Question

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?

Answer 1

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

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 = (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 = (\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)}`

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{(0.5*0.5)/(150)} = 0.0408`

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

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

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

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

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

`Z = (0.55 - 0.5)/(0.0408)`

`Z = 1.225`

`Z = 1.225` has a pvalue of 0.8897

X = 0.45

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

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

`Z = (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