37Normal probability, -score
IQ scores are . Using , find .
Standardize: z = (130−100)/15 = 2. Then P(X>130) = P(z>2) = 1 − Φ(2) = 1 − 0.9772 = 0.0228. So about a 2.28% chance.
IQ scores are . Using , find .
Standardize: z = (130−100)/15 = 2. Then P(X>130) = P(z>2) = 1 − Φ(2) = 1 − 0.9772 = 0.0228. So about a 2.28% chance.