Answer to Question #97568 in Statistics and Probability for Juliet Beglaryan

Question

2. A convention of sumo wrestlers is held at a hotel. The weight of sumo wrestlers can be approximated by a normal with mean 540 pounds and standard deviation of 45 pounds. The hotel elevator can accommodate 9 wrestlers at a time. Suppose that a simple random sample of 9 wrestlers enters the elevator. The elevator will fail if the total weight of the occupants exceeds 5000 pounds. What is the chance that the elevator will fail to operate?

Answer 1

If the total weight of the 9 wrestlers exceeds 5000 pounds, then the mean weight $\bar{x}$ of the 9 wrestlers must exceed $5000/9$ pounds.

Change the criterion for failing to operate from "total weight exceeding 5000 pounds" to "mean weight for 9 people exceeding $5000/9$ pounds".

Let $X=$ the mean weight in pounds: $X\sim N(\mu,\sigma^2)$

Then

$$Z={X-\mu \over \sigma}\sim N(0,1)$$

Given that $\mu=540$ pounds, $\sigma=45/\sqrt{9}$ pounds.

$$P(X>{5000 \over 9})=1-P(X\leq{5000 \over 9})=$$

$$=1-P(Z\leq{{5000 \over 9}-540 \over 15})\approx1-0.850141\approx0.1499$$

The probability that the elevator will fail to operate is $0.1499.$