Question

Consider the probability that no less than 95 out of 152 registered voters will vote in the presidential election. Assume the probability that a given registered voter will vote in the presidential election is 61%. Approximate the probability using the normal distribution. Round your answer to four decimal places.

Answer 1

Answer:

0.3821 = 38.21% probability that no less than 95 out of 152 registered voters will vote in the presidential election.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses 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 $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

Assume the probability that a given registered voter will vote in the presidential election is 61%.

This means that $p = 0.61$

152 registed voters:

This means that $n = 152$

Mean and Standard deviation:

$\mu = E(X) = 152*0.61 = 92.72$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{152*0.61*0.39} = 6.01$

Probability that no less than 95 out of 152 registered voters will vote in the presidential election.

This is, using continuity correction, $P(X \geq 95 - 0.5) = P(X \geq 94.5)$, which is 1 subtracted by the pvalue of Z when X = 94.5. So

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

$Z = \frac{94.5 - 92.72}{6.01}$

$Z = 0.3$

$Z = 0.3$ has a pvalue of 0.6179

1 - 0.6179 = 0.3821

0.3821 = 38.21% probability that no less than 95 out of 152 registered voters will vote in the presidential election.