i using `bionrnd()`

function to generate a random vector and Laplace approximation formula to approximate the binomial distribution. but Laplace histogram dose not like the binomial distribution histogram. where is my mistake in coding? help me please and here is my code :

```
clear
clc
close all
n = input ('Please Enter Number of Exams : ');
p = input ('Please Enter Probability : ');
eta=n*p;
sigma=sqrt(n*p*(1-p));
for x=1:n*100;
a=rand(1,n);
b=0;
for i=1:n
if(a(i)>=1-p)
b=b+1;
end
end
c(x)=b;
d(x)=binornd(n , p);
e(x)=(1./(sqrt(2*pi*x*p*(1-p)))).* exp(-(((x+1)*p-(x*p)).^2)./(2*x*p*(1-p)));
end
E=sigma*e+eta;
histfit(c);
axis([0 n 0 inf])
M = mean(c);
varc=var(c);
disp(['var is : ', num2str(varc)]);
disp(['Mean is : ', num2str(M)]);
figure
histfit(d);
axis([0 n 0 inf])
M2 = mean(d);
varc2=var(d);
disp(['var is : ', num2str(varc2)]);
disp(['Mean is : ', num2str(M2)]);
figure
histfit(e,100);
axis([0 n 0 inf])
M3 = mean(e);
varc3=var(e);
disp(['var is : ', num2str(M3)]);
disp(['Mean is : ', num2str(M3)]);
```

the e(x) is formula of Laplace approximation in this link : De Moivre Laplace theorem formula

output is :

the 3rd figure must like to others.

Answer:

Problem #1: `c(x)`

and `d(x)`

are lists of random numbers with a given distribution of probabilities. When you plot a histogram of `c(x)`

or `d(x)`

, you are plotting the frequency of occurrence of each number. This frequency is just equal to the distribution.

`e(x)`

is a totally different object. You have coded it as the probability distribution *itself*, not a set of sample values. Compare the values in `e(x)`

with the values in `c(x)`

or `d(x)`

. The former is a list of floats, while the latter are lists of integers. It doesn't make sense to plot the frequency of occurrence of `e(x)`

as a histogram.

The simplest way around this is probably to rewrite `e`

as a function, generate a set of random samples over the distribution `e`

, then plot them as a histogram. Then you will be treating `e`

the same way as the other distributions. Check out http://au.mathworks.com/matlabcentral/fileexchange/26003-random-numbers-from-a-user-defined-distribution for a way to do this.

Problem #2: I don't think

```
e(y)=(1./(sqrt(2*pi*y*p*(1-p)))).*exp(-(((y+1)*p-(y*p)).^2)./(2*y*p*(1-p)));
```

is a correct translation of the distribution formula. You should have

```
e(y) = (1./(sqrt(2*pi*n*p*(1-p)))).*exp(-((y-n*p)).^2)./(2*n*p*(1-p)));
```

Note that I've written the formula as a function of a new variable `y`

, since you will need to do so anyway for the reasons above.

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