Question

Suppose candidate a for a town council seat receives 43% of the votes in an election. as voters leave the polls they are asked who they voted for. what is the probability that less than 40% of the 80 voters surveyed indicate they voted for candidate a? assume an infinite population.

Answer 1

The probability that less than 40% of the 80 voters surveyed indicate they voted for candidate A is P(phat <0.4) = P((phat-p)/sqrt(p*(1-p)/n) <(0.4-0.43)/sqrt(0.43*(1-0.43)/80)) =P(Z<-0.542) =0.2939 (from standard normal table)

Answer 2

29.46% probability that less than 40% of the 80 voters surveyed indicate they voted for candidate a

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

For a proportion p in a sample of size n, we have
`\mu = p, \sigma = \sqrt{(p(1-p))/(n)}`

In this problem, we have that:

`\mu = 0.43, \sigma = \sqrt{(0.43*0.57)/(80)} = 0.05535`

What is the probability that less than 40% of the 80 voters surveyed indicate they voted for candidate a?

This is the pvalue of Z when X = 0.4. So

`Z = (X - \mu)/(\sigma)`

`Z = (0.4 - 0.43)/(0.05535)`

`Z = -0.54`

`Z = -0.54` has a pvalue of 0.2946

29.46% probability that less than 40% of the 80 voters surveyed indicate they voted for candidate a