U.S. consumers are increasingly viewing debit cards as a convenient substitute f
ID: 3246447 • Letter: U
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 $7,790 (Kiplinger’s, August 2007). Assume that the average amount spent on a debit card is normally distributed with a standard deviation of $500. Use Table 1.
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.)
Compute the 25th percentile of the amount spent on a debit card. (Round "z" value to 2 decimal places.)
Compute the 75th percentile of the amount spent on a debit card. (Round "z" value to 2 decimal places.)
What is the interquartile range of this distribution? (Round "z" value to 2 decimal places.)
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 $7,790 (Kiplinger’s, August 2007). Assume that the average amount spent on a debit card is normally distributed with a standard deviation of $500. Use Table 1.
Explanation / Answer
Solution:-
a) Since = 7790 and = 500 we have:
P ( X > 8000 ) = P ( X > 80007790 ) = P ( X / > 80007790 / 500)
Since Z = x / and 80007790 / 500 = 0.42 we have:
P ( X > 8000 ) = P ( Z > 0.42 )
Using the standard normal table to conclude that:
P (Z > 0.42) = 0.3372
The proportion of consumers who spend over $8,000 is 0.3372.
b). 25th percentile of the amount spent on a debit card.
So we begin by going into the interior of the standard normal distribution table to find the area under the curve closest to 0.25, and from this we can determine the corresponding Z score. Once we have this we can use the equation X= + Z, because we already know that the mean and standard deviation are 7790 and 500, respectively.
The exact Z value holding 25% of the values below it is -0.675 which was determined from a table of standard normal probabilities with precision.
X = 7900 + (-0.675)(500) = 7562.5
Therefore, 25th percentile is 7562.50
c). 75th percentile of the amount spent on a debit card.
The exact Z value holding 75% of the values below it is 0.675 which was determined from a table of standard normal probabilities with precision.
X = 7900 + (0.675)(500) = 8237.5
Therefore, 75th percentile is 8237.50.
d). What is the interquartile range of this distribution?
The IQR is the difference between the 1st quartile and the 3rd quartile (or the 25th percentile and the 75th percentile), thus, if we can find IQR from the 25th and 75th percentiles.
Thus, the interquartile range is: 8237.50 - 7562.50 = 675.
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