Suppose that 20% of the residents in a certain state support an increase in the property tax. An opinion poll will randomly sample 900 state residents and will then compute the proportion in the sample that support a property tax increase. How likely is the resulting sample proportion to be within .02 of the true proportion (i. e., between .18 and .22)? (Hint: Use the sampling distribution of the sample proportion in this case.)

86.64% probability that the resulting sample proportion is within .02 of the true proportion.

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 the sampling distribution of a sample proportion p in a sample of size n, we have that

In this problem:

How likely is the resulting sample proportion to be within .02 of the true proportion (i.e., between .18 and .22)?

This is the pvalue of Z when X = 0.22 subtracted by the pvalue of Z when X = 0.18.

X = 0.22

has a pvalue of 0.9332.

X = 0.18

has a pvalue of 0.0668

0.9332 - 0.0668 = 0.8664

86.64% probability that the resulting sample proportion is within .02 of the true proportion.

its a, i had the same answer, its from study island right

step-by-step explanation: