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.)

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

Part a)

X ~ N ( µ = 7040 , σ = 500 )
P ( X > 8000 ) = 1 - P ( X < 8000 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 8000 - 7040 ) / 500
Z = 1.92
P ( ( X - µ ) / σ ) > ( 8000 - 7040 ) / 500 )
P ( Z > 1.92 )
P ( X > 8000 ) = 1 - P ( Z < 1.92 )
P ( X > 8000 ) = 1 - 0.9726
P ( X > 8000 ) = 0.0274

Part b)
X ~ N ( µ = 7040 , σ = 500 )
P ( X < x ) = 25% = 0.25
To find the value of x
Looking for the probability 0.25 in standard normal table to calculate Z score = -0.674
Z = ( X - µ ) / σ
-0.674 = ( X - 7040 ) / 500
X = 6703
P ( X < 6703 ) = 0.25

Part c)
X ~ N ( µ = 7040 , σ = 500 )
P ( X < x ) = 75% = 0.75
To find the value of x
Looking for the probability 0.75 in standard normal table to calculate Z score = 0.67
Z = ( X - µ ) / σ
0.67 = ( X - 7040 ) / 500
X = 7375
P ( X < 7375 ) = 0.75

Part d)
IQR = Q3 - Q1 =  7375 - 6703 = 672