Question

Consider the probability that at least 88 out of 153 registered voters will vote in the presidential election. Assume the probability that a given registered voter will vote in the presidential election is 63%. Approximate the probability using the normal distribution. Round your answer to four decimal places

Answer 1

0.9319 = 93.19% probability that at least 88 out of 153 registered voters will vote in the presidential election.

Explanation:

Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

`E(X) = np`

The standard deviation of the binomial distribution is:

`√(V(X)) = √(np(1-p))`

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that
`\mu = E(X)`,
`\sigma = √(V(X))`.

153 voters:

This means that
`n = 153`

Assume the probability that a given registered voter will vote in the presidential election is 63%.

This means that
`p = 0.63`

Mean and standard deviation:

`\mu = E(X) = np = 153(0.63) = 96.39`

`\sigma = √(V(X)) = √(np(1-p)) = √(153*0.63*0.37) = 5.97`

Consider the probability that at least 88 out of 153 registered voters will vote in the presidential election.

Using continuity correction, this is:
`P(X \geq 88 - 0.5) = P(X \geq 87.5)`, which is 1 subtracted by the p-value of Z when X = 87.5.

`Z = (X - \mu)/(\sigma)`

`Z = (87.5 - 96.39)/(5.97)`

`Z = -1.49`

`Z = -1.49` has a p-value of 0.0681.

1 - 0.0681 = 0.9319

0.9319 = 93.19% probability that at least 88 out of 153 registered voters will vote in the presidential election.