[MATH HL] Continuous probability question

[appleme]

This is from May 2008 paper 1:

The random variable T has the probability density function

f (t) = pi/4 (cos(pi*t/2)), -1 < t < 1

Find

(a) P(T = 0)

The answer is zero on the markscheme - the working shows that they've integrated with the limits 0 and 0 to get this... but why?

Would you get zero if the question was P(T = 1) ? Is it only because you're trying to find a very specific value/pdf in a cdf function that the probability of getting that very specific number is near to nothing/zero??

Any help would be appreciated!

Edited April 26, 2009 by appleme

[SharkSpider]

Integrate from 0 to 0, and you will get your answer, just like you will when you integrate from 1 to 1. Continuous random variables have an equal probability of picking one of infinitely many values.

[Spriteling]

In a continuous probability function, the probability of something happening is the area under the curve. Thus, you would integrate to find the probability. Therefore, you cannot have something like (P=0) for a continuous probability function, because you can't have area under the curve for just one point. Therefore, you should that by trying to integrate between 0 and 0, and showing that you get an answer of 0.

[appleme]

Thank you both so much!