The recruits in the top 387 qualify for advanced training Th

The recruits in the top 38.7% qualify for advanced training. The scores are found to be normally distributed with a mean of 88 and a standard deviation of 6.2 (these are not percent scores). Determine the lowest whole number score for an applicant who qualifies for advanced training.

Need step by step explanation.

Solution

Mean ( u ) =88
Standard Deviation ( sd )=6.2
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P ( Z > x ) = 0.387
Value of z to the cumulative probability of 0.387 from normal table is 0.29
P( x-u/ (s.d) > x - 88/6.2) = 0.387
That is, ( x - 88/6.2) = 0.29
--> x = 0.29 * 6.2+88 = 89.7794                  
89.78 is the lowest whole number score for an applicant who qualifies for advanced training