Answer to Question #136648 in Statistics and Probability for konke

Question

The long-distance calls made by South Africans are normally distributed with a mean of 16.3 minutes and a standard deviation of 4.2 minutes. What percentage of calls last more than 18 minutes (round off to two decimal places)?

Answer 1

Let $X$ be a random variable, which denotes the call duration in minutes: $X\sim N(\mu, \sigma ^2).$

Then $Z=\dfrac{X-\mu}{\sigma}\sim N(0,1).$

Given $\mu=16.3\ min, \sigma=4.2\ min$

$P(X>18)=1-P(X\leq18)=$

$=1-P(Z\leq\dfrac{18-16.3}{4.2})\approx1-P(Z\leq0.4047619)\approx$

$\approx1-0.657174=0.342826$

$34.28\%$