Question

Suppose 46% of politicians are lawyers. If a random sample of size 662 is selected, what is the probability that the proportion of politicians who are lawyers will differ from the total politicians proportion by less than 4%

Answer 1

Answer:

0.9606 = 96.06% probability that the proportion of politicians who are lawyers will differ from the total politicians proportion by less than 4%

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

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 $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Suppose 46% of politicians are lawyers.

This means that $p = 0.46$

Sample of size 662

This means that $n = 662$

Mean and standard deviation:

$\mu = p = 0.46$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.46*0.54}{662}} = 0.0194$

What is the probability that the proportion of politicians who are lawyers will differ from the total politicians proportion by less than 4%?

p-value of Z when X = 0.46 + 0.04 = 0.5 subtracted by the p-value of Z when X = 0.46 - 0.04 = 0.42. So

X = 0.5

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.5 - 0.46}{0.0194}$

$Z = 2.06$

$Z = 2.06$ has a p-value of 0.9803

X = 0.42

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.42 - 0.46}{0.0194}$

$Z = -2.06$

$Z = -2.06$ has a p-value of 0.0197

0.9803 - 0.0197 = 0.9606

0.9606 = 96.06% probability that the proportion of politicians who are lawyers will differ from the total politicians proportion by less than 4%