Using the normal distribution, it is found that there is a 16% probability that a randomly selected score is greater than 590.
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, respectively, given by [tex]\mu = 510[/tex] and [tex]\sigma = 80[/tex].
The probability that a randomly selected score is greater than 590 is one subtracted by the p-value of Z when X = 590, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{590 - 510}{80}[/tex]
Z = 1
Z = 1 has a p-value of 0.84.
1 - 0.84 = 0.16.
0.16 = 16% probability that a randomly selected score is greater than 590.
More can be learned about the normal distribution at https://brainly.com/question/24663213