Question

The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.2 minutes and standard deviation 1.5 minutes. Suppose that a random sample of n-49 independent customers is observed. Find the approximate probability that the average time waiting in line for these customers is (a) Greater than 10 minutes (b) Between 6 and 10 minutes (c) Less than 6 minutes

Answer 1

Given that, mean $(\mu)$ = 8.2 minutes and

standard deviation $(\sigma)$ = 1.5 minutes

sample size ( n ) = 49

We want to find, the following probabilities,

a)

$P(\bar x > 10) = P(\frac { \bar x -\mu}{\frac {\sigma}{\sqrt n }} > \frac {10-8.2}{\frac{1.5}{\sqrt { 49}}} ) = P(Z > 8.4) = 0.0000$

Probability = 0.0000

b)

$\\ P(6< \bar x < 10) \\\\ = P(\frac {6 -8.2}{\frac { 1.5}{\sqrt { 49}}} < \frac { \bar x -\mu}{\frac {\sigma}{\sqrt n }} < \frac {10-8.2}{\frac{1.5}{\sqrt { 49}}} ) \\\\ = P(-10.27 < Z < 8.4) \\\\ P(Z < 8.4) - P(Z < -10.27) \\\\ = 1.0000 - 0.0000 \\\\ = 1.0000$

Probability = 1.0000

c)

$P(\bar x < 6) = P(\frac { \bar x -\mu}{\frac {\sigma}{\sqrt n }} < \frac {6-8.2}{\frac{1.5}{\sqrt { 49}}} ) = P(Z < -10.27) = 0.0000$

Probability = 0.0000