# Central limit theorem question

#### Jaket1

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.

#### romsek

Math Team
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

$n = 121$

#### Jaket1

Hi Romsek, where does the number 200 come from?

#### romsek

Math Team
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.

1 person

haha thank you!

#### romsek

Math Team
$n = \left \lfloor \dfrac{240 \mu +p^2 \sigma ^2-\sqrt{p^4 \sigma ^4+480 \mu p^2 \sigma ^2}}{2 \mu ^2} \right \rfloor$

$n = 70$