Question

Some parts of California are particularly earthquake-prone.

Suppose that in one metropolitan area, the chance a homeowner is insured against an earthquake is 0.27. A sample of four homeowners are to be selected at random.

Suppose X is a random variable that is modeled by a binomial distribution which describes the number homeowners out of the four that have earthquake insurance.

(a) Find the probability mass function of X. (Round your answers to four decimal places.) 0 2 4 p(x)

b) What is the most likely value for X?

c) What is the probability that at most 1 of the four selected have earthquake insurance? (Round your answer to four decimal places.)

d) What is the probability that at least two of the four selected have earthquake insurance? (Round your answer to four decimal places.)

e) What is the expected value and standard deviation of X? (Round your answer to two decimal places.)
E(X)= μx = ?
SD(X) = σx = ?

Answer 1

a)
`P(X=0)=(4C0)(0.27)^0 (1-0.27)^(4-0)=0.283982`

`P(X=1)=(4C1)(0.27)^4 (1-0.27)^(4-1)=0.420138`

`P(X=2)=(4C2)(0.27)^5 (1-0.27)^(4-2)=0.23309`

`P(X=3)=(4C3)(0.27)^3 (1-0.27)^(4-3)=0.057474`

`P(X=4)=(4C4)(0.27)^4 (1-0.27)^(4-4)=0.05314`

b) For this case the value with the higher probability is X=1, so then would be the most likely value.

c)
`P(X \leq 1) = P(X=0) +P(X=1) = 0.283982+0.420138=0.704121 \approx 0.7041`

d)
`P(X \leq 2) =1-{(X<2)=1-P(X \leq 1)=1-0.7041 =0.2959`

e) For this case the expected value is given by:

`E(X) = \mu = np= 4*0.27= 1.08`

The variance is given by:

`\sigma^2 = Var(X) = np(1-p) =4*0.27*(1-0.27)= 0.7884`

And the deviation is:

`Sd(X) = \sigma = √(0.7884)=0.89`

Explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

`X \sim Binom(n=4, p=0.27)`

The probability mass function for the Binomial distribution is given as:

`P(X)=(nCx)(p)^x (1-p)^(n-x)`

Where (nCx) means combinatory and it's given by this formula:

`nCx=(n!)/((n-x)! x!)`

Part a

`P(X=0)=(4C0)(0.27)^0 (1-0.27)^(4-0)=0.283982`

`P(X=1)=(4C1)(0.27)^4 (1-0.27)^(4-1)=0.420138`

`P(X=2)=(4C2)(0.27)^5 (1-0.27)^(4-2)=0.23309`

`P(X=3)=(4C3)(0.27)^3 (1-0.27)^(4-3)=0.057474`

`P(X=4)=(4C4)(0.27)^4 (1-0.27)^(4-4)=0.05314`

Part b

For this case the value with the higher probability is X=1, so then would be the most likely value.

Part c

For this case we want this probability:

`P(X \leq 1)`

And we can find this probability like this:

`P(X \leq 1) = P(X=0) +P(X=1) = 0.283982+0.420138=0.704121 \approx 0.7041`

Part d

For this case we want this probability:

`P(X \leq 2)`

And we can find this probability using the complement rule like this:

`P(X \leq 2) =1-{(X<2)=1-P(X \leq 1)=1-0.7041 =0.2959`

Part e

For this case the expected value is given by:

`E(X) = \mu = np= 4*0.27= 1.08`

The variance is given by:

`\sigma^2 = Var(X) = np(1-p) =4*0.27*(1-0.27)= 0.7884`

And the deviation is:

`Sd(X) = \sigma = √(0.7884)=0.89`