# The p.d.f. of r.v. X is given by f (x) = 12a , for 0 < x < 2a and = 0, otherwise. Show that P[X<a2]=P[X>3a2] . - Mathematics and Statistics

Sum

The p.d.f. of a continuous r.v. X is given by

f (x) = 1/ (2a) , for 0 < x < 2a and = 0, otherwise. Show that P [X < a/ 2] = P [X >( 3a)/ 2] .

#### Solution

P (X< a /2) =  int_(-∞)^(a/2) f (x) dx

= int_(-∞)^0 f (x) dx+ int_(-∞)^(a/2) f (x) dx

= 0+  int_(0)^(a/2) 1/(2a) dx

= 1/(2a) int_(0)^(a/2) 1 dx = 1/(2a)[x]_0^(a/2)

= 1/(2a) [a/2-0] = 1/4 .......(1)

P [X >( 3a)/ 2] = int_((3a)/2)^(∞) f(x) dx

= int_((3a)/2)^(*2a) f(x) dx + int_(2a)^(∞) f(x) dx

= int_((3a)/2)^(2a) 1/(2a) dx + 0

= 1/(2a) int_((3a)/2)^(2a) 1 dx = 1/(2a)[x]_((3a)/2)^(2a)

= 1/(2a)[2a-(3a)/2] = 1/(2a)(a/2) = 1/4 ......(2)

From (1) and (2), we get

P [X < a/ 2] = P [X >( 3a)/ 2].

Is there an error in this question or solution?
Chapter 7: Probability Distributions - Miscellaneous Exercise 2 [Page 244]

#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 7 Probability Distributions
Miscellaneous Exercise 2 | Q 14 | Page 244

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