The credit card debt of college seniors follows a normal dis

The credit card debt of college seniors follows a normal distribution with mean $3,262 and standard deviation $1100. Show your work or calculator commands to answer the following questions.

A) What percent of college seniors owe less than $2000 to credit card companies?

B) Ninety percent of college seniors owe less than what amount of debt to credit card companies?

C) What is the probability that a randomly selected college senior owes more than $4200 to credit card companies?

D) For a random sample of 25 college seniors, what is the probability that their sample mean credit card debt is more than $4200?

E) Determine if any of the probabilities you found in parts C) and D) is an unusual probability. Justify your answer.

Solution

Mean ( u ) =3262
Standard Deviation ( sd )=1100
Normal Distribution = Z= X- u / sd ~ N(0,1)                  

a)
P(X < 2000) = (2000-3262)/1100
= -1262/1100= -1.1473
= P ( Z <-1.1473) From Standard Normal Table
= 0.1256                  
b)
P(X > 4200) = (4200-3262)/1100
= 938/1100 = 0.8527
= P ( Z >0.853) From Standard Normal Table
= 0.1969                  
c)
P ( Z < x ) = 0.9
Value of z to the cumulative probability of 0.9 from normal table is 1.282
P( x-u/s.d < x - 3262/1100 ) = 0.9
That is, ( x - 3262/1100 ) = 1.28
--> x = 1.28 * 1100 + 3262 = 4672.2                  
d)
P(X > 4200) = (4200-3262)/1100/ Sqrt ( 25 )
= 938/220= 4.2636
= P ( Z >4.2636) From Standard Normal Table
= 0