Question

In a survey of 1368 ​people, 989 people said they voted in a recent presidential election. Voting records show that 70% of eligible voters actually did vote. Given that 70​% of eligible voters actually did​ vote, (a) find the probability that among 1368 randomly selected​ voters, at least 989 actually did vote.​ (b) What do the results from part​ (a) suggest?

Answer 1

a)

n=	1368	p=	0.7000
here mean of distribution=μ=np=		957.6
and standard deviation σ=sqrt(np(1-p))=		16.9493

for normal distribution z score =(X-μ)/σx
therefore from normal approximation of binomial distribution and continuity correction:

probability that among 1368 randomly selected​ voters, at least 989 actually did vote :

probability =

P(X>988.5)

=

P(Z>1.82)=

1-P(Z<1.82)=

1-0.9656=

0.0344

b)

as the probability is less then 0.05 level we conclude that some people lie that they have voted while in actual they do not, therefore the survey overestimate the voting percentage,