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Answer:
99.36% probability that the sample proportion will differ from the population proportion by less than 6%
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
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.
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 sample proportion p in a sample of size n, we have that the sampling distribution of the sample proportions has .
In this question:
.
What is the probability that the sample proportion will differ from the population proportion by less than 6%
This is the pvalue of Z when X = 0.18 + 0.06 = 0.24 subtracted by the pvalue of Z when X = 0.18 - 0.06 = 0.12. So
X = 0.24
By the Central Limit Theorem
has a pvalue of 0.9968
X = 0.12
has a pvalue of 0.0032
0.9968 - 0.0032 = 0.9936
99.36% probability that the sample proportion will differ from the population proportion by less than 6%