calculate the z score = (x - mean)/sd = (132=105)/15 = 27/15 = 9/5 = 18/10 = 1.8 which is close to 2, but a little less. An online z table calculator shows Pr(<z)= 0.96407 when z=1.8

had z been 2, 90% of the adults would have been between -2 and 2 standard deviations. But -1.8 and 1.8 standard deviations is a little less than 90%

the other 10% would be divided as 5% above 132 and 5% below 105-27= 78.

For z=2, P(<z) = 0.95, for z=1.8 P(<z) = 0.96407