find the zscores that bound the middle 90 of the area under

find the z-scores that bound the middle 90% of the area under the standard normal curve.

Solution

To find P(a < = Z < = b) = F(b) - F(a)
P(X < -1.645) = (-1.645-0)/1
= -1.645/1 = -1.645
= P ( Z <-1.645) From Standard Normal Table
= 0.04998
P(X < 1.645) = (1.645-0)/1
= 1.645/1 = 1.645
= P ( Z <1.645) From Standard Normal Table
= 0.95002
P(-1.645 < X < 1.645) = 0.95002-0.04998 = 0.9                  

0.9000 is exactly half-way between 0.9495 and 0.9505. Therefore, z = 1.645.
Using symmetry, z = -1.645 and z = 1.645 bound the middle 90% of a normal distribution.