Question

U.S. consumers are increasingly viewing debit cards as a convenient substitute for cash and checks. The average amount spent annually on a debit card is $7,400 (Kiplinger’s, August 2007). Assume that the average amount spent on a debit card is normally distributed with a standard deviation of $500. [You may find it useful to reference the z table.]

a. A consumer advocate comments that the majority of consumers spend over $8,000 on a debit card. Find a flaw in this statement. (Round "z"value to 2 decimal places and final answer to 4 decimal places.)

b. Compute the 25th percentile of the amount spent on a debit card. (Round "z" value to 3 decimal places and final answer to 1 decimal place.)

c. Compute the 75th percentile of the amount spent on a debit card. (Round "z" value to 3 decimal places and final answer to 1 decimal place.)

d. What is the interquartile range of this distribution? (Round "z" value to 3 decimal places and final answer to 1 decimal place.)

Answer 1

(a)

= 7400

= 500

To find P(X>8000):

Z = (8000 - 7400)/500 = 1.20

Table of Area Under Standard Normal Curve gives area = 0.3849

So,

P(X>8000) = 0.5 - 0.3849 = 0.1151 = 11.51%

Since 11.51% of values are greater than 8000, it is incorrect to comment that the majority of consumers spend over $8000 on debit card.

(b) For area = 0.25, Table gives Z score = 0.675

So,

Z = - 0.675 = (X - 7400)/500

So,

X = 7400 - (0.675X 500) = 7400 - 337.5 = 7062.50

So,

25th Percentile is given by:

7062.50

(c)

(b) For area = 0.25, Table gives Z score = 0.675

So,

Z = 0.675 = (X - 7400)/500

So,

X = 7400 - (0.675X 500) = 7400 + 337.5 = 7737.50

So,

75th Percentile is given by:

7737.50

(d)

Interquartile range = 7737.50 - 7062.50 = 675

So,

Interquartile range is given by:

675