Question

5.
The number of days a group of 200 homes is on the market is normally
distributed with a mean of 50 and a standard deviation of 12. Label the
normal distribution curve, then answer the questions.
a. What percent of the homes are on the market between 14 and 86 days?
b. What is the probability that a home is on the market for 62 days or more?
C. Approximately how many homes were on the market between 26 and 50
days?

Answer 1

Using the normal distribution, given the graph at the end of this problem, we have that:

a. 99.74% of the homes are on the market between 14 and 86 days.

b. 0.1587 = 15.87% probability that a home is on the market for 62 days or more.

c. Approximately 95 homes were on the market between 26 and 50 days.

Normal Probability Distribution

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

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

• The mean is of 50, hence $\mu = 50$.

• The standard deviation is of 12, hence $\sigma = 12$.

Item a:

The proportion is the p-value of Z when X = 86 subtracted by the p-value of Z when X = 14, hence:

X = 86:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{86 - 50}{12}$

$Z = 3$

$Z = 3$ has a p-value of 0.9987.

X = 14:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{14 - 50}{12}$

$Z = -3$

$Z = -3$ has a p-value of 0.0013.

0.9987 - 0.0013 = 0.9974.

0.9974 = 99.74% of the homes are on the market between 14 and 86 days.

Item b:

The probability is 1 subtracted by the p-value of Z when X = 62, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{62 - 50}{12}$

$Z = 1$

$Z = 1$ has a p-value of 0.8413.

1 - 0.8413 = 0.1587.

0.1587 = 15.87% probability that a home is on the market for 62 days or more.

Item c:

The proportion is the p-value of Z when X = 50 subtracted by the p-value of Z when X = 26, hence:

X = 50:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{50 - 50}{12}$

$Z = 0$

$Z = 0$ has a p-value of 0.5.

X = 26:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{26 - 50}{12}$

$Z = -2$

$Z = -2$ has a p-value of 0.0228.

0.5 - 0.0228 = 0.4772.

Out of 200 homes:

0.4772 x 200 = 95.4

Approximately 95 homes were on the market between 26 and 50 days.

You can learn more about the normal distribution at https://brainly.com/question/24663213