silemus.

37Normal probability, zz-score

IQ scores are N(100,152)\displaystyle N(100, 15^2). Using Φ(2)=0.9772\displaystyle \Phi(2) = 0.9772, find P(score>130)\displaystyle P(\text{score} > 130).

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.

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