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Hey,

We have been given some probability density functions to plot. So far I have found this pretty easy, however I have now come across one I can't do.

f(x) = 20x3(1x) 20x^3 (1-x) for 0 < x < 1

I have tried plotting this however it gives values of y which are more than 1 which I thought a pdf couldn't do?

Thanks
Reply 1
Big_Sam
Hey,

We have been given some probability density functions to plot. So far I have found this pretty easy, however I have now come across one I can't do.

f(x) = 20x3(1x) 20x^3 (1-x) for 0 < x < 1

I have tried plotting this however it gives values of y which are more than 1 which I thought a pdf couldn't do?

Thanks

It can have values greater than 1, but the total area must add up to 1, which it does. What are you trying to do with it?
A pdf can have a value greater than 1. The restriction is that it is >= 0 and the integral over +/- infinity is 1.

Consider a continuous uniform distribution on the interval 0 to 0.1

Its value is 10 over than interval.
Reply 3
ah right so when you are plotting a pdf there is no limit on the y axis as to the value as long as it is positive?
Reply 4
but I thought an event couldn't have a probability greater than 1? and if its pdf gives a y value greater than one then the probability is greater than 1?
Reply 5
What "event" do you think corresponds to any particular value of the PDF?

For example, with your p.d.f., when x = 1/2, the value of the PDF is 5/4. What *event* do you think corresponds to this value?
Reply 6
I thought that if the pdf is 5/4 when x=1/2 that meant that the probability of x=1/2 is 5/4 which is greater than 1?
Reply 7
No. If f is the p.d.f. , the value of f doesn't tell you directly about the probability of any particular event.

All it tells you is that the probability that axba \leq x \leq b is abf(t)dt\int_a^b f(t) \, dt.
Reply 8
ah ok I see what you mean.

thanks for the help guys

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