Continuous Random Variables we are working with commonly use

Continuous Random Variables

we are working with commonly used discrete and continuous distribution

If X is a normal random variable with a mean of 17 and a variance of 9, what is P(12 LE X LE 25)? If Y is a normal random variable with a mean 25 and unknown variance. If P(X LE 29,9) = 0.9192, what is the variance sigma^2

Solution

Q1.
Mean ( u ) =17
Standard Deviation ( sd )=9
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 12) = (12-17)/9
= -5/9 = -0.5556
= P ( Z <-0.5556) From Standard Normal Table
= 0.28926
P(X < 25) = (25-17)/9
= 8/9 = 0.8889
= P ( Z <0.8889) From Standard Normal Table
= 0.81297
P(12 < X < 25) = 0.81297-0.28926 = 0.5237                  
                  
Q2.
P ( Z < X ) = 0.9192
Value of z to the cumulative probability of 0.9192 from normal table is 1.4
P( X-u/s.d < X - 25/ s.d ) = 0.9192
That is, ( 29 - 25/ s.d ) = 1.3997
--> s.d = ( 29 - 25 ) / 1.3997
--> s.d = ( 4 ) / 1.3997
--> s.d = 2.8577                  
--> Variance = 2.8577^2 = 8.1664