Question

in a survey of 1327 people, 871 people said they voted in a recent presidential election. voting records show that 63% of eligible voters actually did vote. given that 63% of eligible voters actually did vote, (a) find the probability that among 1327 randomly selected voters, at least 871 actually did vote. (b) what do the results from part (a) suggest? (a) p(x ≥ 871) = (round to four decimal places as needed.)

Explanation:

Step1: Identify distribution parameters

This is a binomial - to - normal approximation problem. Let \(n = 1327\) (number of trials/surveyed people) and \(p=0.63\) (probability of an eligible voter voting). The mean of the binomial distribution is \(\mu = np\) and the standard deviation is \(\sigma=\sqrt{np(1 - p)}\).
\(\mu=np=1327\times0.63 = 836.01\)
\(\sigma=\sqrt{np(1 - p)}=\sqrt{1327\times0.63\times(1 - 0.63)}=\sqrt{836.01\times0.37}\approx\sqrt{309.3237}\approx17.5876\)

Step2: Apply continuity correction

For \(P(X\geq871)\) in a binomial distribution approximated by a normal distribution, with continuity correction, we find \(P(X\geq870.5)\). We standardize \(x = 870.5\) using the formula \(z=\frac{x-\mu}{\sigma}\).
\(z=\frac{870.5 - 836.01}{17.5876}=\frac{34.49}{17.5876}\approx1.96\)

Step3: Find the probability

We want \(P(Z\geq1.96)\). Since the total area under the standard - normal curve is 1, and \(P(Z\geq z)=1 - P(Z < z)\). From the standard - normal table, \(P(Z < 1.96)=0.9750\). So \(P(Z\geq1.96)=1 - 0.9750 = 0.0250\)

Answer:

\(0.0250\)

(b) The results from part (a) suggest that the proportion of people who claimed to have voted (\(\frac{871}{1327}\approx0.6565\)) is higher than the actual proportion of eligible voters who voted (\(0.63\)). The probability that at least 871 people actually voted out of 1327 randomly - selected voters is only \(0.0250\) (a relatively small probability). This indicates that it is unlikely that the number of people who said they voted is accurate based on the known actual voting rate, and there may be some over - reporting of voting behavior in the survey.