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

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

Step-by-step 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:

The standard deviation of the binomial distribution is:

Normal probability distribution

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

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

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 , .

153 voters:

This means that

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

This means that

Mean and standard deviation:

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

Using continuity correction, this is: , which is 1 subtracted by the p-value of Z when X = 87.5.

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.

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