The time spent waiting in the line is approximately normally distributed. The mean waiting time is 7 minutes and the variance of the waiting time is 9. Find the probability that a person will wait for more than 13 minutes.

Answer :

Given info :

The population mean $(μ) = 7$

The population variance, $(σ^2) = 9$

∴ $σ = 3$

Solution :

The probability that a person will wait for more than 13 minutes :

$$\text{P}(x > 13) = \text{P}\left(\frac{x – \mu}{\sigma} > \frac{13 – 7}{3}\right)$$ $$= \text{P}(z > 2)$$ $$= 1 – \text{P}(z < 2)$$ $$= 1 – 0.9772$$ $$= 0.0228$$