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Using the normal distribution, it is found that the score needed to qualify is of 88.32.
Normal Probability Distribution
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]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
In this problem, the mean and the standard deviation are given, respectively, by [tex]\mu = 80[/tex] and [tex]\sigma = 8[/tex].
The top 15% of the applicants are selected, hence the score needed to qualify is the 85th percentile, which is X when Z = 1.04, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.04 = \frac{X - 80}{8}[/tex]
X - 80 = 1.04(8)
X = 88.32.
The score needed to qualify is of 88.32.
More can be learned about the normal distribution at https://brainly.com/question/24663213
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