Mathematics, 19.03.2020 21:36 nathanphelps2468

The mean per capita income is 21,604 dollars per annum with a standard deviation of 727 dollars per annum. What is the probability that the sample mean would differ from the true mean by more than 36 dollars if a sample of 193 persons is randomly selected? Round your answer to four decimal places.

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Answer from: jeniferfayzieva2018

0.4902 = 49.02% probability that the sample mean would differ from the true mean by more than 36 dollars if a sample of 193 persons is randomly selected

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 $\mu$ 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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 21604, \sigma = 727, n = 193, s = \frac{727}{\sqrt{193}} = 52.33$

What is the probability that the sample mean would differ from the true mean by more than 36 dollars if a sample of 193 persons is randomly selected?

Eiher it differs by 36 or less dollars, or it differs by more than 36 dollars. The sum of the probabilities of these events is decimal 1. So

Probability it differs by 36 or less dollars.

pvalue of Z when X = 21604 + 36 = 21640 subtracted by the pvalue of Z when X = 21604 - 36 = 21568.

X = 21640

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

By the Central Limit Theorem

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

$Z = \frac{21640 - 21604}{52.33}$

Z = 0.69

Z = 0.69 has a pvalue of 0.7549

X = 21568

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

$Z = \frac{21568 - 21604}{52.33}$

Z = -0.69

Z = -0.69 has a pvalue of 0.2451

0.7549 - 0.2451 = 0.5098

What is the probability that the sample mean would differ from the true mean by more than 36 dollars if a sample of 193 persons is randomly selected?

p + 0.5098 = 1

p = 0.4902

0.4902 = 49.02% probability that the sample mean would differ from the true mean by more than 36 dollars if a sample of 193 persons is randomly selected

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