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 $7,070 (Kiplinger’s, August 2007). Assume that the average amount spent on a debit card is normally distributed with a standard deviation of $500.

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.)

The proportion of consumers who spend over $8,000

b.	Compute the 25th percentile of the amount spent on a debit card. (Round "z-value" to 2 decimal places.)

25th percentile

c.	Compute the 75th percentile of the amount spent on a debit card. (Round "z-value" to 2 decimal places.)

75th percentile

d.	What is the interquartile range of this distribution? (Round "z-value" to 2 decimal places.)

Answer 1

Solution: We are given:

Let X be the amount spent annually on a debit card

a. The proportion of consumers who spend over $8,000

Answer: We have to find

Find we need to find the z score corresponding to

We know that:

  

  

  

Now we have to find . Using the standard normal table, we have:

Therefore, the proportion of of consumers who spend over $8,000 is

b. Compute the 25th percentile of the amount spent on a debit card.

Answer: We need to first find the z value corresponding to probability 0.25

Using the standard normal table, we have:

Now using the z score formula, we have:

Therefore, 25th percentile is

c. Compute the 75th percentile of the amount spent on a debit card.

Answer:We need to first find the z value corresponding to probability 0.75

Using the standard normal table, we have:

Now using the z score formula, we have:

Therefore, 75th percentile is

d. What is the interquartile range of this distribution?

Answer: Interquartile range = 75th percentile - 25th percentile

  

  

Therefore, the interquartile range of this distribution is