# Find k if the following function represent p.d.f. of r.v. X. f (x) = kx (1 – x), for 0 < x < 1 and = 0 otherwise, Also find P (14<x<12),P(x<12). - Mathematics and Statistics

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Find k if the following function represent p.d.f. of r.v. X.

f (x) = kx (1 – x), for 0 < x < 1 and = 0 otherwise, Also find P (1 /4 < x < 1 /2) , P (x < 1 /2).

#### Solution

Since, the function f is the p.d.f. of X,

 int_(-∞)^∞ f (x) dx = 1

∴  int_(-∞)^0 f (x) dx+ int_(0)^1 f (x) dx+ int_(1)^∞f (x) dx = 1

∴ 0+ int_(0)^1kx (1-x) dx + 0 = 1

∴ k int_(0)^1 (x - x^2) dx = 1

∴ k [x^2/2-x^3/3]_0^1 = 1

∴ k (1/2 - 1/3-0) = 1

∴ k/6 = 1

∴ k = 6

P (1 /4 < x < 1 /2) =  int_(1/4)^(1/2) f (x) dx

=  int_(1/4)^(1/2) kx( 1-x) dx

= k int_(1/4)^(1/2) (x-x^2) dx

= 6[x^2/2 - x^3/3]_(1/4)^(1/2)..........[∵ k = 6]

= 6 [(1/8-1/24)-(1/32-1/192)]

= 6[2/24-5/192]

= 6 (11/192)

∴ P (1 /4 < x < 1 /2) = 11/32

P (x < 1 /2) =  int_(0)^(1/2) f (x) dx

= 0 + int_(0)^(1/2) kx(1 - x) dx

= k int_(0)^(1/2) (x - x^2) dx

= k[x^2/2-(x^3)/3]_0^(1/2)

=k [1/8-1/24-0]

= k (2/24)

= 6 (1/12) ............[∵ k = 6]

∴ P (x <  1/2)

= 1/2

Concept: Probability Distribution of Discrete Random Variables
Is there an error in this question or solution?
Chapter 7: Probability Distributions - Exercise 7.2 [Page 239]

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