Mathematics, 13.07.2020 19:01 dianamunoz580

In 2016, the median weekly earnings for people employed full-time in the United States was $837. (20 points) a) What proportion of full-time employees had weekly earnings of more than $837? b) A sample of 150 full-time employees is chosen. What is the probability that more than 55% of them earned more than $837 per week? c) What is the probability that less than 60% of the sample of 150 employees earned more than $837 per week? d) What is the probability that between 45% and 55% of the sample of 150 employees earned more than $837 per week? e) Would it be unusual if less than 45% of the sample of 150 employees earned more than $755 per week?

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Answer from: kritalewis

a) p = 0.5

b) 11.03% probability that more than 55% of them earned more than $837 per week.

c) 99.29% probability that less than 60% of the sample of 150 employees earned more than $837 per week

d) 77.94% probability that between 45% and 55% of the sample of 150 employees earned more than $837 per week

e) 45% is within 2 standard deviations of the mean, which means it would not be unusual if less than 45% of the sample of 150 employees earned more than $755 per week

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 $\mu$ 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.

If X is two or more standard deviations from the mean, it is considered unusual.

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 2016, the median weekly earnings for people employed full-time in the United States was $837.

This means that 50% of employees earn more than $837 and 50% below.

So we use p = 0.5

a) What proportion of full-time employees had weekly earnings of more than $837?

From above, p = 0.5

b) A sample of 150 full-time employees is chosen. What is the probability that more than 55% of them earned more than $837 per week?

n = 150, so $s = \sqrt{\frac{0.5*0.5}{150}} = 0.0408$

This is 1 subtracted by the pvalue of Z when X = 0.55. So

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

By the Central Limit Theorem

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

$Z = \frac{0.55 - 0.5}{0.0408}$

Z = 1.225

Z = 1.225 has a pvalue of 0.8897

1 - 0.8897 = 0.1103

11.03% probability that more than 55% of them earned more than $837 per week.

c) What is the probability that less than 60% of the sample of 150 employees earned more than $837 per week?

This is the pvalue of Z when X = 0.6.

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

$Z = \frac{0.6 - 0.5}{0.0408}$

Z = 2.45

Z = 2.45 has a pvalue of 0.9929

99.29% probability that less than 60% of the sample of 150 employees earned more than $837 per week.

d) What is the probability that between 45% and 55% of the sample of 150 employees earned more than $837 per week?

This is the pvalue of Z when X = 0.55 subtracted by the pvalue of Z when X = 0.45.

X = 0.55

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

$Z = \frac{0.55 - 0.5}{0.0408}$

Z = 1.225

Z = 1.225 has a pvalue of 0.8897

X = 0.45

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

$Z = \frac{0.45 - 0.5}{0.0408}$

Z = -1.225

Z = -1.225 has a pvalue of 0.1103

0.8897 - 0.1103 = 0.7794

77.94% probability that between 45% and 55% of the sample of 150 employees earned more than $837 per week.

e) Would it be unusual if less than 45% of the sample of 150 employees earned more than $755 per week?

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

$Z = \frac{0.45 - 0.5}{0.0408}$

Z = -1.225

So 45% is within 2 standard deviations of the mean, which means it would not be unusual if less than 45% of the sample of 150 employees earned more than $755 per week

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