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:

$P(x14)_{}=1-P(x\leq14)$

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

$z=\frac{(14-12)}{2}=1$

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