Question

Assume that a randomly selected subject is given a bone density test. Bone density test scores are normally distributed with a mean of 0 and a standard deviation of 1. Find the bone density test scores that can be used as cutoff values separating the most extreme 1% of all scored. mulitiple choice a) -0.542 and 2.646 b.) -1.324 and 1.324 c.) -3.724 and 1.653 d) -0.786 and 0.786 d.) -2.575 and 2.575

Answer 1

Answer 2

Normal Distribution
Mean ( u ) =0
Standard Deviation ( sd )=1
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P ( Z < x ) = 0.01
Value of z to the cumulative probability of 0.01 from normal table is -2.326
P( x-u/s.d < x - 0/1 ) = 0.01
That is, ( x - 0/1 ) = -2.33
--> x = -2.33 * 1 + 0 = -2.326