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 = 50
We 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%.
