let x be a continuous random variable with density function...?

Question

let x be a continuous random variable with density function

f(x) = (1/(x^2)) for x>=1
and
f(x) = 0 everywhere else.

how do you find P(X=2)?

i know how to find inequalities for these type of problems but i can't figure out equality. I know the answer is 0 but how do we get that?

Answer 1

If you know how to deal with inequalities, you can do equalities too. X=2 means the same as the double inequality 2=

Answer 2

For a continuous random variable, the probability is defined as the area under a segment of the probability curve. This is why for an equality the probabilty is zero; the area under a point is zero. this is why for a continuous random variable we don't ask for the probability of a single value equality, we ask for the probabilty of a range, either within the neighborhood of a single value, or from a value to the middle (as in how the normal distribution table is set up) or from the value to a tail.

Answer 3

the probability that x=2 is the area above that point and below the function. the point x=2 has no length so there is no area in the region in question, so P(x=2)=0

Answer 4

This is found by the definite integral int(x from 2 to 2) f(x)dx.

Look at the bounds of the integral. See why the answer is zero?