In a survey of 1466 people, 1049 people said they voted in a recent presidential election....
In a survey of 1127 people, 739 people said they voted in a recent presidential election. Voting records show that 63% of eligible voters actually did vote. Given that 63% of eligible voters actually did vote, (a) find the probability that among 1127 randomly selected voters, at least 739 actually did vote. (b) What do the results from part (a) suggest? (a) P(x≥739) = b) What does the result from part (a) suggest? A.People are being honest because the probability...
In a survey of 1145 people, 862 people said they voted in a recent presidential election. Voting records show that 73% of eligible voters actually did vote. Given that 73% of eligible voters actually did vote, (a) find the probability that among 1145 randomly selected voters, at least 862 actually did vote. (b) What do the results from part (a) suggest? (a) P(X>862)=
In a survey of 1425 people, 1045 people said they voted in a recent presidential election. Voting records show that 71% of eligible voters actually did vote. Given that 71% of eligible voters actually did vote, (a) find the probability that among 1425 randomly selected voters, at least 1045 actually did vote. (b) What do the results from part (a) suggest?
In a survey of 13181318 people, 877877 people said they voted in a recent presidential election. Voting records show that 6464% of eligible voters actually did vote. Given that 6464% of eligible voters actually did vote, (a) find the probability that among 13181318 randomly selected voters, at least 877877 actually did vote. (b) What do the results from part (a) suggest?
1107 people, 816 people said they voted in a recent presidential election. Voting records show that 71% of eligible voters actually did vote. Given that 71% of eligible voters actually did vote, (a) find the probability that among 1107 randomly selected voters, at least 816 actually did vote. (b) What do the results from part (a) suggest? (a) P(X ≥816) ? round 4 decimals
31. Assume that women's heights are normally distributed with a mean given by μ=62.6 in, and a standard deviation given by σ=2.8 in. a. If 1 woman is randomly selected, find the probability that her height is between 62.2 in and 63.2 in.The probability is approximately (b) If 49 women are randomly selected, find the probability that they have a mean height less than 63 in .43. In a survey of 1345 people, 1029 people said they voted in a...
In a past election, the voter turnout was 69%. In a survey, 934 subjects were asked if they voted in the election. a. Find the mean and standard deviation for the numbers of voters in groups of 934. b. In the survey of 934 people, 695 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 69%? Why or why not? c. Based on these...
In a survey of 1000 eligible voters selected at rando found that 80% of those who had a college degree voted in the last presidential election, whereas 55% of who did not have a college degree voted in the last presidential election. m, it was found that 80 had a college degree. Additionally, it was 2. the people Draw a tree diagram to represent this scenario. a. b. Assuming that the poll is representative of all eligible voters, find the...
In the US, 60 percent of eligible voters vote in presidential election years. For a random sample of 8 eligible voters in the US, find the probability that all of them voted in the last presidential election.
1.) Voting records show that 61% of eligible voters actually did vote in a recent presidential election. In a survey of 1002 people, 70% said that they voted in that election. Use the survey results to test the claim that the percentage of all voters who say that they voted is equal to 61%. Test the claim by constructing an appropriate confidence interval. What are the null and alternative hypotheses? What is the value of = significance level? Is the...