The time spent waiting in the line is approximately normally

The time spent waiting in the line is approximately normally distributed. The mean waiting time is 4 minutes and the variance of the waiting time is 4. Find the probability that a person will wait for more than 2 minutes. Round answer to 4 decimal places.

Thanks!

Solution

So the probability that a person will wait for more than 2 minutes is

P(X>2) = P((X-mean)/s >(2-4)/2)

=P(Z>-1) = 0.8413 (from standard normal table)