Assume that adults have IQ scores that are normally distributed with a mean 105 of and a standard deviation 20 . Find the probability that a randomly selected adult has an IQ between 88 and 122 .

0.6048

Step-by-step explanation:

Given the following :

Assume a normal distribution :

Mean (m) = 105

Standard deviation (sd) = 20

Find the probability that a randomly selected adult has an IQ between 88 and 122 .

Z = (x - m) / sd

For IQ score of 88:

Z = (88 - 105) / 20

Z = (-17) / 20

Z = - 0.85

For IQ score of 122:

Z = (122 - 105) / 20

Z = (17) / 20

Z = 0.85

P(-0.85<Z<0.85) = P(Z < 0.85) - P(Z < - 0.85)

P(-0.85<Z<0.85) = (0.8023 - 0.1977)

P(-0.85<Z<0.85) = 0.6048

ou would have 12 of each