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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%

User Adib Faramarzi
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1 Answer

17 votes
17 votes

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%

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 = (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 = (\sigma)/(√(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{(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{(p(1-p))/(n)} = \sqrt{(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 = (X - \mu)/(\sigma)

By the Central Limit Theorem


Z = (X - \mu)/(s)


Z = (0.5 - 0.46)/(0.0194)


Z = 2.06


Z = 2.06 has a p-value of 0.9803

X = 0.42


Z = (X - \mu)/(s)


Z = (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%

User Nagy Ervin
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