Answer:
A a house in that market that sells for $300,000 is unusual.
Step-by-step explanation:
Let the random variable X denote the price of a house and the random variable Y denote the living area of a house.
The number of houses surveyed by the real estate agent is, n = 1057.
Assume that both the random variables, X and Y are approximately normally distributed.
That is,
[tex]X\sim N(\$169400,\ \$68438)\\\\Y\sim N(2058\ \text{sq. ft.},\ 790\ \text{sq. ft.})[/tex]
To compute the probability of a Normal distribution we first need to convert the raw scores to z-scores.
[tex]z=\frac{\text{Raw score}-\mu}{\sigma}[/tex]
A z-score higher than 1.96 and lower than -1.96 are considered unusual. The values having these z-scores are considered as outliers.
(1)
Compute the z-score for X = 300000 as follows:
[tex]z=\frac{X-\mu}{\sigma}\\\\=\frac{300000-169400}{68438}\\\\=2.70[/tex]
(2)
Compute the z-score for Y = 3000 as follows:
[tex]z=\frac{Y-\mu}{\sigma}\\\\=\frac{3000-2058}{790}\\\\=1.19[/tex]
The z-score for a house in that market that sells for $300,000 is more than 1.96.
This implies, that the price $300,000 is unusually high.
The complete statement is:
The house that sells for $300 comma 000 has a z-score of 2.70 and the house with 3000 sq ft has a z-score of 1.19.