Answer:
The z-score for the trainee is of 2.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 p-value, we get the probability that the value of the measure is greater than X.
The mean of the test scores is 72 with a standard deviation of 5.
This means that [tex]\mu = 72, \sigma = 5[/tex]
Find the z-score for a trainee, given a score of 82.
This is Z when X = 82. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{82 - 72}{5}[/tex]
[tex]Z = 2[/tex]
The z-score for the trainee is of 2.
