Chance an elevator of 16 people is overloaded

Question

Passenger weights average 75 kg with a standard deviation of 15 kg. Weights are independent, so by the central limit theorem the total weight of a full car is approximately normal. With n = 16 passengers, what is the probability the total weight exceeds the posted capacity of 1350 kg?

Accepted Answer

We want the total weight of 16 passengers, so mean and variance add. The total is approximately normal with 	ext{mean} = n\mu = 16 	imes 75 = 1200 	ext{ kg}, 	ext{SD} = \sigma\sqrt{n} = 15 	imes \sqrt{16} = 15 	imes 4 = 60 	ext{ kg}. Find the z-score of the 1350 kg capacity: z = \frac{1350 - 1200}{60} = \frac{150}{60} = 2.5. From the normal table: P(	ext{total} 1350) = P(Z 2.5) \approx 0.0062, about 0.6\%. The expected load is 1200 kg with a 60 kg spread, so hitting 1350 kg is a 2.5-standard-error event, uncommon but not negligible for a safety margin. Takeaway. For a sum, mean = n\mu…

Chance an elevator of 16 people is overloaded

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