Adults have IQ scres that are normally disributed with a mean of 100 and a standard deviation of 15.Find the probability that a randomly selected...

The problem gives us that adults have an IQ score which is
randomly distributed and has a mean of 100 and a standard deviation of 15. We need to
find the probability that a randomly selected adult has an IQ between 110 and
120.

To do this we find the z-value which is given by z =
(variable - mean)/standard deviation.

So for the IQ = 110,
z = (110-100) / 15 = 2/3; and for IQ= 120, z = (120-100) / 15 =
4/3

Now we need to use a normal distribution table like the
one provided here: href="http://www.mathsisfun.com/data/standard-normal-distribution-table.html">Normal
distribution table

For z = 4/3 we get the
cumulative probability as 0.2454 and for z = 2/3 we get the cumulative probability as
0.4082. Therefore the probability between z=4/3 and z= 2/3 or the IQ between 120 and 110
is 0.4082 - 0.2454 = 0.1628

Therefore the
required probability is 0.1628