Assume the random variable X is normally distributed with me

Assume the random variable X is normally distributed, with mean =42 and standard deviation =6. Find the 5th percentile.

Solution

Mean ( u ) =42
Standard Deviation ( sd )= 6
Normal Distribution = Z= X- u / sd ~ N(0,1)                  

P ( Z < x ) = 0.05
Value of z to the cumulative probability of 0.05 from normal table is -1.645
P( x-u/s.d < x - 42/6 ) = 0.05
That is, ( x - 42/6 ) = -1.64
--> x = -1.64 * 6 + 42 = 32.13