# Solved (Free): The times that a cashier spends processing individual customer’s order are independent and identically distributed random variables with mean

#### ByDr. Raju Chaudhari

Mar 28, 2021

The times that a cashier spends processing individual customer's order are independent and identically distributed random variables with mean 2.5 minutes and standard deviation 2 minutes. What is the approximate probability that it will take more than 4 hours to process the orders of 100 people?

#### Solution

Let $Y$ denote the times that a cashier spends processing individual customer's order.

Let $Y_i, i=1,2,\cdots, 100$ be independent and identically distributed random variables with mean $2.5$ minutes and standard deviation $2$ minutes.

Then $E(\overline{Y}) = 2.5$ and $sd(\overline{Y})=\frac{2}{\sqrt{100}}= \frac{2}{10}=0.2$.

Since $n=100 > 30$ is large, we use Central Limit Theorem. That is $\frac{\overline{Y}-E(\overline{Y})}{sd(\overline{Y})}\sim N(0,1)$.

 \begin{aligned} P\bigg(\sum_{i=1}^{100} Y_i>4\times 60\bigg) &=P\bigg(\sum_{i=1}^{100} Y_i>240\bigg)\\ &=P\bigg(\overline{Y}>\frac{240}{100}\bigg)\\ &=P\big(\overline{Y}>2.4\big)\\ &=P\big(\overline{Y}>2.4\big)\\ &=P\bigg(\frac{\overline{Y}-E(\overline{Y})}{sd(\overline{Y})}>\frac{2.4-2.5}{0.2}\bigg)\\ &\approx P(Z>-0.5)\\ & \approx 1- P(Z\leq -0.5)\\ &\approx 1 - 0.3085\\ &\approx 0.6915 \end{aligned}