Using the normal distribution, it is found that:
a) The z-score associated with a 60th percentile score is of Z = 0.253.
b) The score associated with the 60th percentile is of 51.265.
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 mean and the standard deviation are given, respectively, by:
[tex]\mu = 50, \sigma = 5[/tex]
The 60th percentile is X when Z has a p-value of 0.6, so X when Z = 0.253, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.253 = \frac{X - 50}{5}[/tex]
X - 50 = 5(0.253)
X = 51.265.
More can be learned about the normal distribution at https://brainly.com/question/24663213
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