A sample of 45 approved loans of bank XWB shows that clients on average are borr
ID: 3303333 • Letter: A
Question
A sample of 45 approved loans of bank XWB shows that clients on average are borrowing the amount of 172 000 euro, with standard deviation of 52 000 euro.
a) For investment planning and corresponding capital reserves, bank XWB would like to be 95% confident inside which values on average loan amounts may be borrowed by the bank's clients. Calculate this confidence interval.
b) Bank XWB can be 90% confident that customers borrow an average of more than what amount?
c) Suppose that the amounts of approved loans for bank XWB are distributed according the normal distribution with mean 170 000 euro and standard deviation 45 000 euro. Suppose that based on the loan evaluations of its clients, the bank XWB is not able to approve loans amount bigger than 220 000 euro. Calculate the probability that the bank XWB will not be able to approve the loan to its client.
Explanation / Answer
Sol:
n=45
mean=172000
sd=52000
SolutionA:
since population std deviation is known
use z crit.
z crit for 95% =1.96
95% confidence interval for truepopulation mean is
sample mean-z crit stddev/sqrt(n),sample mean+z crit stddev/sqrt(n)
172000-1.96(52000/sqrt(45)),172000+1.96(52000/sqrt(45))
=156806.7,187193.3
lower limit=156806.7
upper limit=187193.3
Solutionb:
zcrit=1.645
90% confidenceinterval is:
172000-1.645(52000)/sqrt(45),172000+1.645(52000)/sqrt(45)
lower limit=159248.4
uopper limit=184751.6
Solutionc:
mean=170000
stddev=45000
P(X>220000)
z=220000-170000/45000
z=1.11
P(Z>1.11)
1-P(Z<1.11)
=1-0.8665
=0.1335
probability that the bank XWB will not be able to approve the loan to its client.
=0.1335
ANSWER 0.1335
159248.4 184751.6