Using the normal distribution, it is found that 5 students get scores that are above 96.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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.
In this problem, the mean and the standard deviation are given, respectively, by:
[tex]\mu = 76, \sigma = 10[/tex].
The proportion of students who scored above 96 is one subtracted by the p-value of Z when X = 96, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{96 - 76}{10}[/tex]
Z = 2
Z = 2 has a p-value of 0.9772.
1 - 0.9772 = 0.0228
Out of 230 students:
0.0228 x 230 = 5 students.
More can be learned about the normal distribution at https://brainly.com/question/24663213
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