"the national association of realtors estimates that 23% of all homes purchased in 2004 were considered investment properties. if a sample of 800 homes sold in 2004 is obtained what is the probability that at most 200 homes are going to be used as investment property?"

The probability is 0.9162.

The probability for this is given by:

[tex]P(X \leq 200) = \Sigma_{n=0}^{200} (0.23)^x(0.77)^{800-x}[/tex]

Using a graphing calculator, we get the probability is 0.9162.

Answer Link

MrRoyal MrRoyal · Answer 2 · 2021-12-09T18:13:31+01:00

The probability that at most 200 homes are going to be used as investment property is 0.91

The given parameter is:

[tex]p = 23\%[/tex] --- the mean of the distribution

[tex]n = 800[/tex] --- the population

[tex]x = 200[/tex] ---- the sample size

Start by calculating the standard deviation

[tex]\sigma = \sqrt {\frac{p (1-p)}{n}}[/tex]

So, we have:

[tex]\sigma = \sqrt {\frac{23\% \times (1-23\%)}{800}}[/tex]

[tex]\sigma =0.0149}[/tex]

Next, calculate the sample proportion (p-hat)

[tex]\bar p = \frac xn[/tex]

So, we have:

[tex]\bar p = \frac{200}{800}[/tex]

[tex]\bar p = 0.25[/tex]

Next, calculate the z-score as follows:

[tex]z = \frac{\bar p- p}{\sigma}[/tex]

So, we have:

[tex]z = \frac{0.25- 23\%}{0.0149}[/tex]

[tex]z = 1.34[/tex]

The required probability is then calculated as:

[tex]P(x \le 200) = P(z \le 1.34)[/tex]

Using z-scores of probability, we have:

[tex]P(x \le 200) = 0.90988[/tex]

Approximate

[tex]P(x \le 200) = 0.91[/tex]

Hence, the probability is 0.91