42. Many families in California are using backyard structures for home offices,

Question

42. 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 custom- ized wooden, shingled backyard structure is $3100. 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. If the cost for a backyard shed-office combination built in Albany, California, is $13,000, should this structure be considered an outlier? Explain. d.

Explanation / Answer

here as z score =(X-mean)/std deviation

a) z score =(2300-3100)/1200=-0.67

b)z score =(4900-3100)/1200=1.5

c)as z score is measure of distance of a score from mean of distribution in terms of standard deviation.

theefore from a) and b) 2300 is 0.67 standard deviation below the mean and 4900 is 1.5 standard deviation above the mean.

as both scores are with in 3 standard deviation from mean ; therefore none should be considered an outlier.

d)

for 13000 ; zscore =(13000-3100)/1200

=8.25

as this score is far away from mean ; and z score is greater than 3 ; therefore it should be considered an outlier.

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