Question

The average loan amount issued by a small short-term lender is $861 with a standard deviation of $148. Determine the probabilities, assuming that the population data is normally distributed.

a) What is the probability that the lender issues more than $900 to a random borrower?

b) What is the probability that the lender issues at most $900, on average, to a random sample of 25 borrowers?

c) What is the probability that the lender issues between $900 and $950, on average, to a random sample of 25 borrowers?

Answer 1

for normal distribution z score =(X-μ)/σx
mean μ=	861
standard deviation σ=	148

a)

probability that the lender issues more than $900 to a random borrower :

probability =P(X>900)=P(Z>(900-861)/148)=P(Z>0.26)=1-P(Z<0.26)=1-0.6026=0.3974

b)

std error=σx̅=σ/√n=

29.6000

probability that the lender issues at most $900, on average, to a random sample of 25 borrowers :

probability =P(X<900)=(Z<(900-861)/29.6)=P(Z<1.32)=0.9066

c) probability that the lender issues between $900 and $950, on average, to a random sample of 25 borrowers :

probability =P(900<X<950)=P((900-861)/29.6)<Z<(950-861)/29.6)=P(1.32<Z<3.01)=0.9987-0.9066=0.0921