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 $7,030 (Kiplinger’s, August 2007). Assume that the average amount spent on a debit card is normally distributed with a standard deviation of $400. [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

Solution :

Given that,

mean = = 7,030

standard deviation = = 400

a ) P ( x > 8,000 )

= 1 - P (x < 8,000 )

= 1 - P ( x -  / ) < (8,000 - 7,030 /400)

= 1 - P ( z < 970 / 400 )

= 1 - P ( z < 2.43 )

Using z table

= 1 - 0.9925

= 0.0075

Probability = 0.0075

b ) The 25^th percentile

P(Z < z) = 25%

P(Z < z) = 0.25

Using standard normal table,

P(Z < - 0.6745) = 0.99

z = -0.67

Using z-score formula,

x = z * +

x = - 0.67 * 400 + 7030

= 6762

The 25^th percentile = 6762

C ) The 75^th percentile

P(Z < z) = 75%

P(Z < z) = 0.75

Using standard normal table,

P(Z < 0.6745) = 0.99

z =0.67

Using z-score formula,

x = z * +

x = 0.67 * 400 + 7030

= 7298

The 75^th percentile = 7298

d ) The interquartile range

= Q_{3 -}  Q₁

= 7298 - 6762

= 536

The interquartile range = 536