Question

the long-distance calls made by employees of a company are normally distributed with a mean of 6.3 minutes and a standard deviation of 2.2 minutes. Find the probability that a call
a. last between 5 and 10 minutes
b. last more than 7 minutes
c. last less than 4 minutes

Answer 1

Let $X$ be a random variable which denotes the call duration in minutes. Our random variable $X$ is normal

$X\sim N(\mu,\sigma^2)$. Then

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

Given that $\mu=6.3 \ minutes, \sigma=2.2 \ minutes.$

a.

$$P(5<X<10)=P\bigg({5-6.3 \over 2.2}<Z<{10-6.3 \over 2.2}\bigg)=$$ $$=P\bigg(Z<{3.7 \over 2.2}\bigg)-P\bigg(Z<{-1.3 \over 2.2}\bigg)\approx$$ $$\approx0.953698-0.277291\approx0.6764$$

b.

$$P(X>7)=P\bigg(Z>{7-6.3 \over 2.2}\bigg)=1-P\bigg(Z<{0.7\over 2.2}\bigg)\approx$$ $$1-0.624826\approx0.3752$$

c.

$$P(X<4)=P\bigg(Z<{4-6.3\over 2.2}\bigg)=P\bigg(Z<{-2.3\over 2.2}\bigg)\approx$$ $$\approx0.1479$$