The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.0 minutes and standard deviation 1.3 minutes. Suppose that a random sample of customers is observed. Find the probability that the average time waiting in line for these customers is

P (5 < X < 10) = 1

Step-by-step explanation:

Given:-

- Sample size n = 49

- The sample mean u = 8.0 mins

- The sample standard deviation s = 1.3 mins

Find:-

Find the probability that the average time waiting in line for these customers is between 5 and 10 minutes.

Solution:-

- We will assume that the random variable follows a normal distribution with, then its given that the sample also exhibits normality. The population distribution can be expressed as:

X ~ N (u, s / √n)

Where

s / √n = 1.3 / √49 = 0.2143

- The required probability is P (5 < X < 10) minutes. The standardized values are:

P (5 < X < 10) = P ((5 - 8) / 0.2143 < Z < (10-8) / 0.2143)

= P (-14.93 < Z < 8.4)

- Using standard Z-table we have:

P (5 < X < 10) = P (-14.93 < Z < 8.4) = 1