Mathematics, 30.03.2021 08:20 student10190

5) The National Association of Realtors estimatest that 23% of all homes purchased in 2004 were considered investment properties. If a random sample of 800 homes sold in 2004 is obtained, what is the probability that al
most 25% of those are used as investment property?

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Answer from: coreybro

0.9099 = 90.99% probability that at most 25% of those are used as investment property.

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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

23% of all homes purchased in 2004 were considered investment properties.

This means that p = 0.23

Sample of 800 homes

This means that n = 800

Mean and Standard deviation:

$\mu = p = 0.23$

$\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.23*0.77}{800}} = 0.0149$

What is the probability that at most 25% of those are used as investment property?

This is the pvalue of Z when X = 0.25. So

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

$Z = \frac{0.25 - 0.23}{0.0149}$

Z = 1.34

Z = 1.34 has a pvalue of 0.9099

0.9099 = 90.99% probability that at most 25% of those are used as investment property.

Answer from: Quest

q: 92 degrees

hope i ^-^

Answer from: Quest

show us the graphs dude

step-by-step explanation:

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