Respuesta :
Answer:
Norma performed best on the aptitude test and should be offered the job.
Step-by-step explanation:
We are given that three potential employees took an aptitude test. Each person took a different version of the test. The scores are reported below. Tobias got a score of 74.7; this version has a mean of 61.7 and a standard deviation of 13. Norma got a score of 351; this version has a mean of 291 and a standard deviation of 25. Vincent got a score of 7.38; this version has a mean of 6.9 and a standard deviation of 0.4.
Now, to find which of the applicants should be offered the job, we have to find the z-score of each of the applicants, and the one who gets the highest z-score will get the job.
The z-score probability distribution for the normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean and [tex]\sigma[/tex] = standard deviation
Firstly, we will find the z-score for Tobias;
The z-score of 74.7 = [tex]\frac{X-\mu}{\sigma}[/tex]
= [tex]\frac{74.7-61.7}{13}[/tex] = 1
Now, we will find the z-score for Norma;
The z-score of 351 = [tex]\frac{X-\mu}{\sigma}[/tex]
= [tex]\frac{351-291}{25}[/tex] = 2.4
Now, we will find the z-score for Vincent;
The z-score of 7.38 = [tex]\frac{X-\mu}{\sigma}[/tex]
= [tex]\frac{7.38-6.9}{0.4}[/tex] = 1.2
As we can clearly see that the z-score is highest for Norma which means that Norma performed best on the aptitude test and should be offered the job.
