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The p.d.f. of r.v. of X is given by f (x) = kx , for 0 < x < 4 and = 0, otherwise. Determine k . Determine c.d.f. of X and hence P (X ≤ 2) and P(X ≤ 1). - Mathematics and Statistics

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Question

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

f (x) = `k /sqrtx` , for 0 < x < 4 and = 0, otherwise. Determine k .

Determine c.d.f. of X and hence P (X ≤ 2) and P(X ≤ 1).

Sum

Solution

Since, f is p.d.f. of the r.v. X,

` int_(-∞)^∞ f(x) dx = 1`

∴ `int_(-∞)^∞ f(x) dx + int_(-0)^4 f(x) dx + int_(4)^∞ f(x) dx = 1`

∴`0 + int_(0)^4 k /sqrtx dx + 0 = 1`

∴ `k int_(0)^4 x ^(-1/2) dx = 1`

∴ `k[x^(1/2)/(1/2)]_0^4 =1`

∴ 2k[2 - 0] = 1

∴ 4k = 1

∴ k = `1/4`

Let F(X) be the c.d.f. of X.

∴ F(X) = P(X ≤ x) = `int_(-∞)^x f(x) dx`

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

= `0 + int_(0)^x k /sqrtx dx`

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

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

= `2k sqrtx = 2(1/4) sqrtx` .....[∵ k = `1/4`]

∴ F(X) = `sqrtx /2`

P(X ≤ 2) = F (2) = `sqrt2/2 = 1/sqrt2`

P(X ≤ 1) = F (1) = `sqrt1/2 = 1/2`

Hence, `k 1/ 4 ,P (X ≤ 2) = 1/sqrt2 , P (X ≤ 1) = 1/2`

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Chapter 7: Probability Distributions - Miscellaneous Exercise 2 [Page 244]

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Balbharati Mathematics and Statistics 2 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board
Chapter 7 Probability Distributions
Miscellaneous Exercise 2 | Q 15 | Page 244

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