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?

Respuesta :

Answer:

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 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 [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.

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

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

This means that [tex]p = 0.23[/tex]

Sample of 800 homes

This means that [tex]n = 800[/tex]

Mean and Standard deviation:

[tex]\mu = p = 0.23[/tex]

[tex]\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.23*0.77}{800}} = 0.0149[/tex]

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

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

[tex]Z = \frac{0.25 - 0.23}{0.0149}[/tex]

[tex]Z = 1.34[/tex]

[tex]Z = 1.34[/tex] has a pvalue of 0.9099

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

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