Question

In a survey of 1453 people, 1033 people said they voted in a recent presidential election. Voting records show that 69 % of eligible voters actually did vote. Given that 69 % of eligible voters actually did vote, (a) find the probability that among 1453 randomly selected voters, at least 1033 actually did vote. (b) What do the results from part (a) suggest?

(a) P(x ?1033 )= _______ (Round to four decimal places as needed)

Answer 1

The number of people who voted could be modelled here as binomial distribution such that:

This can be approximated to a normal distribution as:

Now the required probability here is computed as:

Applying the continuity correction factor, we get here:

Converting this to a standard normal variable, we get:

Getting the above probability from the standard normal tables, we get:

Therefore 0.0397 is the required probability here.

b) The result from part a) suggest that there is a very low probability that the value of X would be greater than 1033, which shows that the given value is not significant at the 5% level of significance.