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

Answer:

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:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

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:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

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:

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

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{153*0.63*0.37} = 5.97$

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.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{87.5 - 96.39}{5.97}$

$Z = -1.49$

$Z = -1.49$ 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.