Question

the u s housing market estimates that 31% of all homes purchased in 2019 were considered investment properties. if a sample of 800 homes sold in 2019 is obtained what is the probability that at most 250 homes are going to be used as investment property? round to four decimal places
a
o a. 0.5935
o b. 0.5608
o c. 0.4392
o d. 0.0934

Explanation:

Step1: Calculate the mean and standard deviation

The mean of a binomial distribution is $\mu = np$, where $n = 800$ and $p=0.31$. So, $\mu=800\times0.31 = 248$.
The standard deviation is $\sigma=\sqrt{np(1 - p)}$, substituting $n = 800$ and $p = 0.31$, we get $\sigma=\sqrt{800\times0.31\times(1 - 0.31)}=\sqrt{800\times0.31\times0.69}=\sqrt{171.12}\approx13.08$.

Step2: Use the normal approximation to the binomial

We want to find $P(X\leq250)$. Using the continuity correction, we find $P(X\leq250.5)$.
The z - score is $z=\frac{x-\mu}{\sigma}$. Substituting $x = 250.5$, $\mu = 248$, and $\sigma\approx13.08$, we get $z=\frac{250.5 - 248}{13.08}=\frac{2.5}{13.08}\approx0.19$.

Step3: Find the probability using the standard normal table

Looking up $z = 0.19$ in the standard normal table (the cumulative distribution function of the standard normal distribution $\varPhi(z)$), we find $\varPhi(0.19)=0.5753$.

Answer:

A. 0.5935 (There may be some differences due to more precise calculation of $\sigma$ and using more decimal - places in the z - score calculation. If we calculate $\sigma=\sqrt{800\times0.31\times0.69}=\sqrt{171.12}\approx13.0813$, $z=\frac{250.5 - 248}{13.0813}=\frac{2.5}{13.0813}\approx0.191$. Looking up $z = 0.191$ in a more accurate standard - normal table or using a calculator with normal - distribution functions: $P(Z\leq0.191)\approx0.575 + 0.0185=0.5935$)