Answer to Question #271894 in Statistics and Probability for Tony

Question

The amount of time required at a company’s customer care desk has been found to be approximately normally distributed with mean of 3 minutes and a variance of 2500 square seconds. What is the probability that a randomly selected customer will: (i) Spend more than 6 minutes. (ii) Spend between 1 minute and 4 minutes.

Answer 1

Let the random variable $X$ represent the amount of time spent by a customer at a customer care desk. Therefore, $X\sim N(3,(25/36))$

The mean $\mu=3$,

The variance is first converted into minutes square. Since 3600s² =1m² then, 2500s²=(25/36)m². Thus, variance $\sigma^2=2500s^2=(25/36)m^2$ and $\sigma=5/6$.

$i)$

 The probability that a customer spends more than 6 minutes is given as,

$p(X\gt6)=p(Z\gt(6-3)/(5/6))=p(Z\gt 3.6)=1-p(Z\lt 3.6)=1-.99984=0.00016$

Therefore, the probability that a randomly selected customer will spend more than 6 minutes in the customer care desk is 0.00016.

$ii)$

The probability that a randomly selected customer will spend  between 1 minute and 4 minutes is given as,

$p(1\lt X\lt 4)=p((1-\mu)/\sigma\lt Z\lt (4-\mu)/\sigma)=p((1-3)/(5/6)\lt Z\lt (4-3)/(5/6)=p(-2.4\lt Z\lt 1.2)$

This can also be written as,

$p(-2.4\lt Z\lt 1.2)=\phi(1.2)-\phi(-2.4)=0.8849-0.0082=0.8767$

Therefore, the probability that a randomly selected customer will spend between 1 minute and 4 minutes is 0.8767.