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.

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)

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

Step-by-step explanation:

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

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

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

In this problem, we have that:

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

has a pvalue of 0.2946

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

its not b

step-by-step explanation:

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)