Question

1. Based on past experience, a bank believes that 7% of the people who receive loans will not make payments on time. The bank takes a random sample of 200 recently approved loans.

2. 1. Find the probability that between 5% and 10% of these clients will not make timely payments.

   2. What is the 90th percentile of the sample proportions of clients who will not make timely payments.

   3. Find the quartiles of the sample proportions of clients who will not make timely payments.

Answer 1

for normal distribution z score =(p̂-p)/σp
here population proportion=     p=	0.070
sample size       =n=	200
std error of proportion=σp=√(p*(1-p)/n)=sqrt(0.07*0.93/200) =	0.0180

a)

probability that between 5% and 10% of these clients will not make timely payments:

probability =P(0.05<X<0.1)=P((0.05-0.07)/0.018)<Z<(0.1-0.07)/0.018)=P(-1.11<Z<1.66)=0.9515-0.1335=0.8180

b)

for 90th percentile critical z=	1.28
Hence corresponding value=mean+z*std deviation=0.07+1.28*0.0180 =			0.0931

c)

for 1st quartile at 25th percetile , crtiical z = -0.67

corresponding value=mean+z*std deviation=0.07-0.67*0.180 =

0.0579

for 3rd quartile at 75th percentile , critical z = 0.67

corresponding value=mean+z*std deviation=0.07+0.67*0.180 =

0.0821