Using the normal distribution, it is found that the expression for the z-score of her test score is given by:
\(Z = \frac{72 - 79}{7} = -1\)
Normal Probability Distribution
In a normal distribution with mean \(\mu\) and standard deviation \(\sigma\), the z-score of a measure X is given by:
\(Z = \frac{X - \mu}{\sigma}\)
- It 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, which is the percentile of X.
In this problem:
- The mean is of 79, hence \(\mu = 79\).
- The standard deviation is of 7, hence \(\sigma = 7\).
- Her score is of 72, hence \(X = 72\).
Then, her z-score is given by:
\(Z = \frac{X - \mu}{\sigma}\)
\(Z = \frac{72 - 79}{7} = -1\)
You can learn more about the normal distribution at https://brainly.com/question/24663213
