Question

We collected a sample of the prices of new homes. The mean of our sample is $155,000, with a standard deviation of $15,000. Calculate the z-scores of each of the given prices. Determine if the they are usual or unusual. Round up each z-score to two decimals. (a) $200, 000 (b) $55,000 (c) $175,000 (d) $122,000

Answer 1

Answer:

(a) $200, 000, z-score= 3 and it is unusual.

(b) $55,000, z-score= -6.67 and it is unusual.

(c) $175,000, z-score= 1.33 and it is usual.

(d) $122,000, z-score= -2.2 and it is unusual

Explanation:

Given: Mean of sample= $155000

           Standard deviation= $15000.

Now, calculating z-score of each given prices.

z-score= $\frac{x-mean}{standard\ deviation}$

(a) Price= $200000

$z-score = \frac{200000-155000}{15000}$

⇒$z-score= \frac{\ 45000}{\ 15000} = 3$

It is unusual as score is very high.

b) $ 55000

$z-score = \frac{55000-155000}{15000}$

⇒$z-score = \frac{-100000}{15000}$

∴ $z-score= -6.67$

It is unusual again as score it very low.

c) $ 175000

$z-score = \frac{175000-155000}{15000}$

⇒ $z-score = \frac{20000}{15000}= 1.33$

It is usual as score is in the top 0.30

d) $122000

$z-score = \frac{122000-155000}{15000}$

⇒ $z-score = \frac{33000}{15000}$

∴$z-score= -2.2$

It is unusual as score is too low