A lift and a weight limit

Topic: Functions of random variables and limit theorems · more questions on this topic
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A lift carries at most 650 kg. A person’s weight is \operatorname{N}(75, 10^2), and weights are independent. Let Z_n be the total weight of n people.

1. Find \mathbb{P}(Z_8\ge 650). Use \Phi(1.77)\approx 0.962.

2. Find the largest n such that \mathbb{P}(Z_n\ge 650)\le 0.01.

Worked answer (click to show)

1. Z_8\sim\operatorname{N}(600, 800).

\mathbb{P}(Z_8\ge 650)=1-\Phi\bigl((650-600)/\sqrt{800}\bigr)=1-\Phi(1.768)\approx 1-0.962=0.038.

2. Z_n\sim\operatorname{N}(75n, 100n), so \mathbb{P}(Z_n\ge 650)=1-\Phi\bigl((650-75n)/(10\sqrt{n})\bigr). At n=8 the probability is about 0.038. At n=7 the standardised point is (650-525)/(10\sqrt{7})\approx 4.72, and 1-\Phi(4.72) is far below 0.01. The largest such n is 7.