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38Normal probability, 40<X<6040 < X < 60

X∼N(50,102)\displaystyle X \sim N(50, 10^2). Using Φ(1)=0.8413\displaystyle \Phi(1) = 0.8413, find P(40<X<60)\displaystyle P(40 < X < 60).

Standardise 40 and 60 into z-scores: z = (x−50)/10 gives z = −1 and z = 1. Then P(40<X<60) = 2Φ(1) − 1 = 2(0.8413) − 1 = 0.6826, matching the 68% rule within one standard deviation.

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