Answer:
0.0668 = 6.68% probability that the worker earns more than $8.00
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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.
The average hourly wage of workers at a fast food restaurant is $7.25/hr with a standard deviation of $0.50.
This means that ![\mu = 7.25, \sigma = 0.5](https://tex.z-dn.net/?f=%5Cmu%20%3D%207.25%2C%20%5Csigma%20%3D%200.5)
If a worker at this fast food restaurant is selected at random, what is the probability that the worker earns more than $8.00?
This is 1 subtracted by the pvalue of Z when X = 8. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{8 - 7.25}{0.5}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B8%20-%207.25%7D%7B0.5%7D)
![Z = 1.5](https://tex.z-dn.net/?f=Z%20%3D%201.5)
has a pvalue of 0.9332
1 - 0.9332 = 0.0668
0.0668 = 6.68% probability that the worker earns more than $8.00
In order to figure out what N is you have to isolate it by dividing by 50 on both sides as shown below:
It would be 1 fjjffjdj fifth because the ending would replace the 8
Answer:
x = 1
Step-by-step explanation: