Question

.A study of 800 homeowners in a certain area showed that the average value of the homes was $82,000, and the standard deviation was $5000. If 50 homes are for sale, find the probability that the mean of the values of these homes is greater than $83,500.

Answer 1

There is 1.70% that 50 homes have a value greater than $83,500.

Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

$z=\frac{x-\mu}{\sigma/\sqrt{n} } \\\\where\ x=raw\ score,\mu=mean,\sigma=standard \ deviation,n=sample\ size$

μ = $82000, σ = 5000, n = 50. Hence for x > 83500:

$z=\frac{83500-82000}{5000/\sqrt{50} }\\\\\\z=2.12$

From the normal distribution table, P(x > 83500) = P(z > 2.12) = 1 - P(z < 2.12) = 1 - 0.9830 = 0.0170 = 1.70%

Find out more at: https://brainly.com/question/7449180