Questions: An elevator has a placard stating that the maximum capacity is 4500 lb-29 passengers. So, 29 adult male passengers can have a mean weight of up to 4500 / 29=155 pounds. Assume that weights of males are normally distributed with a mean of 180 lb and a standard deviation of 33 lb.
a. Find the probability that 1 randomly selected adult male has a weight greater than 155 lb.
b. Find the probability that a sample of 29 randomly selected adult males has a mean weight greater than 155 lb.
c. What do you conclude about the safety of this elevator?
Transcript text: An elevator has a placard stating that the maximum capacity is $4500 \mathrm{lb}-29$ passengers. So, 29 adult male passengers can have a mean weight of up to $4500 / 29=155$ pounds. Assume that weights of males are normally distributed with a mean of 180 lb and a standard deviation of 33 lb .
a. Find the probability that 1 randomly selected adult male has a weight greater than 155 lb .
b. Find the probability that a sample of 29 randomly selected adult males has a mean weight greater than 155 lb .
c. What do you conclude about the safety of this elevator?
Solution
Solution Steps
Step 1: Calculating the probability for 1 randomly selected adult male
Given a maximum capacity of 4500lb for 29 passengers, each passenger's mean weight limit is 155.1724lb.
Using the Z-score formula for a single observation: $Z = \frac{X - \mu}{\sigma} = \frac{155.172 - 1}{33} = -0.752.$
The probability that 1 randomly selected adult male has a weight greater than 155.1724lb is approximately 0.774.
Step 2: Calculating the probability for a sample of N randomly selected adult males
Using the Z-score formula for a sample mean: $Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{N}} = \frac{155.172 - 1}{33 / \sqrt{29}} = -4.051.$
The probability that a sample of 29 randomly selected adult males has a mean weight greater than 155.1724lb is approximately 1.
Final Answer:
Based on the calculated probabilities, the elevator may not be considered safe due to a high probability of overload.