Question

An insurance company found that 25% of all insurance policies are terminated before their maturity date. Assume that 10 polices are randomly selected from the company’s policy database. Out of the 10 randomly selected policies;

a) What is the expected number of policies to be terminated before maturing? [1]

b) What is the standard deviation for the number of policies terminated before maturity? [1]

c) What is the probability that no policy will be terminated before maturity? [1]

d) What is the probability that all policies will be terminated before maturing? [2]

e) What is the probability that at least two policies will be terminated? [3]

f) What is the probability that more than 5 but less than eight policies will be terminated? [3] g) What is the probability that at most eight policies will be terminated? [3]

Answer 1

An insurance company found that 25% of all insurance policies are terminated before maturity.

So, the chance of a randomly selected insurance policy terminating before maturity is 0.25.

10 policies are randomly selected.

Let, X be the random variable denoting the number of policies out of these 10, that terminate before maturity.

So, X follows binomial distribution with parameters n=10, and p=0.25.

We also know that if a random variable X follows binomial distribution with parameters n and p, then its expectation is np, and standard deviation is sqrt(n*p*(1-p)).

We also know that in this case

P(X=r)

=(n C r)(p^r)(q^(n-r))

Where, q=1-p, and (n C r)=n!/(r!(n-r)!).

(a) To find the expected number of policies that terminate before maturing.

Here, n=10, and p=0.25.

So, the expected number of policies that terminate before maturing is

=10*0.25

=2.5.

So, the expectation is 2.5.

(b) To find the standard deviaton of the number of policies that terminate before maturing

=sqrt(n*p*(1-p))

=sqrt(10*0.25*0.75)

=1.369.

So, the sd is 1.369.

(c) To find the probability that no policy will terminate before maturity.

P(X=0)

=(10 C 0)(0.25^0)(0.75^10)

=0.0563

So, the required probability is 0.0563.

(d) To find the probability that all policies will be terminated before maturing.

P(X=10)

=(10 C 10)(0.25^10)(0.75^0)

=0.000001

So, the required probability is 0.000001.

(e) To find the chance that atleast two policies will be terminated before maturing.

P(X>=2)

=1-P(X=0)-P(X=1)

=1-(10 C 0)(0.25^0)(0.75^10)-(10 C 1)(0.25^1)(0.75^9)

=0.7559.

So, the required probability is 0.7559.

(f) To find the probability that more than 5 but less than 8 policies will be terminated.

P(5<X<8)

=P(X=6)+P(X=7)

=(10 C 6)(0.25^6)(0.75^4)+(10 C 7)(0.25^7)(0.75^3)

=0.016222+0.003089

=0.019311.

So, the required probability is 0.019311.

(g) To find the probabilty that at most 8 policies will be terminated.

P(X<=8)

=

Where i runs from 0 to 8.

=0.9999.

so, the required probability is 0.9999.