A large manufacturing firm tests job applicants who recently graduated from college. The test scores are normally distributed with a mean of 500 and a standard deviation of 50. Management is considering placing a new hire in an upper level management position if the person scores in the upper 6 percent of the distribution. What is the lowest score a college graduate must earn to qualify for a responsible position? A. 50 B. 625 C. 460 D. 578

Answer:The lowest score a college graduate must be 577.75 or greater to qualify for a responsible position and lie in the upper 6%.Step-by-step explanation:We are given the following information in the question: Mean, μ = 500 Standard Deviation, σ = 50We are given that the distribution of test score is a bell shaped distribution that is a normal distribution. Formula: [tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]We have to find the value of x such that the probability is 0.06. P(X > x)  = 6% = 0.06[tex]P( X > x) = P( z > \displaystyle\frac{x - 500}{50})=0.06[/tex]  [tex]= 1 - P( z \leq \displaystyle\frac{x - 500}{50})=0.06 [/tex]  [tex]=P( z \leq \displaystyle\frac{x - 500}{50})=1-0.06=0.94 [/tex]  Calculation the value from standard normal z table, we have,  [tex]P(z < 1.555) = 0.94[/tex][tex]\displaystyle\frac{x - 500}{50} = 1.555\\x = 577.75[/tex]  Hence, the lowest score a college graduate must be 577.75 or greater to qualify for a responsible position and lie in the upper 6%.