Juries should have the same racial distribution as the surrounding communities. According to the U. S. Census Bureau, 18% of residents in Minneapolis, Minnesota, are African Americans. Suppose a local court will randomly sample 100 state residents and will then observe the proportion in the sample who are African American. How likely is the resulting sample proportion to be between 0.066 and 0.294 (i. e., 6.6% to 29.4% African American)?

a. There is roughly a 95% chance that the resulting sample proportion will be between 0.066 and 0.294 of the true proportion.
b. There is roughly a 99.7% chance that the resulting sample proportion will be between 0.066 and 0.294 of the true proportion.
c. There is roughly a 68% chance that the resulting sample proportion will be between 0.066 and 0.294 of the true proportion.
d. It is certain that the resulting sample proportion will be between 0.066 and 0.294 of the true proportion.

b. There is roughly a 99.7% chance that the resulting sample proportion will be between 0.066 and 0.294 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.

In this problem, we have that:

Proportion

A proportion has

How likely is the resulting sample proportion to be between 0.066 and 0.294 (i.e., 6.6% to 29.4% African American)?

This is the pvalue of Z when X = 0.294 subtracted by the pvalue of Z when X = 0.066. So

X = 0.294

has a pvalue of 0.9985

X = 0.066

has a pvalue of 0.0015

0.9985 - 0.0015 = 0.9970

So the correct answer is:

b. There is roughly a 99.7% chance that the resulting sample proportion will be between 0.066 and 0.294 of the true proportion.

6mm^3 or .6cm^3

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factor trees are nice.