38Normal probability,
. Using , find .
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.
. Using , find .
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.