In a survey of

11511151

​people,

782782

people said they voted in a recent presidential election. Voting records show that

6565​%

of eligible voters actually did vote. Given that

6565​%

of eligible voters actually did​ vote, (a) find the probability that among

11511151

randomly selected​ voters, at least

782782

actually did vote.​ (b) What do the results from part​ (a) suggest?

Question content area bottom

Part 1

​(a)

​P(Xgreater than or equals≥782782​)equals=

Let's solve this step by step.

### Part (a): Find the probability that at least 782 people voted

We are asked to calculate \( P(X \geq 782) \), where:

- \( X \) represents the number of people who actually voted out of 1151 surveyed people.
- According to the problem, 65% of eligible voters actually voted.
  
Since this is a binomial distribution problem, where each individual either votes or doesn’t, we can model \( X \) as a binomial random variable with:

- \( n = 1151 \) (the total number of people in the sample),
- \( p = 0.65 \) (the probability of voting), and
- \( X \sim \text{Binomial}(n = 1151, p = 0.65) \).

We need to find the probability that at least 782 people voted, which means:

\[
P(X \geq 782) = 1 - P(X < 782)
\]

To approximate this binomial probability, we can use the normal approximation to the binomial distribution since \( n \) is relatively large. The normal approximation is applied when:

\[
X \sim N(\mu, \sigma^2)
\]

Where:
- \( \mu = np = 1151 \times 0.65 = 748.15 \)
- \( \sigma = \sqrt{np(1 - p)} = \sqrt{1151 \times 0.65 \times 0.35} \approx 15.85 \)

Now, convert \( X \geq 782 \) to a z-score for the normal distribution:

\[
z = \frac{782 - \mu}{\sigma} = \frac{782 - 748.15}{15.85} \approx 2.14
\]

Using a z-table or a standard normal distribution calculator, the probability of \( Z \geq 2.14 \) is approximately 0.0162.

Thus, the probability that at least 782 people voted is approximately:

\[
P(X \geq 782) \approx 0.0162
\]

### Part (b): Interpretation of the results

The results from part (a) suggest that it is quite unlikely (about a 1.62% chance) to observe at least 782 people voting in a sample of 1151 individuals, given that the actual voting rate is 65%. This suggests that the 782 respondents who claimed they voted might include some who over-reported their voting behavior.

### Summary:
- (a) \( P(X \geq 782) \approx 0.0162 \)
- (b) The results suggest possible over-reporting of voting behavior in the survey, as the probability of observing such a high number of voters in the sample is quite low.

---

Would you like a detailed explanation on any part of this?  
Here are 5 related questions you might find interesting:
1. How is the binomial distribution used in probability?
2. What is the normal approximation to the binomial distribution, and when is it used?
3. How do you interpret the z-score in hypothesis testing?
4. Why might people over-report voting in surveys?
5. How does sample size affect the accuracy of normal approximation?

**Tip**: When using normal approximation, always check if \( np \) and \( n(1-p) \) are greater than 5 for accuracy!

In a survey of 1151 people, 782 people said they voted in a recent presidential election. Voting records show that 65% of eligible voters actually did vote. Given that 65% of eligible voters actually did vote, (a) find the probability that among 1151 randomly selected voters, at least 782 actually did vote. (b) What do the results from part (a) suggest?

Explore how to find the probability that at least 782 people out of 1151 voted in a survey, given a 65% voting rate. This problem applies binomial distribution, normal approximation, and z-scores, with step-by-step calculations for accurate results. Suitable for high school and introductory college statistics students.

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