The high costs in the California real estate market have caused families who cannot afford to buy bigger homes to consider backyard sheds as an alternative form of housing expansion. Many are using the backyard structures for home offices, art studios, and hobby areas as well as for additional storage. The mean price of a customized wooden, shingled backyard structure is \$3100 (Newsweek, September 29, 2003). Assume that the standard deviation is \$1200.

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. The Newsweek article described a backyard shed-office combination built in Albany, California, for \$13,000. Should this structure be considered an outlier? Explain.

#### Solution

Let $X$ = Backyard structure cost.

The $Z$ score formula is `$Z= \dfrac{X-\mu}{\sigma}$`

.

$\mu = 3100$ and $\sigma =1200$.

(a) The z-score for a backyard structure costing \$2300 is

` $$ \begin{aligned} Z&= \frac{2300-3100}{1200}\\ &= -0.667 \end{aligned} $$ `

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

` $$ \begin{aligned} Z&= \frac{4900-3100}{1200}\\ &= 1.5 \end{aligned} $$ `

(c) As the $Z$-score in part (a) and (b) is not greater than 3.5, so either should not be considered as an outlier.

(d) The z-score for a backyard structure costing \$13000 is

` $$ \begin{aligned} Z&= \frac{13000-3100}{1200}\\ &= 8.25 \end{aligned} $$ `

As the value of $Z$-score is far more than 3.5, backyard structure costing \$ 13000 should be considered as an outlier.

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators