Question

An elevator has a placard stating that the maximum capacity is 1580 lb-10 passengers.​ So, 10 adult male passengers can have a mean weight of up to 1580/10 = 158 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 158 lb.​ (Assume that weights of males are normally distributed with a mean of 160 lb and a standard deviation of ​35 lb.) Does this elevator appear to be​ safe?

Answer 1

a) 0.5714 = 57.14% probability that it is overloaded because they have a mean weight greater than 158 lb

b) No, because the probability of being overloaded is considerably high(57.14%). Ideally, it should be under 5%, which would be considered an unusual event.

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Assume that weights of males are normally distributed with a mean of 160 lb and a standard deviation of ​35 lb.

This means that
`\mu = 160, \sigma = 35`

Sample of 10:

This means that
`n = 10, s = (35)/(√(10)) = 11.07`

a) Find the probability that it is overloaded because they have a mean weight greater than 158 lb.​

This is 1 subtracted by the pvalue of Z when X = 158. So

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (158 - 160)/(11.07)`

`Z = -0.18`

`Z = -0.18` has a pvalue of 0.4286

1 - 0.4286 = 0.5714

0.5714 = 57.14% probability that it is overloaded because they have a mean weight greater than 158 lb.

b.) Does this elevator appear to be​ safe?

No, because the probability of being overloaded is considerably high(57.14%). Ideally, it should be under 5%, which would be considered an unusual event.