Mathematics, 21.01.2021 22:10 brooklynunderwood46

A pollster believes that 20% of the voters in a certain area favor a bond issue. If 81 voters are randomly sampled from the large number of voters in this area, approximate the probability that the sampled fraction of voters favoring the bond issue will not differ from the true fraction by more than 0.02. (Round your answer to four decimal places.)

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0.3472 = 34.72% probability that the sampled fraction of voters favoring the bond issue will not differ from the true fraction by more than 0.02.

Step-by-step explanation:

To solve this question, we use 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 $\mu$ and standard deviation $\sigma$ , the zscore 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 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 $\mu$ 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 $s = \sqrt{\frac{p(1-p)}{n}}$

In this question:

Proportion p = 0.2 , sample of n = 81 . So

$\mu = p = 0.2$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.2*0.8}{81}} = 0.0444$

Approximate the probability that the sampled fraction of voters favoring the bond issue will not differ from the true fraction by more than 0.02.

Probability of the sampled proportion being between 0.2 - 0.02 = 0.18 and 0.2 + 0.02 = 0.22, which is the pvalue of Z when X = 0.22 subtracted by the pvalue of Z when X = 0.18. So

X = 0.22

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

By the Central Limit Theorem

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

$Z = \frac{0.22 - 0.2}{0.044}$

Z = 0.45

Z = 0.45 has a pvalue of 0.6736

X = 0.18

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

$Z = \frac{0.18 - 0.2}{0.044}$

Z = -0.45

Z = -0.45 has a pvalue of 0.3264

0.6736 - 0.3264 = 0.3472 = 34.72% probability that the sampled fraction of voters favoring the bond issue will not differ from the true fraction by more than 0.02.

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