The rent for a one-bedroom apartment in Southern California follows the normal distribution with a mean of $2,350 per month and a standard deviation of $260 per month. The distribution of the monthly costs 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 $2,260 per month

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Answer:

0.9974 = 99.74% probability of selecting a sample of 65 one-bedroom apartments and finding the mean to be at least $2,260 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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

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

The rent for a one-bedroom apartment in Southern California follows the normal distribution with a mean of $2,350 per month and a standard deviation of $260 per month.

This means that [tex]\mu = 2350, \sigma = 260[/tex]

Sample of 65 one-bedroom apartments

65>30, which means that the sampling distribution of the sample means will be approximately normal. Also, [tex]n = 65, s = \frac{260}{\sqrt{65}} = 32.249[/tex]

What is the probability of selecting a sample of 65 one-bedroom apartments and finding the mean to be at least $2,260 per month?

This is 1 subtracted by the pvalue of Z when X = 2260.

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

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{2260 - 2350}{32.249}[/tex]

[tex]Z = -2.79[/tex]

[tex]Z = -2.79[/tex] has a pvalue of 0.0026

1 - 0.0026 = 0.9974

0.9974 = 99.74% probability of selecting a sample of 65 one-bedroom apartments and finding the mean to be at least $2,260 per month.