Answer:
Step-by-step explanation:
Given that a loan company knows that 5% of its loans will be delinquent.
since each loan is independent of the other p , probability for any random loan to be delinquent is constant 0.05
X no of delinquent loan accounts is binomial with n =400 and p = 0.05
Since n is very large and also np = 20 and nq >5 we can approximate to normal
Mean = np = 20 : Variance = npq = 19
Std dev = 4.36
X is N(1, 4.36)
With continuity correcton we calculate the prob
a) that exactly 25 accounts will be delinquent?
=
b) that fewer than 30 accounts will be delinquent?
=P(X<29.5)
= 0.9854
c) more than 24 accounts will be delinquent
=P(X>24,5)
=0.1509