Respuesta :
Answer:
The minimum score required for the job offer is 751.
Step-by-step explanation:
When the distribution is normal, we use 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 question, we have that:
[tex]\mu = 587, \sigma = 152[/tex]
What is the minimum score required for the job offer?
Top 14%, so the minimum score is the 100-14 = 86th percentile, which is X when Z has a pvalue of 0.86. So X when Z = 1.08.
Then
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.08 = \frac{X - 587}{152}[/tex]
[tex]X - 587 = 1.08*152[/tex]
[tex]X = 751.16[/tex]
Rounding to the nearest whole number:
The minimum score required for the job offer is 751.
