Answer:
Keenan's z-score was of 0.61.
Rachel's z-score was of 0.81.
Step-by-step explanation:
Z-score:
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.
Keenan scored 80 points on an exam that had a mean score of 77 points and a standard deviation of 4.9 points.
This means that [tex]X = 80, \mu = 77, \sigma = 4.9[/tex]
So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{80 - 77}{4.9}[/tex]
[tex]Z = 0.61[/tex]
Keenan's z-score was of 0.61.
Rachel scored 78 points on an exam that had a mean score of 75 points and a standard deviation of 3.7 points.
This means that [tex]X = 78, \mu = 75, \sigma = 3.7[/tex]. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{75 - 78}{3.7}[/tex]
[tex]Z = 0.81[/tex]
Rachel's z-score was of 0.81.
