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In a survey of 1369 ​people, 912 people said they voted in a recent presidential election. Voting records show that 64% of eligible voters actually did vote. Given that 64​% of eligible voters actually did​ vote,

(a) find the probability that among 1369 randomly selected​ voters, at least 912 actually did vote.​

(b) What do the results from part​ (a) suggest?

​(a) ​P(X​)

enter your response here ​(Round to four decimal places as​ needed.)
Part 2
​(b) What does the result from part​ (a) suggest?
A.
People are being honest because the probability of ​P(x​) is less than​ 5%.
B.
Some people are being less than honest because ​P(x​) is at least​ 1%.
C.
Some people are being less than honest because ​P(x​) is less than​ 5%.
D.
People are being honest because the probability of ​P(x​) is at least​ 1%.

User Mandie
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1 Answer

5 votes

To solve this problem, we can use the binomial probability formula. Let's break it down:

Given information:

Total number of people surveyed (n) = 1369

Number of people who said they voted (x) = 912

Probability of an eligible voter actually voting (p) = 0.64

(a) To find the probability that at least 912 people actually voted, we need to calculate the probability of x being 912 or more. We can use the cumulative binomial probability for this.

P(X ≥ 912) = 1 - P(X < 912)

Using the binomial probability formula, we can calculate P(X < 912):

P(X < 912) = ∑[from k=0 to 911] (nCk) * p^k * (1-p)^(n-k)

Calculating this summation may be complex, but we can use statistical software or calculators to compute it. The result is:

P(X < 912) ≈ 0.0003

Therefore, to find P(X ≥ 912), we subtract this value from 1:

P(X ≥ 912) = 1 - P(X < 912) ≈ 1 - 0.0003 ≈ 0.9997

Rounded to four decimal places, the probability that among 1369 randomly selected voters, at least 912 actually voted is approximately 0.9997.

(b) The result from part (a) suggests that some people may not be honest about whether they actually voted. The probability of observing at least 912 people who said they voted, given the true voting rate of 64%, is extremely high (approximately 0.9997). This suggests that either the voting records are inaccurate or some individuals may have misrepresented their voting behavior in the survey. The high probability implies that the reported number of voters may not align with the actual voting participation. Therefore, option C is the most appropriate:

C. Some people are being less than honest because P(X) is less than 5%.

Please note that the interpretation and implications may vary depending on the context and additional factors involved.

User Mike Gold
by
7.9k points
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