Fixing fzero in standard normal distribution

I am trying to solve a variable n used in this equation for normal distribution

Z = (X - mean) / std

mean = 500*n

std = sqrt(500*n*(1-n))

X = 440

Z = -2.33

This is the code i am trying to use to solve it

 N = @(n) (((440 - 500*n)/(sqrt(500*n*(1-n))))+2.33)

N = function_handle with value:
    @(n)(((440-500*n)/(sqrt(500*n*(1-n))))+2.33)

 fzero(N,0)

Error using fzero
Initial function value must be finite and real.

How do you suggest I change it so I get a real value from n?

Answer 1

[Star Strider]

Change the initial parameter estimate to something other than 0. The fzero function is a root-finding algorithm, so it looks for zero-crossings.

Perhaps something like this —

 N = @(n) (((440 - 500*n)/(sqrt(500*n*(1-n))))+2.33)

N = function_handle with value:
    @(n)(((440-500*n)/(sqrt(500*n*(1-n))))+2.33)

nv = fzero(N,rand)

nv = 0.9098

.

[MADISON RAGONE]

Thank you! Why does rand work? If I remember correctly rand is just random numbers?

[Star Strider]

As always, my pleasure!

It is just that. However it is a random number between 0 and 1, so limited in its amplitude.

Answer 2

[Matt J]

Note that your problem has a closed-form solution, i.e. fzero is not necessary:

X = 440; Z = -2.33;

syms n; assume(X-500*n<=0)

n=double( solve(  (X - 500*n)^2 == 500*n*(1-n)*Z^2 ) )

n = 0.9098