Answer:
95.44% probability that the sample proportion will differ from the population proportion by less than 0.03.
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:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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 ![s = \sqrt{\frac{p(1-p)}{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D)
In this question:
![p = 0.05, n = 212, \mu = 0.05, s = \sqrt{\frac{0.05*0.95}{212}} = 0.015](https://tex.z-dn.net/?f=p%20%3D%200.05%2C%20n%20%3D%20212%2C%20%5Cmu%20%3D%200.05%2C%20s%20%3D%20%5Csqrt%7B%5Cfrac%7B0.05%2A0.95%7D%7B212%7D%7D%20%3D%200.015)
What is the probability that the sample proportion will differ from the population proportion by less than 0.03?
This is the pvalue of Z when X = 0.03 + 0.05 = 0.08 subtracted by the pvalue of Z when X = 0.05 - 0.03 = 0.02. So
X = 0.08
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{0.08 - 0.05}{0.015}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B0.08%20-%200.05%7D%7B0.015%7D)
![Z = 2](https://tex.z-dn.net/?f=Z%20%3D%202)
has a pvalue of 0.9772
X = 0.02
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{0.02 - 0.05}{0.015}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B0.02%20-%200.05%7D%7B0.015%7D)
![Z = -2](https://tex.z-dn.net/?f=Z%20%3D%20-2)
has a pvalue of 0.0228
0.9772 - 0.0228 = 0.9544
95.44% probability that the sample proportion will differ from the population proportion by less than 0.03.