Home
Class 12
MATHS
If the p.d.f of a random variable X is ...

If the p.d.f of a random variable X is
`F(x)={(0.5x, 0lexle2), (0, "otherwise"):}`
Then the value of `P(0.5 le X le 1.5)` is

A

`1/4`

B

`1/2`

C

`3/4`

D

`1`

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Topper's Solved these Questions

  • PROBABILITY DISTRIBUTION

    TARGET PUBLICATION|Exercise Critical Thinking|33 Videos
  • PROBABILITY DISTRIBUTION

    TARGET PUBLICATION|Exercise Competitive Thinking|9 Videos
  • PLANE

    TARGET PUBLICATION|Exercise EVALUATION TEST|14 Videos
  • THREE DIMENSIONAL GEOMETRY

    TARGET PUBLICATION|Exercise Evaluation Test|9 Videos

Similar Questions

Explore conceptually related problems

The p.d.f. of a random variable X is (f)x=1/5, 0lexle5 =0 , otherwise Then the value of P(1 lt X lt 3) is

The p.d.f. of a random variable X is (f)x=2x, 0lexle1 =0 , otherwise Then the value of P(1/3 lt X lt 1/2) is

The p.d.f of a continous random variable X is f(x)=x/8, 0 lt x lt 4 =0 , otherwise Then the value of P(X gt 3) is

The p.d.f. of a r.v. X is f(x)={{:(0.5x","0ltxlt2),(0", otherwise"):} , then P(X le1) =

The p.d.f. of a r.v. X is f(x)={{:(0.5x", "0ltxlt2),(0", otherwise"):} , then P(X gt1.5) =

If the p.d.f of a continuous random variable X is f(x)=kx^(2)(1-x), 0 lt x lt 1 = 0 otherwise Then the value of k is

The p.d.f. of a r.v. X is f(x)={{:(0.5x", "0ltxlt2),(0", otherwise"):} , then P(0.5leXle1.5) =

The pdf of a discrete random variable is defined as f(x)={{:(kx^2","0lexle6),(0", ""elsewhere"):} Then the value of F(4) is

The p.d.f. of a r.v. X is f(x)={{:(kx","0ltxlt2),(0", otherwise"):} , then k =

The p.d.f. of a r.v. X is f(x)={{:(kx^(2)(1-x)","0ltxlt1),(0", otherwise"):} , then k =