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 3000. Assume that the standard deviation is 1100.

a. What is the -score for a backyard structure costing 2300 (to 2 decimals)? If your answer is negative, enter minus (-) sign.

b. What is the -score for a backyard structure costing 4800 (to 2 decimals)?

c. Interpret the -scores in parts (a) and (b). Comment on whether either should be considered an outlier. 2300 is ____standard deviations (select answer: below or above) the mean. 4800 is _____standard deviations (select answer: below or above) the mean.

Selct answer:

- the z score in part A is an outlier

- the z score in part B is an outlier

- both are outliers

- neither are outliers

d. If the cost for a backyard shed-office combination built in Albany, California, is 13000, should this structure be considered an outlier? Explain. 13000 is_____ (to 2 decimals) standard deviations (select answer: above or below) the mean. This cost (select answer: is or is not ) an outlier.

Answer 1

Any value below or above 2 standard deviation from the mean is considered as outlier

mean = 3000 and standard deviation is 1100

(A) score for a backyard structure costing 2300

z = (x-mean)/(standard deviation)

= (2300-3000)/1100

= -700/1100

= -0.64

(B) score for a backyard structure costing 4800

z = (x-mean)/(standard deviation)

= (4800-3000)/1100

= 1800/1100

= 1.64

(C) 2300 is __0.64__standard deviations below the mean. 4800 is ___1.64__standard deviations above the mean.

both are less than 2 standard deviations, so none of these is outlier

option D is correct

neither are outliers

(D) score for a backyard structure costing 13000

z = (x-mean)/(standard deviation)

= (13000-3000)/1100

= 10000/1100

= 9.09

It is 9.09 standard deviation above the mean, so it is an outlier.