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:

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:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

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

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

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

This means that [tex]p = 0.61[/tex]

152 registed voters:

This means that [tex]n = 152[/tex]

Mean and Standard deviation:

[tex]\mu = E(X) = 152*0.61 = 92.72[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{152*0.61*0.39} = 6.01[/tex]

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

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{94.5 - 92.72}{6.01}[/tex]

[tex]Z = 0.3[/tex]

[tex]Z = 0.3[/tex] 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.

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