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An employer has applicants take a short exam when applying to a job opening. One of their criteria is that the applicant must score in the top 2.5% of all applicants who took the exam. The exam is constructed such that the scores are normally distributed with an average of 58 and a standard deviation of 7.

Answer :

Answer:

Applicants need a score of at least 71.72.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 58, \sigma = 7[/tex]

Top 2.5%.

Scores that are X when Z has a pvalue of 1-0.025 = 0.975 and higher. So it is scores that are X when Z = 1.96 and higher.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.96 = \frac{X - 58}{7}[/tex]

[tex]X - 58 = 7*1.96[/tex]

[tex]X = 71.72[/tex]

Applicants need a score of at least 71.72.

Applicants need a score of at least 71.72.

Given information:

The exam is constructed such that the scores are normally distributed with an average of 58 and a standard deviation of 7.

Calculation of the score:

[tex]Z = x - \mu \div \sigma\\\\1.96 = x - 58 \div 7\\\\x - 58 = 7 \times 1.96\\\\[/tex]

x = 71.72

learn more about the standard deviation here: https://brainly.com/question/2748723

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