Mathematics, 12.03.2021 21:50 ctyrector

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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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 $\mu$ 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.

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 $\mu = 2350, \sigma = 260$

Sample of 65 one-bedroom apartments

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

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.

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

By the Central Limit Theorem

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

$Z = \frac{2260 - 2350}{32.249}$

Z = -2.79

Z = -2.79 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.

Answer from: Quest

i used to know but i learned it a long time ago sorry

step-by-step explanation:

Answer from: Quest

step-by-step explanation:

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