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

n is approximately 29 customers

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 = (( \sum X_i)/(n) - \mu_(\overline x))/((\sigma)/(√(n) ) )`

Therefore, we get;

`z = ((120)/(n) - 4.5)/((4)/(√(n) ) ) = 0.53983`

`(4)/(√(n) ) * 0.53983 = \frac{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