Why does this probability density function looks off?

I have a random variable x which has a mean of 10 and a variance of 16. I used the following code to generate the array and the PDF:

x= randn(100,1) * sqrt(16)+10;
mu = 10;
sigma = 4;
pd = makedist('Normal',mu,sigma);
y=pdf(pd,x);
hist(y)

Should i be plotting with something other than hist or is x itself wrong? Thank you.

Answer 1

[the cyclist]

The mistake is in using hist. Plot x vs. y instead, because y is the value of the pdf itself.

x= randn(100,1) * sqrt(16)+10;
mu = 10;
sigma = 4;
pd = makedist('Normal',mu,sigma);
y=pdf(pd,x);

figure
plot(x,y,'.')

[Mohannad Abboushi]

Ok but it doesn't look like the probabilities add to 1?

[the cyclist]

Remember, it is not the values of pdf(x) that sum to 1. It is dx*pdf(x) that sums to one.

You have not chosen evenly spaced x values, and they are not sorted, so it is just a bit tricky to prove it.

[sorted_x,sortingIndex] = sort(x);
dx = [0; diff(sorted_x)];
probabilitySum = sum(dx.*y(sortingIndex))

This gives 0.9866 in the case I tested, which is about as close as I would expect for N = 100.

[Mohannad Abboushi]

Ok cool, but for instance when I try to figure out the probability that x>5 I get an oddly low value of around 7. When I feel like it should be closer to 100. Here's how I coded it:

P=x>5;
Probability_x_greater_than_5=sum(P.*y);

[the cyclist]

You fell into the same trap of not multiplying by dx. I'm not certain, but you may also failed to account for the fact that your values of y are not ordered. So, I think you may have gotten a more-or-less random answer.

There is no need to choose your x at random. It would be much better to choose an evenly spaced array of x from the beginning, and your whole problem becomes simpler.

Like this ...

mu = 10;
sigma = 4;

dx = 0.01;

x = mu - 5*sigma : dx : mu + 5*sigma;

pd = makedist('Normal',mu,sigma);
y=pdf(pd,x);
figure
plot(x,y)

probability_total = sum(dx*y)

Probability_x_greater_than_5 = sum(dx.*y(x>5))

I got 0.8941 for that probability.

Notice that these are probabilities, so they lie between 0 and 1. If you want them represented as percentages, then divide by 0.01.

[Mohannad Abboushi]

Awesome thank you very much that was a good explanation!