Question

Assume a normal distribution and that the average phone call in a certain town lasted 12 min with a standard deviation of 2 min What percentage of the calls lasted lessthan 10 min?

Answer 1

Notice that:

`P(x<10)=P(x>14)_{}=1-P(x\leq14)`

Now, we use the formula for the z-score:

`z=((14-12))/(2)=1`

Then the percentage of the calls that lasted less than 10 min is 15.87%