Question

In class there was mention of Air Force jobs which required ASVAB scores to fall between a minimum and maximum score. Assuming that scores are drawn from a Gaussian distribution with mean 50 and standard deviation 10, what proportion of recruits would qualify for a job with min=45 and max=60?

Answer 1

Solution:

Given, the Gaussian (Normal) distribution with,

   = 50

= 10

Find P(45 < X < 60)

= P(X < 60) - P(X < 45)

= P[(X - )/ <  (60 - )/] - P[(X - )/ <  (45 - )/]

= P[Z <  (60 - 50)/10] - P[Z <  (45 - 50)/10]

= P(Z < 1.00) - P(Z < -0.50)

= 0.8413 - 0.3085

= 0.5328