Question

In a certain town, 22% of voters favor a given ballot measure. For groups of 21 voters, find the variance(??) for the number who favor the measure.

Answer 1

Solution :

Let X be a random variable which represents that out of 21 voters the number of voters who favor the ballot measure.

Given that, 22% voters favor a given ballot measure. Hence, the probability that a voter favors the measure = 22/100 = 0.22

Let us consider "getting a voter who favor the ballot measure" as success. Hence, we have only two mutually exclusive outcomes (success and failure).

Probability of success (p) = 0.22

Number of trials (n) = 21

Since, probability of success remains constant in each of the trials, for each trials we have only two mutually exclusive outcomes, outcomes are independent and number of trials are finite, therefore we can consider that X follows binomial distribution with parameters n = 21 and p = 0.22.

The variance of binomial distribution is given as follows :

Variance (σ²) = np(1 - p)

We have, n = 21 , p = 0.22

Hence,

Variance (σ²) = 21 × 0.22 × (1 - 0.22)

Variance (σ²) = 3.6036

The variance for the number who favor the measure is 3.6036.

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