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 `$\frac{\overline{Y}-E(\overline{Y})}{sd(\overline{Y})}\sim N(0,1)$`

.

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

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators