Based on past experience, a bank believes that 12% of the people who receive loa
ID: 3324717 • Letter: B
Question
Based on past experience, a bank believes that
12%
of the people who receive loans will not make payments on time. The bank has recently approved 200 loans. Answer the following questions.
a) What are the mean and standard deviation of the proportion of clients in this group who may not make timely payments?
mu left parenthesis ModifyingAbove p with caret right parenthesispequals=nothing
SD left parenthesis ModifyingAbove p with caret right parenthesisSDpequals=nothing
(Round to three decimal places as needed.)
b) What assumptions underlie your model? Are the conditions met?
A.
With reasonable assumptions about the sample, all the conditions are met.
B.
The randomization and 10% conditions are not met.
C.
The 10% and success/failure conditions are not met.
D.
The success/failure condition is not met.
E.
The randomization and success/failure conditions are not met.
F.
The randomization condition is not met.
G.
The 10% condition is not met.
H.
Without unreasonable assumptions, none of the conditions are met.
c) What is the probability that over
1515%
of these clients will not make timely payments?
Upper P left parenthesis ModifyingAbove p with caret greater than 0.15 right parenthesisPp>0.15equals=nothing
(Round to three decimal places as needed.)
Explanation / Answer
a.
12?%
of the people who receive loans will not make payments on time
The bank has recently approved 200 loans
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/? * ?2? * e ^ -(x-u)^2/ 2?^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
proportion ( p ) = 0.12
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.12*0.88/200)
=0.023
mean = n*p = 200*0.12 = 24
standard Deviation ( sd ) = 0.023
b.
With reasonable assumptions about the? sample, all the conditions are met.
c.
the probability that over 15?%
P(X > 0.15) = (0.15-0.12)/0.023
= 0.03/0.023 = 1.304
= P ( Z >1.304) From Standard Normal Table
= 0.096
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