Question

Many families in California are using backyard structures for home offices, art studios, and hobby areas as well as for additional storage. Suppose that the mean price for a customized wooden, shingled backyard structure is $3,100. Assume that the standard deviation is $1,400.

Required:
a. What is the z-score for a backyard structure costing $2300?
b. What is the z-score for a backyard structure costing $4900?
c. Interpret the z-scores in parts (a) and (b). Comment on whether either should be considered an outlier.
d. If the cost for a backyard shed-office combination built in Albany, California, is $13,000, should this structure be considered an outlier? Explain.

Answer 1

Answer:

a) The z-score for a backyard structure costing $2300 is -0.57.

b) The z-score for a backyard structure costing $4900 is 1.29

c) A backyard structure costing $2300 costs 0.57 standard deviations below the mean, while a backyard structure costing $4900 costs 1.29 standard deviations above the mean. Since both are within 2 standard deviations of the mean, none is an outlier.

d) Since this combination costs more than 2 standard deviations from the mean, yes, it should be considered an outlier.

Step-by-step explanation:

Z-score:

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If the z-score is more than two standard deviations from the mean(lesser than -2 or more than 2), the score X is considered an outlier.

In this question, we have that:

$\mu = 3100, \sigma = 1400$

a. What is the z-score for a backyard structure costing $2300?

We have to find Z when $X = 2300$. So

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

$Z = \frac{2300 - 3100}{1400}$

$Z = -0.57$

The z-score for a backyard structure costing $2300 is -0.57.

b. What is the z-score for a backyard structure costing $4900?

We have to find Z when $X = 2300$. So

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

$Z = \frac{4900 - 3100}{1400}$

$Z = 1.29$

The z-score for a backyard structure costing $4900 is 1.29

c. Interpret the z-scores in parts (a) and (b). Comment on whether either should be considered an outlier.

A backyard structure costing $2300 costs 0.57 standard deviations below the mean, while a backyard structure costing $4900 costs 1.29 standard deviations above the mean. Since both are within 2 standard deviations of the mean, none is an outlier.

d. If the cost for a backyard shed-office combination built in Albany, California, is $13,000, should this structure be considered an outlier?

We have to find the z-score when X = 13000. So

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

$Z = \frac{13000 - 3100}{1400}$

$Z = 7.07$

Since this combination costs more than 2 standard deviations from the mean, yes, it should be considered an outlier.