The verbal part of the Graduate Record Exam GRE has a m of 5

The verbal part of the Graduate Record Exam (GRE) has a m of 500 and s = 100. Use normal distribution and the formulas and steps necessary to answer the following questions:

What is the z score for a GRE score of 597?

What is the percentile rank of this z score? (4 pts)

Work (required):

Work (required):

What GRE score corresponds to a percentile rank of 18%? (2 pts)

Work (required):

If you wanted to select only students at or above the 87th percentile, what GRE score would you use as a cutoff score (i.e. what GRE score corresponds to this percentile)? (2 pts)

Work (required):

Solution

Normal Distribution

Mean ( u ) =500
Standard Deviation ( sd )=100
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X < 597) = (597-500)/100
= 97/100= 0.97

The z score will be = 0.97

= P ( Z <0.97) From Standard Normal Table
= 0.834  
      
b)
P ( Z < x ) = 0.18
Value of z to the cumulative probability of 0.18 from normal table is -0.915
P( x-u/s.d < x - 500/100 ) = 0.18
That is, ( x - 500/100 ) = -0.92
--> x = -0.92 * 100 + 500 = 408.5                  
c)
P ( Z > x ) = 0.87
Value of z to the cumulative probability of 0.87 from normal table is -1.13
P( x-u/ (s.d) > x - 500/100) = 0.87
That is, ( x - 500/100) = -1.13
--> x = -1.13 * 100+500 = 387.4