if X has a Normal Distrabution with a mean 452 and standard

if X has a Normal Distrabution with a mean 45.2 and standard deviation 3.6 then

P(X>47.1)=

P(44<X<46)=

P95=

Solution

Normal Distribution
Mean ( u ) =45.2
Standard Deviation ( sd )=3.6
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X > 47.1) = (47.1-45.2)/3.6
= 1.9/3.6 = 0.5278
= P ( Z >0.528) From Standard Normal Table
= 0.2988                  
b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 44) = (44-45.2)/3.6
= -1.2/3.6 = -0.3333
= P ( Z <-0.3333) From Standard Normal Table
= 0.36944
P(X < 46) = (46-45.2)/3.6
= 0.8/3.6 = 0.2222
= P ( Z <0.2222) From Standard Normal Table
= 0.58793
P(44 < X < 46) = 0.58793-0.36944 = 0.2185                  
c)
P ( Z < x ) = 0.95
Value of z to the cumulative probability of 0.95 from normal table is 1.645
P( x-u/s.d < x - 45.2/3.6 ) = 0.95
That is, ( x - 45.2/3.6 ) = 1.64
--> x = 1.64 * 3.6 + 45.2 = 51.122