The scores on the wechsler adult intelligence scale are approximately normal with μ = 100 and σ = 15. the proportion of adults with scores between 80 and 120 is closest to:

Given that the mean is 100, and standard deviation is 15, the z-score of 80 will be:

z-score = (80-100) / 15=-1.33333

the z-score of 120 will be:

z-score = (120-100) / 15=1.3333

The probabilities associated with the z-score will be:

P (-1.33333) = 0.0918

also

P (1.3333) = 0.9082

therefore the proportion of adults with scores between 80 and 120 is:

0.9082-0.0918

=0.8164

=8.164%

=0.082