Questions: Homework Section 6.6 Question 5.6.6.19 In a survey of 1014 people, 729 people said they voted in a recent presidential election. Within reasonable doubt, what is the probability that among 1014 randomly selected voters, at least 729 actually did vote? (a) P(X ≥ 729) = □ (Round to four decimal places as needed.)

Solution Steps

We are tasked with finding the probability that among 1014 randomly selected voters, at least 729 actually voted. This can be expressed mathematically as \( P(X \geq 729) \), where \( X \) follows a binomial distribution with parameters \( n = 1014 \) and \( p = \frac{729}{1014} \).

Using the binomial probability formula:

\[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

we find that:

\[ P(X = 729) = 0.0279 \]

The mean \( \mu \), variance \( \sigma^2 \), and standard deviation \( \sigma \) of the binomial distribution are calculated as follows:

\[ \mu = n \cdot p = 729.0 \]

\[ \sigma^2 = n \cdot p \cdot q = 204.8964 \]

\[ \sigma = \sqrt{npq} = 14.3142 \]

To find \( P(X \geq 729) \), we can use the normal approximation. First, we calculate the z-score for \( x = 729 \):

\[ z = \frac{x - \mu}{\sigma} = \frac{729 - 729}{14.3142} = 0 \]

Using the standard normal distribution, we find:

\[ P(X \geq 729) = 1 - P(Z < 0) = 1 - 0.5 = 0.5000 \]

Final Answer