A group of realtors estimates that 23% of all homes purchased last year were considered investment properties. If a sample of 800 homes sold last year is obtained, what is the probability that at least 175 homes are going to be used as investment property? Round to four decimal places.

Using the normal approximation to the binomial, it is found that there is a 0.7881 = 78.81% probability that at least 175 homes are going to be used as investment property.

In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:

In this problem:

The mean and the standard deviation for the approximation are given by:

Using continuity correction, the probability that at least 175 homes are going to be used as investment property is , which is 1 subtracted by the p-value of Z when X = 174.5.

has a p-value of 0.2119.

1 - 0.2119 = 0.7881

0.7881 = 78.81% probability that at least 175 homes are going to be used as investment property.

A similar problem is given at link

Probability = 0.77488

Step-by-step explanation:

Since a sample was obtained, this problem is working with the sampling distribution of the sample proportion (p').l which we will need to find the probability.

A sampling distribution of the sample proportion has a mean equal to the population proportion (p).

The standard deviation of the distribution is equal to √((p(1-p))/n).

The sampling distribution in this problem will be normal because the sample size is large.

For this question, the sample proportion(p) is; 23% = 0.23

The sample size(n) is 800, so the standard deviation is;

SD = √((p(1-p))/n) = √((0.23(1-0.23))/800)

SD = 0.0149

The sample proportion is;

(p') = 175/800

p' = 0.21875

Now, for the Z-score;

z = (p' - mean)/SD

z = (0.21875 - 0.23)/ 0.0149

z = - 0.755

Now for the probability, since at least 175,it will be right tail. Thus from the table i attached;

P(z > -0.755) = 1 - 0.22663 = 0.77488

answer: a.   one is a factor of every whole number since every number is divisible by itself.