackpool Posts: 95 Thanks: 2 
Hi Romsek, where does the number 200 come from?

December 11th, 2017, 06:49 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 1,977 Thanks: 1026  
December 12th, 2017, 03:01 AM  #5 
Member Joined: Jan 2016 From: Blackpool Posts: 95 Thanks: 2 
haha thank you!

December 12th, 2017, 11:23 AM  #6 
Senior Member Joined: Sep 2015 From: USA Posts: 1,977 Thanks: 1026 
$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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