Question

The times that a cashier spends processing individual customer’s order are independent 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?

Answer 1

Answer:

0.6915 is the probability that it will take more than 4 hours to process the orders of 100 people.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 2.5 minutes

Standard Deviation, σ = 2 minutes

Since the sample size is large, by central limit theorem, the distribution of sample means is approximately normal.

$P(\sum x_{i} > 4)\\P(\sum x_i > 4\times 60\text{ minutes})\\\\P(\displaystyle\frac{1}{100}\sum x_i > \frac{4\times 60}{100}\text{ minutes})\\\\P(\bar{x} > 2.4)$

Formula:  

$z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$  

P(it will take more than 4 hours to process the orders of 100 people)  

P(x > 2.4)  

$P( x > 2.4) = P( z > \displaystyle\frac{2.4-2.5}{\frac{2}{\sqrt{100}}}) = P(z > -0.5)$  

Calculation the value from standard normal z table, we have,  $P(x > 2.4) = 1 - 0.3085 = 0.6915= 69.15\%$

0.6915 is the probability that it will take more than 4 hours to process the orders of 100 people.