Suppose that a team of campaign volunteers surveyed 200 likely voters to gauge s

Question

Suppose that a team of campaign volunteers surveyed 200 likely voters to gauge support of a school levy that will be on the ballot in an upcoming election. Respondents were asked whether they support the levy and whether they have children attending school in the district. The results of this survey are shown below. Support opposed to levy levy Have children attending school 67 28 Have no children attending school in district 42 63 What is the probability that a randomly selected respondent supports the levy given that he or she has children in the district? Give your answer as a percentage, precise to one decimal place. Percentage What is the probability that a randomly selected respondent has children attending school in the district given that he or she supports the levy? Give your answer as a percentage, rounded to one decimal place percentage

Explanation / Answer

Following table shows the row total and column total:

From table we have

P(have chidren in district and support levy) = 67 /200

P(have chidren in district) = 95/200

So the probability that a randomly selected respondents supports the levy given that he or she has the chidren in the district is

P(supoort levy|have chidren in district) = P(have chidren in district and support levy) / P(have chidren in district) = 67 /95 = 0.7053

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From table we have

P(have chidren in district and support levy) = 67 /200

P(support levy) = 109/200

So the probability that a randomly selected respondents supports the levy given that he or she support levy is

P(have chidren in district|supoort levy) = P(have chidren in district and support levy) / P(support levy) = 67 / 109 = 0.6147

Support levy Opposed to levy Total Have children attending school in district 67 28 95 Have no children attending school in district 42 63 105 Total 109 91 200

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