Question

Find the z-score such that 35% of all observations (from a standard Normal distribution) are greater than z

Answer 1

Solution:

We have to find the z-score such that 35% of all observations are greater than z .

That is: find z such that:

P( Z > z ) = 35%

P( Z > z ) = 0.35

Area above z is 0.35

then area below z would be : 1 - 0.35 = 0.65

Thus find z such that:

P( Z < z ) = 0.65

So look in z table for area = 0.6500 or its closest area and find z value.

Area 0.6517 is closest to 0.6500, and it corresponds to 0.3 and 0.09

thus z = 0.39

Thus required z value is 0.39

P( Z < 0.39) =0.6500

That is: P(Z > 0.39) =0.35