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May 11th, 2014, 06:07 AM   #1
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Question about Probability Density Functions

Hello everyone (:

Can somebody please explain this to me? 👉 In a continuous probability distribution, why is P(X = x) = 0? i.e., why is the probability that the continuous random variable, X, takes a discrete value, x (say, x = 1) equal to 0? Why can the PDF only be used to calculate the probability of a continuous RV take a value less than OR greater than but NOT equal to a certain discrete value?

For eg. : Consider the (continuous) probability distribution of the heights of 20 individuals. According to what I've learnt, the probability that the height of a randomly selected individual out of those 20 individuals is, say 180cm = 0. How is that possible? Cos, 180cm seem to be a likely possibility/outcome intuitively; so why does probability theory prove otherwise?

Thank you for the help (:
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May 11th, 2014, 09:28 AM   #2
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continuous probability distribution is a distrubution for a random variable which takes continuous value (anologue type of value).
Example : there are infinite numbers between 0 and 1. Like 0.01, 0.0068, 0.563234 etc. this is true for any two numbers a and b. If a random variable is to take any one value between a and b it can assume value among those infinite number of values. This kind of probability bistribution cannot be represented in discrete points so continuous distribution is used.

A continuous random variable cannot take a fixed value like X = 0 because the probability for the RV to take the value is 0
Example : there are infinite number of points between two numbers a and b among which c may be one.
the probability that the RV may take the value c
P(X=c) = number of c / total number of possible points
= 1 / infinity
= 0
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May 11th, 2014, 02:19 PM   #3
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180 cm is a likely possibility as an approximation, but not exact,
i.e. 180.0000000000...... has probability 0.
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May 11th, 2014, 10:53 PM   #4
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Quote:
Originally Posted by mathman View Post
180 cm is a likely possibility as an approximation, but not exact,

i.e. 180.0000000000...... has probability 0.

Oh alright; so it's just that if you substitute x = 180 in the PDF of the concerned continuous distribution, you will get some probability value but if you substitute x = 180 in the PMF of the distribution you get 0 (since P(X = x) = 0 for continuous RV distributions).

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