Question

MUST use the NORM.DIST() or NORM.INVO) Normal Distribution functions in Excel. Provide your final answer rounded to 4 decimal places in the yellow highlight cell. Problem (i): The average credit card debt for college seniors is $3262. The debt has a normal distribution with a standard deviation of $1100. (4 points) a) What is the probability that a randomly selected college senior owes more than $4000? b) What is the probability that a randomly selected college senior owes between $4000 and $4500? c) What is the probability that a randomly selected college senior owes at most $3000? d) Find the 85th percentile of the credit card debt for college seniors.

Answer 1

Note: Final answers are highlighted in colour.

Mean=3262, Stttandard Deviation= 1100

a)Probability that a randomly selected college senior owes more than $4000.

P(X>4000)=1-P(X<=4000)= 1- NORM.DIST(4000,3262,1100,TRUE)=1-748861= 0.2511

b) Probability that a randomly selected college senior owes between $4000 and $4500

P(4000<X<4500)= P(X<4500)-P(X<4000)

= NORM.DIST(4500,3262,1100,TRUE)- NORM.DIST(4000,3262,1100,TRUE)

= 0.869802- 748861

= 0.1209

c)Probability that a randomly selected college senior owes at most $3000.

P(X<=3000)

= NORM.DIST(3000,3262,1100,TRUE)

= 0.4059

d)85th percentile for the credit card debt for college seniors

is X such than P(Debt<X)= 0.85

= NORM.INV(0.85,3262,1100)

= 4402.0767