Questions: An elevator has a placard stating that the maximum capacity is 1690 lb-10 passengers. So, 10 adult male passengers can have a mean weight of up to 1690 / 10=169 pounds. If the elevator is loaded with 10 adult male passengers, find the probability that it is overloaded because they have a mean weight greater than 169 lb. (Assume that weights of males are normally distributed with a mean of 177 lb and a standard deviation of 31 lb.) Does this elevator appear to be safe? The probability the elevator is overloaded is (Round to four decimal places as needed.)

An elevator has a placard stating that the maximum capacity is 1690 lb-10 passengers. So, 10 adult male passengers can have a mean weight of up to 1690 / 10=169 pounds. If the elevator is loaded with 10 adult male passengers, find the probability that it is overloaded because they have a mean weight greater than 169 lb. (Assume that weights of males are normally distributed with a mean of 177 lb and a standard deviation of 31 lb.) Does this elevator appear to be safe?

The probability the elevator is overloaded is
(Round to four decimal places as needed.)

Transcript text: An elevator has a placard stating that the maximum capacity is $1690 \mathrm{lb}-10$ passengers. So, 10 adult male passengers can have a mean weight of up to $1690 / 10=169$ pounds. If the elevator is loaded with 10 adult male passengers, find the probability that it is overloaded because they have a mean weight greater than 169 lb . (Assume that weights of males are normally distributed with a mean of 177 lb and a standard deviation of 31 lb .) Does this elevator appear to be safe? The probability the elevator is overloaded is $\square$ (Round to four decimal places as needed.)

Solution

Solution Steps

Step 1: Define the Problem

We need to find the probability that the mean weight of 10 adult male passengers exceeds $ 169 \, \text{lb} $. The weights of adult males are normally distributed with a mean $ \mu = 177 \, \text{lb} $ and a standard deviation $ \sigma = 31 \, \text{lb} $.

Step 2: Calculate the Z-scores

To determine the probability of exceeding the mean weight of $ 169 \, \text{lb} $, we first calculate the Z-score for $ 169 \, \text{lb} $ using the formula:

\[ Z = \frac{X - \mu}{\sigma / \sqrt{n}} \]

Where:

$ X = 169 \, \text{lb} $
$ \mu = 177 \, \text{lb} $
$ \sigma = 31 \, \text{lb} $
$ n = 10 $

Calculating the Z-score:

\[ Z = \frac{169 - 177}{31 / \sqrt{10}} \approx -0.8161 \]

Step 3: Calculate the Probability

The probability that the mean weight exceeds $ 169 \, \text{lb} $ is given by:

\[ P(X > 169) = 1 - P(X \leq 169) = 1 - \Phi(Z_{start}) \]

Where $ \Phi $ is the cumulative distribution function (CDF) of the standard normal distribution. Since $ Z_{end} = \infty $, we have:

\[ P(X > 169) = \Phi(\infty) - \Phi(-0.8161) = 0.7928 \]

Step 4: Conclusion on Safety

The calculated probability of the elevator being overloaded is $ P = 0.7928 $. Since this probability is significantly greater than $ 0.5 $, we conclude that the elevator does not appear to be safe.