Why is norm pdf 0 .3989?

Wouldn't normpdf(0) be .5? The probability of getting below 0 given a mean of 0 for a normal standard distribution function should be .5.

[Adam]

What code are you using?

pdf( 0 )

is not valid syntax unless you have some 3rd party pdf function.

[the cyclist]

They mean normpdf(0), not the norm of pdf(0). It's a function from the Statistics and Machine Learning Toolbox.

Answer 1

[the cyclist]

You want

normcdf(0)

not

normpdf(0)

Answer 2

[Chase Weinberg]

Oh I see, so normcdf(0)= .5000

So what does normpdf(0) mean?

I see that we plug 0 into the function 1/σsqrt(2π) * e^-(x−μ)^/(2σ^2)

that comes out to .3989, but what does this mean practically? The probability of getting 0 within the standard normal is 39.89%? this seems high.

[the cyclist]

It's the value of the probability density function (not the cumulative probability, as you wanted) of a standard normal distribution, evaluated at x=0.

[Chase Weinberg]

Whats the practical explanation. We are saying there is a 39.89% chance of this value occurring.

Wikipedia gave an example of estimating the chance of a bacteria dying between 4 and 6 hours. They say there's a 0% chance of it dying at exactly hour 5. But within that time frame say 5 hours exactly to 5 hours and one nanosecond, there is a much greater chance of it dying.

They're saying that the chance of the bacteria dying at 5 hours is 0 but the chance it dies in the time frame is much more significant because you would take the probability divided by the time frame.

I'm still confused on the practical explanation of normpdf(0)=.3989

[the cyclist]

I'm not sure this is the best forum for a probability lesson. :-)

One has to be careful not to confuse the probability density with the probability itself.

Assuming a normal distribution, the probability that the event happens in the interval [x1 x2] is given by the integral of

normpdf(x) * dx

over the interval [x1 x2].

normpdf is not the probability. It's the probability density. You have to integrate the density over an interval to get a probability.

So, the "practical explanation" of normpdf(0) is that if you wanted to know the probability of something happening in the interval, say, x = 0.00 to x = 0.01, it would be approximately

normpdf(0) * ((0.01) - (0.00))
= 0.3989 * 0.01
= 0.003989

This is only approximate, because the pdf itself actually also changes a tiny bit from 0.00 to 0.01.