52.5k views
2 votes
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.

User Simon Fox
by
8.1k points

1 Answer

7 votes

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.

Explanation:

Z-score:

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


Z = (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 = (X - \mu)/(\sigma)


Z = (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 = (X - \mu)/(\sigma)


Z = (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 = (X - \mu)/(\sigma)


Z = (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.

User La Bla Bla
by
8.4k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.