Using the normal distribution, it is found that there are 68 students with scores between 72 and 82.
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]
In this problem, the mean and the standard deviation are given, respectively, by:
[tex]\mu = 72, \sigma = 10[/tex]
The proportion of students with scores between 72 and 82 is the p-value of Z when X = 82 subtracted by the p-value of Z when X = 72.
X = 82:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{82 - 72}{10}[/tex]
Z = 1
Z = 1 has a p-value of 0.84.
X = 72:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{72 - 72}{10}[/tex]
Z = 0
Z = 0 has a p-value of 0.5.
0.84 - 0.5 = 0.34.
Out of 200 students, the number is given by:
0.34 x 200 = 68 students with scores between 72 and 82.
More can be learned about the normal distribution at https://brainly.com/question/24663213
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