Chance 100 packages exceed a weight limit

Question

A courier's packages weigh 2 kg on average with a standard deviation of 0.5 kg each. Weights are independent, so by the central limit theorem the total weight of many packages is approximately normal. Over n = 100 packages, what is the probability the total weight exceeds 210 kg?

Accepted Answer

We want the total, so mean and variance both add across the 100 packages. The total weight is approximately normal with 	ext{mean} = n\mu = 100 	imes 2 = 200 	ext{ kg}, 	ext{SD} = \sigma\sqrt{n} = 0.5 	imes \sqrt{100} = 0.5 	imes 10 = 5 	ext{ kg}. Find the z-score of the 210 kg threshold: z = \frac{210 - 200}{5} = 2. From the normal table: P(	ext{total} 210) = P(Z 2) \approx 0.0228, about 2.3\%. The total sits near 200 kg with only a 5 kg spread, so exceeding 210 kg is a 2-standard-error event. Takeaway. For a sum, use mean = n\mu and SD = \sigma\sqrt{n}: the total's spread of 5 kg ma…

Chance 100 packages exceed a weight limit

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