Question

In a survey of

1301

​people,

842

people said they voted in a recent presidential election. Voting records show that

62​%

of eligible voters actually did vote. Given that

62%

of eligible voters actually did​ vote, (a) find the probability that among

1301

randomly selected​ voters, at least

842

actually did vote.​ (b) What do the results from part​ (a) suggest?

​(a)

​P(Xgreater than or equals≥842​)equals=nothing

​(Round to four decimal places as​ needed.)

​(b) What does the result from part​ (a) suggest?

A.Some people are being less than honest because

​P(xgreater than or equals≥842842​)

is less than​ 5%.

B.Some people are being less than honest because

​P(xgreater than or equals≥842842​)

is at least​ 1%.

C.People are being honest because the probability of

​P(xgreater than or equals≥842842​)

is less than​ 5%.

D.People are being honest because the probability of

​P(xgreater than or equals≥842842​)

is at least​ 1%.

Answer 1

Given that the true proportion is 0.62, the distribution of the sample proportion of people who actually did vote is modelled here as:

The probability that at least 842 actually did vote is computed here as:

P(X >= 842)

Converting it into a proportion, we get here:
P(p >= 842 / 1301)

P(p >= 0.6472)

Converting it to a standard normal variable, we have here:

Getting it from the standard normal tables, we have here:

Therefore 0.0216 is the required probability here.

b) We examine an unusual event with a probability cutoff of 5%. As we can see the probability in the above part as 0.0216 < 0.05, therefore

A.Some people are being less than honest because ​P(x >= 842​) is less than​ 5%. Therefore A is the correct conclusion here.