The problem involving normal distribution

The topic is Normal(Gaussian) distribution

I need to find the value of x for such expression:

mean is 5. standard deviation is 4

P(x<X<9)=0.2

I tried to evaluate the result by Z factor using:

phi((x-5)/4)<X<phi((9-5)/4)=0.2

But obtained different result.

the answer must be 6.44

Re: The problem involving normal distribution

What "result" are you trying to get? In other words, what question are you asking? To find "x" such that P(x< X< 9)= .2? Yes, when x= 9, z= (9- 5)/4= 1. What does your Normal Table give for the probability that z< 1? What must be subtracted from that to get 0.2? What z value corresponds to **that** normal probability? And what x value is that?

You say you "obtained different result". What result did you obtain?

Re: The problem involving normal distribution

The probability of z<1 is 0.84134. phi(1)=1-phi(-1)=1-0.15866.

0.84134-P(X<x)=0.2

P(X<x)=0.64134 But such probibilty doesn't exist in Normal table. What is the mistake here?