Question

2. Some IQ tests are standardized to a Normal model with a mean of 100 and a standard deviation of 16. a. What score would begin the interval for the top 16% of all scores? Use the Empircal Rule (68/95/99.7%) to find your answer. Validate this answer using the Z-scores table. b. The top 10% of all scores represent the label of "genius". What is the range of scores for anyone who qualifies as a genius? C. What proportion of test takers score a 130 or higher?

Answer 1

Normal distribution: P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = 100

Standard deviation = 16

a) According to the empirical rule, 68% of data values lie within 1 standard deviation of mean.

Therefore, 100 - 38 = 32% values are either below 1 standard deviation or above 1 standard deviation.

32/2 = 16% is above 1 standard deviation of mean

100 + 1x16 = 116

Top 16% values are above 116

Verification using z score

Let T be the value that represent the lowest score in top 16%

P(X > T) = 0.16

P(X < T) = 0.84

P(Z < (T - 100)/16 = 0.84

Take the z score corresponding to 0.84 from standard normal distribution table.

(T - 100)/16 = 0.99

T = 115.84

b) Let G be the least score to get the label of "Genius"

P(X > G) = 0.10

P(X < G) = 1 - 0.10 = 0.90

P(Z < (G - 100)/16) = 0.90

(G - 100)/16 = 1.28

G = 120.48

A score of 120.48 or above will be labeled as "Genius"

c) P(X > 130) = 1 - P(X < 130)

= 1 - P(Z < (130 - 100)/16)

= 1 - P(Z < 1.875)

= 1 - 0.9696

= 0.0304