b) Explanation: Let we should replace the bulb in X hours, if we want not more than 2% of the bulbs. P(X≤x) = 2% P(X-" < = 0.02 P(Z < = 0.02 X-500 50 Using MS Excel: X-500 50 X-500 50 =NORM.S.INV(0.02) NORM.S.INV(probability) -2.054 P(Z < -2.054) = = - 2.054 X = 397.3 0.02 Hence, We should replace the bulb within 397.3 hours

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