Probability that at least 175 homes are investment property

Expert intro: This is a binomial probability problem based on a sample proportion. Because the sample size is large, a normal approximation is appropriate, with a continuity correction to improve accuracy.

Detailed walkthrough

Let $X$ be the number of homes in the sample that are investment properties.

Since 23% are estimated to be investment properties,

$X\sim \text{Bin}(800,0.23)$

We want:

$P(X\ge 175)$

Use a normal approximation:

$\mu=np=800(0.23)=184$

$\sigma=\sqrt{np(1-p)}=\sqrt{800(0.23)(0.77)}\approx 10.60$

Apply a continuity correction:

$P(X\ge 175)\approx P(X\ge 174.5)$

Standardize:

$z=\frac{174.5-184}{10.60}\approx -0.90$

So

$P(X\ge 175)\approx P(Z\ge -0.90)$

$=P(Z\le 0.90)\approx 0.816$

Therefore, the probability is approximately

$\boxed{0.816}$

💡 Pitfall guide

A common mistake is to use the binomial model without checking whether a normal approximation is reasonable. Here $np=184$ and $n(1-p)=616$, so the approximation is valid. Another error is forgetting the continuity correction; using 175 instead of 174.5 can slightly change the result.

🔄 Real-world variant

If the question asked for exactly 175 homes, you would approximate

$P(174.5<X<175.5)$

instead of a one-sided tail. If the sample size changed, you would recalculate both the mean and standard deviation using the new $n$ while keeping $p=0.23$ the same.

🔍 Related terms

binomial distribution, normal approximation, continuity correction