Question

1. What z-score value separates 20% of the distribution in the tail on the left (i.e., the bottom 20% of the distribution) from the rest of the distribution?

2. What z-score value separates 40% of the distribution in the tail on the right (i.e., the top 40% of the distribution) from the rest of the distribution?

3. IQ scores are standardized to produce a normal distribution with a mean of µ = 100 and a standard deviation of σ = 15. Find the proportion of the population in the following IQ category:

Genius or near genius: IQ greater than 140.

Very superior intelligence: IQ between 120 and 140

Average of normal intelligence: IQ between 90 and 109

Answer 1

(1) z score value that separates the bottom 20%, i.e. 20th percentile of the distribution is -0.842 [From z percentile table]

(2) z score value that separates the top 40% of the distribution, i.e. 60th percentile is 0.253 [From z percentile table]

(3) mean of µ = 100 and a standard deviation of σ = 15

(i) P(X>140) = P(z>(x-µ)/σ)

= P(z>(140-100)/15)

= P(z>2.67)

=1 -P(z<2.67)

= 1-0.9962

= 0.0038 [From z distribution table]

(ii) P(120<x<140) = P((x-µ)/σ<z<(y-µ)/σ)

= P((120-100)/15<z<(140-100)/15)

= P(1.33<z<2.67)

=P(z<2.67)-P(z<1.33)

= 0.9962 - 0.9082

= 0.0880 [From z distribution table]

(iii) P(90<x<109) = P((x-µ)/σ<z<(y-µ)/σ)

= P((90-100)/15<z<(109-100)/15)

= P(-0.67<z<0.6)

=P(z<0.6)-P(z<-0.67)

= 0.7257-0.2514

= 0.4743 [From z distribution table]