Answer:
0.997 = 99.7% probability that the resulting sample proportion to be between 0.066 and 0.294
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use 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.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean and standard deviation
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.
This means that
So, by the Central Limit Theorem:
How likely is the resulting sample proportion to be between 0.066 and 0.294?
This is the pvalue of Z when X = 0.294 subtracted by the pvalue of Z when X = 0.066. So
X = 0.294
By the Central Limit Theorem
has a pvalue of 0.9985
X = 0.066
has a pvalue of 0.0015
0.9985 - 0.0015 = 0.997
0.997 = 99.7% probability that the resulting sample proportion to be between 0.066 and 0.294