Question

The chance a home owners is insured against an earthquake is .35. A sample of four homeowners is selected at random. Suppose X is a random variable that is modeled by a binomial distribution which describes the number of homeowners out of four that have insurance. Find the probability mass function of X when x = 0, 1, 2, 3, and 4. Then find what is the most likely value of x. Find the probability that at most 1 of the four selected have earthquake insurance. Find the probability that at least two of the 4 selected have earthquake insurance. Find the expected value of x and find the standard deviation of x.

Answer 1

since here we have to use binomial distribution with n=4 and p=0.35

now we need to know the formula of the theorem 1st

P(X=x) = nCx * p^x * q^(n-x)

Where p is the prob. of success

q is the prob. of failure

Here in this question p=0.35 q=1-p=0.65 n=4

Substitute x=0 to 4 and we get PMF

P(X=0)= 4C0 * 0.35^0 *0.65^4 =0.1785063

Same way we get

PMF

X=0	X=1	X=2	X=3	X=4
0.1785063	0.384475	0.31053750	0.11147500	0.01500625

Most likely value = Pi*Xi =0*0.17850625 +1*0.38447500+2*0.31053750+3*0.11147500+4*0.01500625 =1.4

probability that at most 1 of the four selected have earthquake insurance.

= P(X=0)+P(X=1)

=0.17850625 + 0.38447500

=0.5629812

the probability that at least two of the 4 selected have earthquake insurance

=1-[P(X=0)+P(X=1)]

=1-(0.5629812)

=0.4370188

std deviation =sqrt( Pi*(Xi-X')^2)

=sqrt(0.1785*(0-1.4)^2+0.384475*(1-1.4)^2+0.31053750*(2-1.4)^2+0.11147500*(3-1.4)^2+0.01500625*(4-1.4)^2)

=0.9539328