Question

In a certain section of Southern California, the distribution of monthly rent for a one-bedroom apartment has a mean of $2,275 and a standard deviation of $290. The distribution of the monthly rent does not follow the normal distribution. In fact, it is positively skewed. What is the probability of selecting a sample of 65 one-bedroom apartments and finding the mean to be at least $2,095 per month

Answer 1

Answer:

100% probability of selecting a sample of 65 one-bedroom apartments and finding the mean to be at least $2,095 per month

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

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, 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.

Mean of $2,275 and a standard deviation of $290.

This means that $\mu = 2275, \sigma = 290$

Sample of 65:

This means that $n = 65, s = \frac{290}{\sqrt{65}}$

Finding the mean to be at least $2,095 per month

This is 1 subtracted by the p-value of Z when X = 2095. So

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

By the Central Limit Theorem

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

$Z = \frac{2095 - 2275}{\frac{290}{\sqrt{65}}}$

$Z = -5$

$Z = -5$ has a p-value of 0.

1 - 0 = 1

100% probability of selecting a sample of 65 one-bedroom apartments and finding the mean to be at least $2,095 per month