Question

F(x) = x² , 0<x<1  then P(1/2 < x< 3/4) is

Answer 1

F(x) = x² 

P(1/2 < x< 3/4) = \(\frac{\int_\frac{1}{2}^\frac{3}{4}x^2dx}{\int_o^1x^2dx}\)  = \(\frac{19/64}{1}\) = \(\frac{19}{ 64}\)

Hope it helps

Ask if any doubt

Comments

Why integrated?

---

The probability of a continuous density function f(x), here = x^2, is calculated as the probability of the area bounded in the range. It's a norm we follow because there is no other fruitful way to calculate the probability of a function especially if its domain can’t be bounded, as in here. I can write a whole bunch of reasons but that would be very tedious so I’m posting some links where you can learn more about it:

https://www.quora.com/Why-the-probability-is-the-area-under-some-curve

https://opentextbc.ca/introbusinessstatopenstax/chapter/properties-of-continuous-probability-density-functions/

https://math.stackexchange.com/questions/506925/why-the-area-under-the-probability-density-function-curve-is-probability