Why is norm pdf 0 .3989?

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Chase Weinberg
Chase Weinberg 2019-11-15
编辑： the cyclist 2019-11-15
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.
  2 个评论
Adam
Adam 2019-11-15
What code are you using?
pdf( 0 )
is not valid syntax unless you have some 3rd party pdf function.
the cyclist
the cyclist 2019-11-15
They mean normpdf(0), not the norm of pdf(0). It's a function from the Statistics and Machine Learning Toolbox.

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回答（2 个）

the cyclist
the cyclist 2019-11-15
You want
normcdf(0)
not
normpdf(0)
Chase Weinberg
Chase Weinberg 2019-11-15
编辑：Chase Weinberg 2019-11-15
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.
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Chase Weinberg
Chase Weinberg 2019-11-15
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
the cyclist 2019-11-15
编辑：the cyclist 2019-11-15
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.

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