Question

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:

Answer 1

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

Answer 2

Answer: The proportion adults with scores between 80 and 120 is closest to 0.8176 or 81.76%.

We follow these steps in order to find the answer:

We first need to the Z scores for X = 80 and X =120

We calculate the Z scores as :

$Z = \frac{x - \mu }{\sigma}$

Substituting we get,

$Z_{1} = \frac{80 - 100 }{15} = -1.33333$

and

$Z_{2} = \frac{120 - 100 }{15} = 1.33333$

The area to the left of Z₁ is 0.09121122 , while the area to the left of Z₂ is 0.90878878

So, we need to find the area between Z₂ and Z₁ in order to arrive at the proportion between adults with scores 80 and 120.

$\mathbf{Z_{2} - Z{1}= 0.90878878-0.09121122 = 0.817577561}$