Question

In the university library elevator there is a sign indicating a 16-person limit as well as a weight limit of 2750 pounds. Suppose that the weight of students, faculty, and staff is approximately Normally distributed with a mean weight of 160 pounds and a standard deviation of 27 pounds. What is the probability that the random sample of 16 people in the elevator will exceed the weight limit? Round answer to three decimal places

Answer 1

Answer:

0.039 = 3.9% probability that the random sample of 16 people in the elevator will exceed the weight limit

Step-by-step explanation:

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

Normal probability distribution

Problems of normally distributed samples are 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:

$Z = \frac{X - \mu}{\sigma}$

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 $s = \frac{\sigma}{\sqrt{n}}$.

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

If n variables are added, the mean is $n\mu$ and the standard deviation is $s = \sqrt{n}\sigma$

In this problem:

$n = 16, \mu = 160*16 = 2560, s = \sqrt{16}*27 = 108$

What is the probability that the random sample of 16 people in the elevator will exceed the weight limit?

This is 1 subtracted by the pvalue of Z when X = 2750. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{2750 - 2560}{108}$

$Z = 1.76$

$Z = 1.76$ has a pvalue of 0.961

1 - 0.961 = 0.039

0.039 = 3.9% probability that the random sample of 16 people in the elevator will exceed the weight limit