Find the Probability Pz 120 01151 03849 04989 08849 05011 F

Find the Probability P(z > -1.20) 0.1151 0.3849 0.4989 0.8849 0.5011 Find the Probability P(-0.45

Solution

Normal Distribution
Mean ( u ) =0
Standard Deviation ( sd )=1
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
1)
P(X > -1.2) = (-1.2-0)/1
= -1.2/1 = -1.2
= P ( Z >-1.2) From Standard Normal Table
= 0.8849                  
2)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < -0.45) = (-0.45-0)/1
= -0.45/1 = -0.45
= P ( Z <-0.45) From Standard Normal Table
= 0.32636
P(X < 2.73) = (2.73-0)/1
= 2.73/1 = 2.73
= P ( Z <2.73) From Standard Normal Table
= 0.99683
P(-0.45 < X < 2.73) = 0.99683-0.32636 = 0.6704                  
3)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 11) = (11-10)/2
= 1/2 = 0.5
= P ( Z <0.5) From Standard Normal Table
= 0.69146
P(X < 14) = (14-10)/2
= 4/2 = 2
= P ( Z <2) From Standard Normal Table
= 0.97725
P(11 < X < 14) = 0.97725-0.69146 = 0.2857
4)
P ( Z < x ) = 0.99
Value of z to the cumulative probability of 0.99 from normal table is 2.326
P( x-u/s.d < x - 100/15 ) = 0.99
That is, ( x - 100/15 ) = 2.33
--> x = 2.33 * 15 + 100 = 134.89 ~ 135