3Find the area to the left of z 104 4Find the area between

3.Find the area to the left of z = -1.04

4.Find the area between z = -0.34 and z = 1.03.

5. Find the area that separate the middle 92% of the data from the area in the tails of the standard normal distribution.

Solution

Normal Distribution
Mean ( u ) =0
Standard Deviation ( sd )=1
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X < -1.04) = (-1.04-0)/1
= -1.04/1= -1.04
= P ( Z <-1.04) From Standard Normal Table
= 0.1492                  
b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < -0.34) = (-0.34-0)/1
= -0.34/1 = -0.34
= P ( Z <-0.34) From Standard Normal Table
= 0.36693
P(X < 1.03) = (1.03-0)/1
= 1.03/1 = 1.03
= P ( Z <1.03) From Standard Normal Table
= 0.84849
P(-0.34 < X < 1.03) = 0.84849-0.36693 = 0.4816