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 $7,040 (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.)

for part b I got 6703 but it says it is wrong?

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

Solution :

Given that ,

mean = = $ 7040

standard deviation = = $ 500

a) P(x > 8000) = 1 - p( x< 8000)

=1- p P[(x - ) / < (8000 - 7040) / 500]

=1- P(z < 1.92)

Using z table,

= 1 - 0.9726

= 0.0274

b) Using standard normal table,

The z dist'n First quartile is,

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.675 ) = 0.25

z = -0.675

Using z-score formula,

x = z * +

x = -0.675 * 500 + 7040

x = 6702.5

First quartile =Q1 = $ 6702.5

c) The z dist'n Third quartile is,

P(Z < z) = 75%

= P(Z < z) = 0.75  

= P(Z < 0.675 ) = 0.75

z = 0.675

Using z-score formula,

x = z * +

x = 0.675 * 500 + 7040

x = 7377.5

Third quartile =Q3 = $ 7377.5

d) IQR = Q3 - Q1

IQR = $ 7377.5 - $ 6702.5

IQR = $ 675.0