Question

In a survey of 1002 people, 701 said that they voted in a recent presidential election (based on data from ICR Research Group). Voting records show that 61% of eligible voters actually did vote. Given that 61% of eligible voters actually did vote, find the probability that among 1002 randomly selected eligible voters, at least 701 actually did vote. What does the result suggest?

[Write your answer as a complete sentence]

Answer 1

Given:n=1002,p=61%=0.61,x=701

So q=1-p=1-0.61=0.39

The two conditions that must be satisfied are np5 and nq5.

So 1002*0.61=611.225

and 1002*0.39=390.785

Thus the two conditions are satisfied to approximate the binomial distribution by normal distribution.

Mean,=np=1002*0.61=611.22

Standard deviation,=npq=1002*0.61*0.39=15.439

Here the value 701 is the discrete whole number 'x' which is approximated with the continuous distribution.

By using continuity correction, the value 0.5 is subtracted from the value of x.

That is, x-0.5=701

x=700.5

The area to the right of the continuous normal distribution is 700.5.

Formula to find z-score, z=(x-)/=(700.5-611.22)/15.439=89.28/15.439=5.78

The corresponding probability is P(Z5.78)=1-0.9999=0.0001

Since the probability is nearly equal to 0, we can say that the observed rate of 61% is correct.