In a past election, the voter turnout was 62%. In a survey, 1174 subjects were a

Question

In a past election, the voter turnout was 62%. In a survey, 1174 subjects were asked if they voted in the election. a. Find the mean and standard deviation for the numbers of voters in groups of 1174. b. In the survey of 1174 people, 671 said that they voted in the election. Is this result consistent with the turnout, or is this result unlikely to occur with a turnout of 62%? Why or why not? results, does it appear at accurate voting results can be obtained by asking voters how they acted? c. Based on hese a. H Round to one decimal place as needed. GE (Round to one decimal place as needed.) b. Is the result of 671 voting in the election usual or unusual? O A. This result is unusual because 671 is within the range of usual values. O B. This result is usual because 671 is within the range of usual values. O C. This result is unusual because 671 is greater than the maximum usual value. O D. This result is unusual because 671 is below the minimum usual value. Click to select your answer(s).

Explanation / Answer

a) mean =np=1174*0.62=727.9

b) std deviation=(np(1-p))1/2 =16.6

c)option D is correct as it is below 2 std deviation from mean

d)option C is correct

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.