Question

In a survey of 1001 ​people, 689 people said they voted in a recent presidential election. Voting records show that 66​% of eligible voters actually did vote. Given that 66​% of eligible voters actually did​ vote, (a) find the probability that among 1001 randomly selected​ voters, at least 689 actually did vote.​ (b) What do the results from part​ (a) suggest?

Answer 1

a)

n=	1001	p=	0.6600
here mean of distribution=μ=np=		660.66
and standard deviation σ=sqrt(np(1-p))=		14.9875
for normal distribution z score =(X-μ)/σx
therefore from normal approximation of binomial distribution and continuity correction:

probability that among 1001 randomly selected​ voters, at least 689 actually did vote :

probability =

P(X>688.5)

=

P(Z>1.86)=

1-P(Z<1.86)=

1-0.9686=

0.0314

b)as probability of that happening is lower than 0.05 level ; therefore this event is unusual and we can conclude that some people say that they voted while in actual they have not,