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
and standard deviation
, the z-score of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
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
and standard deviation ![s = \sqrt{\frac{p(1-p)}{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D)
23% of all homes purchased in 2004 were considered investment properties.
This means that ![p = 0.23](https://tex.z-dn.net/?f=p%20%3D%200.23)
Sample of 800 homes
This means that ![n = 800](https://tex.z-dn.net/?f=n%20%3D%20800)
Mean and Standard deviation:
![\mu = p = 0.23](https://tex.z-dn.net/?f=%5Cmu%20%3D%20p%20%3D%200.23)
![\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.23*0.77}{800}} = 0.0149](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B0.23%2A0.77%7D%7B800%7D%7D%20%3D%200.0149)
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}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{0.25 - 0.23}{0.0149}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B0.25%20-%200.23%7D%7B0.0149%7D)
![Z = 1.34](https://tex.z-dn.net/?f=Z%20%3D%201.34)
has a pvalue of 0.9099
0.9099 = 90.99% probability that at most 25% of those are used as investment property.