Discussion Problem_(20)

[duck]

If the probability density function(pdf) of a random variable X is given by:
f(x) = 1/8 for 0≤x≤8
      = 0 otherwise.

Find P(|X-1| > 2).

Read "|X-1|>2" as mod of X-1 greater than 2.

[MI]

5/8

[Mauli]

same
5/8

[duck]

Cool.. Correct Answer: 5/8

[Erika]

i did integral 0 to 8- integral 0 to 2 for f(x). And I get 3/4.
Plz guide me to the correct method..i seem to not get these intergal questions for PDF right :(

[duck]

Hi Erika.. :)

|X-1|>2 means 3<X<-1.
As X lies between [0,8], so will integrate pdf from 3 to 8.

You took wrong limits.
It should be "integral 0 to 8 - integral 0 to 3 for f(x)"

[Erika]

Thank you. I got that one :)