Question

In a survey of 1466 people, 1049 people said they voted in a recent presidential election. Voting records show that 69% of eligible voters actually did vote. Given that 69% of eligible voters actually did vote, (a) find the probability that among 1466 randomly selected voters, at least 1049 actually did vote. (b) What do the results from part (a) suggest? (a) P(X2 1049) = (Round to four decimal places as needed.) (b) What does the result from part (a) suggest? A. Some people are being less than honest because P(x2 1049) is at least 1%. B. People are being honest because the probability of P(x2 1049) is less than 5%. C. People are being honest because the probability of P(x2 1049) is at least 1%. D. Some people are being less than honest because P(x2 1049) is less than 5%.

Answer 1

Answer: 10 1049 (a) Sample poro pa vion - 0.7155 > 1466 Here 2 P -P 0.7155 -0.69 P(1-P) 0.69 (1-0.69) 1466 Z 2:11 is The required Probability P(2>2.11) 1- P(Z <3.11) 1-0.9826 0.0174 chl (13)