Question

n a survey of 1227 ​people, 828 people said they voted in a recent presidential election. Voting records show that 65​% of eligible voters actually did vote. Given that 65​% of eligible voters actually did​ vote.A) Find the probability that among 1227 randomly selected​ voters, at least 828 actually did vote.B) What do the results from part​ (a) suggest?

Answer 1

Answer: 0.0344

Step-by-step explanation:

For binomial distribution :-

$\mu=np\ \ ;\ \sigma=\sqrt{np(1-p)}$, where p is the proportion of success in each trial and n is the sample size.

Given : Sample size : n=1227

The proportion of people said they voted in a recent presidential election :p=0.95

Then, $\mu=1227(0.65)=797.55\ \ ;\ \sigma=\sqrt{(1227)(0.65)(0.35)}=16.71$

Let X be the binomial variable.

z-score : $z=\dfrac{x-\mu}{\sigma}$

For x= 828

$z=\dfrac{828-797.55}{16.71}=1.82$

Then,  probability that among 1227 randomly selected​ voters, at least 828 actually did vote is given by :-

$P(x\geq828)=P(z\geq1.82)=1-P(z<1.82)-P(z<-3.91)\\\\=1-0.9656205=0.0343795\approx0.0344$

Hence,  the probability that among 1227 randomly selected​ voters, at least 828 actually did vote= 0.0344