Question

According to the Bureau of Labor Statistics, the average weekly pay for a U.S. production worker was $441.84. Assume that available data indicate that production worker wages were normally distributed with a standard deviation of $90. What is the probability that a worker earned between $400 and $500

Answer 1

Answer:

41.94% probability that a worker earned between $400 and $500.

Step-by-step explanation:

Problems of normally distributed samples can be 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.

In this problem, we have that:

$\mu = 441.84, \sigma = 90$

What is the probability that a worker earned between $400 and $500?

This is the pvalue of Z when X = 500 subtracted by the pvalue of Z when X = 400. So

X = 500

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

$Z = \frac{500 - 441.84}{90}$

$Z = 0.65$

$Z = 0.65$ has a pvalue of 0.7422

X = 400

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

$Z = \frac{400 - 441.84}{90}$

$Z = 0.65$

$Z = -0.46$ has a pvalue of 0.3228

So there is a 0.7422 - 0.3228 = 0.4194 = 41.94% probability that a worker earned between $400 and $500.