Question

In a past presidential election, the actual voter turnout was 62%. In a survey, 882 subjects were asked if they voted in the presidential election. Find the mean and standard deviation for the numbers of actual voters in groups of 882. (Round answer to one decimal place.) = 546.84 (Round answer to two decimal places.) 14.42 = Give the interval of usual values for the number of voters in groups of 882. (Enter answer as an interval using square-brackets only with whole numbers.) usual values = In the survey of 882 people, 621 said that they voted in the last presidential election. Is this result consistent with the actual voter turnout, or is this result unlikely to occur with an actual voter turnout of 62%? O this result is unlikely to occur with the actual voter turnout O this result is consistent with the actual voter turnout

Answer 1

here this is binomial with parameter n=882 and p=0.62

mean E(x)=μ=np=546.8

standard deviation σ=√(np(1-p))=14.42

2 standard deviation from mean values are μ -/+ 2*σ=(518,576)

Usual values =[518, 576]

O this result is unlikely to occur with the actual voter turnout