Let #X ~ N(4, 16)#. How do you find #P(X < 6)#?

1 Answer
Jan 18, 2017

Use the function #Phi(x)# and a Normal distribution table

#X~N(4,16) => P(X<4)=Phi(0.5)approx.6915#

Explanation:

If #X# is a Normal Random Variable

#X~N(mu,sigma^2)#

Then its distribution function is the Gaussian Phi function

# => P(X<=x)=F_X(x)=int_RRe^(-((x-mu)/(sigma))^2/2)dx=Phi((x-mu)/sigma)#

In this case #mu=4# and #sigma=4#

So,

#P(X<=x)=Phi((x-4)/4)#

Note:

#{X<=x}={X < x}cup{X=x}#, and #{X < x} nn {X=x}=O/#

So,

#P(X<=x)=P(X < x)+P(X=x)#

and since a normal random variable is continuous #P(X=x)=0#

Therefore

#P(X<=x)=P(X < x)# in this case

Because of this we can say

#P(X<6)=P(X<=6)=Phi((6-4)/4)=Phi(2/4)=Phi(0.5)#

Then we check our normal distribution tables and see that

#P(X<6)=Phi(0.5)approx.6915#