Question

Many utility companies promote energy conservation by offering discount rates to consumers who keep their energy usage below certain established subsidy standards. A recent EPA report notes that 70% of the island residents of Puerto Rico hae reduced their electricity usage sufficiently to qualify for discounted rates. If five residential subscribers are randomly selected from San Juan, Puerto Rico, find the probability of each of the following events.a. All five qualify for the favorable rateb. At least four qualify for the favorable rates

Answer 1

a. 0.1681 = 16.81% probability that all five qualify for the favorable rate.

b. 0.5283 = 52.83% probability that at least four qualify for the favorable rates

Explanation:

For each Puerto Rico resident, there are only two possible outcomes. Either they qualify for discounted rates, or they do not. The probability of a person in the sample qualifying for discounted rates is independent of any other person in the sample. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

70% of the island residents of Puerto Rico have reduced their electricity usage sufficiently to qualify for discounted rates.

This means that
`p = 0.7`

Five residential subscribers are randomly selected from San Juan, Puerto Rico

This means that
`n = 5`

a. All five qualify for the favorable rate

This is P(X = 5). So

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 5) = C_(5,5).(0.7)^(5).(0.3)^(0) = 0.1681`

0.1681 = 16.81% probability that all five qualify for the favorable rate.

b. At least four qualify for the favorable rates

This is

`P(X \geq 4) = P(X = 4) + P(X = 5)`

So

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 4) = C_(5,4).(0.7)^(4).(0.3)^(1) = 0.3602`

`P(X = 5) = C_(5,5).(0.7)^(5).(0.3)^(0) = 0.1681`

`P(X \geq 4) = P(X = 4) + P(X = 5) = 0.3602 + 0.1681 = 0.5283`

0.5283 = 52.83% probability that at least four qualify for the favorable rates