Question

An elevator has a placard stating that the maximum capacity is 23252325 lblong dash—1515 passengers.​ So, 1515 adult male passengers can have a mean weight of up to 2325 divided by 15 equals 155 pounds.2325/15=155 pounds. If the elevator is loaded with 1515 adult male​ passengers, find the probability that it is overloaded because they have a mean weight greater than 155155 lb.​ (Assume that weights of males are normally distributed with a mean of 161 lb161 lb and a standard deviation of 31 lb31 lb​.) Does this elevator appear to be​ safe?

Answer 1

Answer:

0.7734; no, it does not.

Step-by-step explanation:

To find this probability we will use a z-score.  We are dealing with the probability of a sample mean being larger than a given value; this means we use the formula

$z=\frac{\bar{X}-\mu}{\sigma \div \sqrt{n}}$

Our x-bar in this problem is 155, as that is the sample mean we are trying to find the probability of.  Our mean, μ, is 161 and our standard deviation, σ, is 31.  Our sample size, n, is 15.  This gives us

$z=\frac{155-161}{31\div \sqrt{15}}=\frac{-6}{8.0042}=-0.75$

Using a z table, we see that the area under the curve less than this, corresponding with probability less than this value, is 0.2266.  This means that the probability of the sample mean being larger than this is

1-0.2266 = 0.7734, or 77.34%.

Since there is a 77.34% chance of the passengers having a mean weight that is too heavy, this elevator is not safe.