Using the normal distribution, it is found that 8 scores were below 282.
In a normal distribution 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]
In this problem, the mean and the standard deviation are, respectively, given by [tex]\mu = 279, \sigma = 3[/tex].
The proportion of scores below 282 is the p-value of Z when X = 282, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{282 - 279}{3}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a p-value of 0.84.
Out of 10 scores:
0.84 x 10 = 8.4.
Rounding, 8 scores were below 282.
More can be learned about the normal distribution at https://brainly.com/question/24663213