Question

The times that a cashier spends processing individual customers' orders are independent random variables with mean 4.5 minutes and standard deviation 4 minutes. Find the number of customers n such that the probability that the orders of all n customers can be processed in less than 2 hours, is approximately 0.1. (Round your answer to the nearest integer.)

Answer 1

Answer:

n is approximately 29 customers

Step-by-step explanation:

The given parameters are;

The mean duration of an order, μ = 4.5 minutes

The standard deviation, σ = 4

2 hours = 120 minutes

We get;

The z-score when the probability, P = 0.1 is P(z < 0.53983)

We have;

$z = \dfrac{\dfrac{ \sum X_i}{n} - \mu_{\overline x}}{\dfrac{\sigma}{\sqrt{n} } }$

Therefore, we get;

$z = \dfrac{\dfrac{120}{n} - 4.5}{\dfrac{4}{\sqrt{n} } } = 0.53983$

$\dfrac{4}{\sqrt{n} } \times 0.53983 = \dfrac{120}{{n} } -4.5$

With the aid of a graphing calculator, we get;

81·n²-4320·n + 57600 - 18.6506514278·n = 0

Factorizing gives;

(n - 24.3011921023)·(n - 29.2623961869)·81124141329 = 0

Therefore;

n ≈ 24 or n ≈ 29

Therefore, we take the largest number, n ≈ 29, to make the sample approximately normal