assume data is normally distributed give answers to three de

(assume data is normally distributed, give answers to three decimal places)

Solution

Normal Distribution
Mean ( u ) =90
Standard Deviation ( sd )=23.4
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X < 86) = (86-90)/23.4
= -4/23.4= -0.1709
= P ( Z <-0.1709) From Standard Normal Table
= 0.4321                  
b)
P(X > 100) = (100-90)/23.4
= 10/23.4 = 0.4274
= P ( Z >0.427) From Standard Normal Table
= 0.3346                  
c)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 99.5) = (99.5-90)/23.4
= 9.5/23.4 = 0.406
= P ( Z <0.406) From Standard Normal Table
= 0.65762
P(X < 100.5) = (100.5-90)/23.4
= 10.5/23.4 = 0.4487
= P ( Z <0.4487) From Standard Normal Table
= 0.67318
P(99.5 < X < 100.5) = 0.67318-0.65762 = 0.0156