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We know that the value of PDF at a specific time (X = x) is 0 due to the nature of a distribution of a continuous variable.

However, I noticed that when I plug in some test numbers in NORM.DIST() in Excel with CUM = FALSE, it still gives a significant, non-zero answer.

=EXPON.DIST(15,1/20,FALSE)
O/P: 0.023618328

I looked up Microsoft's documentation to see if their formula has any continuity correction factor added (like, for above example, X=15 +/- 0.05) but that doesn't seem to be the case.

Can someone explain why this is the answer and not 0?

Anish
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  • Indeed, the expression for the normal PDF is nonzero for all real inputs. The Excel function yields probability per unit, which is explained in the linked question. – Sycorax Dec 22 '20 at 00:48
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    @Sycorax, excellent - this is eye opening. Thank you! – Anish Dec 22 '20 at 01:04

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