In a survey of 1059 people, 763 people said they voted in a recent presidential election. Voting records show that \( 69 \% \) of eligible voters actually did vote. Given that \( 69 \% \) of eligible voters actually did vote, (a) find the probability that among 1059 randomly selected voters, at least 763 actually did vote. (b) What do the results from part (a) suggest? (a) \( P(X \geq 763)=\square \) (Round to four decimal places as needed.)

To solve this problem, we will use the normal approximation to the binomial distribution since the sample size is large. ### Step 1: Identify the known conditions - Total number of people surveyed, \( n = 1059 \) - Number of people who said they voted, \( k = 763 \) - Proportion of eligible voters who actually voted, \( p = 0.69 \) ### Step 2: Calculate the mean and standard deviation The mean \( \mu \) and standard deviation \( \sigma \) for a binomial distribution can be calculated as follows: \[ \mu = n \cdot p \] \[ \sigma = \sqrt{n \cdot p \cdot (1 - p)} \] ### Step 3: Calculate the mean and standard deviation Let's calculate \( \mu \) and \( \sigma \): 1. Calculate \( \mu \): \[ \mu = 1059 \cdot 0.69 \] 2. Calculate \( \sigma \): \[ \sigma = \sqrt{1059 \cdot 0.69 \cdot (1 - 0.69)} \] ### Step 4: Standardize the value To find \( P(X \geq 763) \), we will standardize the value using the z-score formula: \[ z = \frac{X - \mu}{\sigma} \] Where \( X = 763 \). ### Step 5: Calculate the probability We will find \( P(X \geq 763) \) using the z-score and then use the standard normal distribution to find the probability. Now, let's perform the calculations. Calculate the value by following steps: - step0: Calculate: \(1059\times 0.69\) - step1: Multiply the numbers: \(730.71\) Calculate or simplify the expression \( \sqrt(1059 * 0.69 * (1 - 0.69)) \). Calculate the value by following steps: - step0: Calculate: \(\sqrt{1059\times 0.69\left(1-0.69\right)}\) - step1: Subtract the numbers: \(\sqrt{1059\times 0.69\times 0.31}\) - step2: Multiply the terms: \(\sqrt{226.5201}\) - step3: Rewrite the expression: \(\sqrt{\frac{2265201}{10000}}\) - step4: Use the properties of radicals: \(\frac{\sqrt{2265201}}{\sqrt{10000}}\) - step5: Simplify the expression: \(\frac{3\sqrt{251689}}{100}\) Calculate or simplify the expression \( (763 - 730.71) / 15.050585 \). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(763-730.71\right)}{15.050585}\) - step1: Remove the parentheses: \(\frac{763-730.71}{15.050585}\) - step2: Subtract the numbers: \(\frac{32.29}{15.050585}\) - step3: Convert the expressions: \(\frac{\frac{3229}{100}}{\frac{3010117}{200000}}\) - step4: Multiply by the reciprocal: \(\frac{3229}{100}\times \frac{200000}{3010117}\) - step5: Reduce the numbers: \(3229\times \frac{2000}{3010117}\) - step6: Multiply: \(\frac{3229\times 2000}{3010117}\) - step7: Multiply: \(\frac{6458000}{3010117}\) ### Step 1: Calculate the Mean and Standard Deviation 1. **Mean (\( \mu \))**: \[ \mu = 1059 \cdot 0.69 \approx 730.71 \] 2. **Standard Deviation (\( \sigma \))**: \[ \sigma = \sqrt{1059 \cdot 0.69 \cdot (1 - 0.69)} \approx 15.05 \] ### Step 2: Standardize the Value Now, we will calculate the z-score for \( X = 763 \): \[ z = \frac{763 - 730.71}{15.05} \approx 2.145 \] ### Step 3: Find the Probability To find \( P(X \geq 763) \), we need to find \( P(Z \geq 2.145) \). This can be calculated as: \[ P(Z \geq 2.145) = 1 - P(Z < 2.145) \] Using the standard normal distribution table or a calculator, we can find \( P(Z < 2.145) \). Let's calculate \( P(Z < 2.145) \) and then find \( P(Z \geq 2.145) \). Using a standard normal distribution table or calculator, we find: \[ P(Z < 2.145) \approx 0.9842 \] Thus, \[ P(Z \geq 2.145) = 1 - 0.9842 = 0.0158 \] ### Final Answer (a) The probability that among 1059 randomly selected voters, at least 763 actually did vote is: \[ P(X \geq 763) \approx 0.0158 \] (b) **What do the results from part (a) suggest?** The result suggests that it is quite unlikely (about 1.58% chance) that at least 763 out of 1059 randomly selected voters actually voted, given that 69% of eligible voters typically vote. This indicates that the observed number of voters (763) is significantly higher than what would be expected under normal circumstances, suggesting that the survey may have over-represented the voting population or that there was a higher turnout than usual.