A large company has found that 40% of it employees own their own homes.

a.) If we randomly select a sample of 20 employees, what is the probability that the sample proportion that own their homes is over 50%?

b.) If we randomly select a sample of 40 employees, what is the probability that the sample proportion that own their homes in over 50%?

c.) If we randomly select a sample of 20 employees, what is the probability that the sample proportion that own their homes is between 30% and 50%?

Thanks!

To answer these questions, we can use the theory of sampling distributions and the Central Limit Theorem. The Central Limit Theorem states that if a sample size is sufficiently large, the sampling distribution of the sample proportion will be approximately normal, regardless of the shape of the underlying population.

In this case, the population proportion is known to be 0.40, and we want to find the probability of certain sample proportions.

a.) If we randomly select a sample of 20 employees, we want to find the probability that the sample proportion that own their homes is over 50%.

To calculate this, we need to find the z-score corresponding to a sample proportion of 50% or greater. The z-score is calculated using the formula:

z = (p - P) / sqrt((P * (1 - P)) / n)

where p is the sample proportion, P is the population proportion, and n is the sample size.

In this case, p = 0.50, P = 0.40, and n = 20. Plugging these values into the formula, we get:

z = (0.50 - 0.40) / sqrt((0.40 * (1 - 0.40)) / 20) = 1

Now, we need to find the probability corresponding to z = 1 using a standard normal distribution table or calculator. The probability is the area under the curve to the right of the z-score.

Using a standard normal distribution table, we find that the probability corresponding to z = 1 is approximately 0.1587.

Therefore, the probability that the sample proportion that own their homes is over 50% is approximately 0.1587.

b.) If we randomly select a sample of 40 employees, we can use the same approach to calculate the probability that the sample proportion that own their homes is over 50%.

Again, we need to find the z-score corresponding to a sample proportion of 50% or greater. Using the formula:

z = (p - P) / sqrt((P * (1 - P)) / n)

with p = 0.50, P = 0.40, and n = 40, we get:

z = (0.50 - 0.40) / sqrt((0.40 * (1 - 0.40)) / 40) ≈ 2.24

Finding the probability corresponding to z = 2.24 using a standard normal distribution table or calculator, we get approximately 0.0125.

Therefore, the probability that the sample proportion that own their homes is over 50% when we randomly select a sample of 40 employees is approximately 0.0125.

c.) If we want to find the probability that the sample proportion that own their homes is between 30% and 50% when we randomly select a sample of 20 employees, we can calculate it by finding the area under the curve between the two z-scores corresponding to 30% and 50%.

First, we find the z-score corresponding to a sample proportion of 30%:

z1 = (p1 - P) / sqrt((P * (1 - P)) / n) = (0.30 - 0.40) / sqrt((0.40 * (1 - 0.40)) / 20) ≈ -1

Next, we find the z-score corresponding to a sample proportion of 50%:

z2 = (p2 - P) / sqrt((P * (1 - P)) / n) = (0.50 - 0.40) / sqrt((0.40 * (1 - 0.40)) / 20) ≈ 1

Using a standard normal distribution table or calculator, we find the area to the left of z1 and the area to the left of z2. The probability that the sample proportion is between 30% and 50% is the difference between these two areas.

Let's call this probability P(x). We have:

P(x) = P(z1 < Z < z2) = P(Z < z2) - P(Z < z1)

Using a standard normal distribution table, we find:

P(Z < z2) ≈ 0.8413
P(Z < z1) ≈ 0.1587

Therefore, the probability that the sample proportion that own their homes is between 30% and 50% when we randomly select a sample of 20 employees is approximately 0.8413 - 0.1587 = 0.6826.

I hope this explanation helps!