Question

The amount of time a passenger waits at an airport check-in counter is random variable with mean 10 minutes and standard deviation of 2 minutes. Suppose a random sample of 50 customers is observed. Calculate the probability that the average waiting time waiting in line for this sample is (a) less than 10 minutes (b) between 5 and 10 minutes

Answer 1

(a) less than 10 minutes

= 0.5

(b) between 5 and 10 minutes

= 0.5

Explanation:

We solve the above question using z score formula. We given a random number of samples, z score formula :

z-score is z = (x-μ)/ Standard error where

x is the raw score

μ is the population mean

Standard error : σ/√n

σ is the population standard deviation

n = number of samples

(a) less than 10 minutes

x = 10 μ = 10, σ = 2 n = 50

z = 10 - 10/2/√50

z = 0 / 0.2828427125

z = 0

Using the z table to find the probability

P(z ≤ 0) = P(z < 0) = P(x = 10)

= 0.5

Therefore, the probability that the average waiting time waiting in line for this sample is less than 10 minutes = 0.5

(b) between 5 and 10 minutes

i) For 5 minutes

x = 5 μ = 10, σ = 2 n = 50

z = 5 - 10/2/√50

z = -5 / 0.2828427125

= -17.67767

P-value from Z-Table:

P(x<5) = 0

Using the z table to find the probability

P(z ≤ 0) = P(z = -17.67767) = P(x = 5)

= 0

ii) For 10 minutes

x = 10 μ = 10, σ = 2 n = 50

z = 10 - 10/2/√50

z = 0 / 0.2828427125

z = 0

Using the z table to find the probability

P(z ≤ 0) = P(z < 0) = P(x = 10)

= 0.5

Hence, the probability that the average waiting time waiting in line for this sample is between 5 and 10 minutes is

P(x = 10) - P(x = 5)

= 0.5 - 0

= 0.5