Using the normal distribution, it is found that 0.1841 = 18.41% of students had a higher score than Amanda.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean \(\mu\) and standard deviation \(\sigma\) is given by:
\(Z = \frac{X - \mu}{\sigma}\)
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation for the grades are given as follows:
\(\mu = 850, \sigma = 100\)
The proportion of students that had a higher grade than Amanda is one subtracted by the p-value of Z when X = 940, hence:
\(Z = \frac{X - \mu}{\sigma}\)
\(Z = \frac{940 - 850}{100}\)
Z = 0.9
Z = 0.9 has a p-value of 0.8159.
1 - 0.8159 = 0.1841.
0.1841 = 18.41% of students had a higher score than Amanda.
More can be learned about the normal distribution at https://brainly.com/question/15181104
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