Q.25 The following is the p.d.f. of continuous r.v.: f(x) = x/8;\ 0 lt x lt 4 = 0 , otherwise Find (i) expression for c.d.f. of X . (ii) F(x) at 0.5 and 5

Answer: (i) F(x) = x²/16 for 0 lt x lt 4 (0 for x ≤ 0 , 1 for x ≥ 4 ) (ii) F(0.5) = 1/64 , F(5) = 1 (i) For 0 lt x lt 4 : F(x) = ∫₀^xt/8 dt = [t²/16]₀^x = x²/16 . Also F(x) = 0 for x ≤ 0 and F(x) = 1 for x ≥ 4 . (ii) F(0.5) = (0.5)²/16 = 0.25/16 = 1/64 . Since 5 ≥ 4 , F(5) = 1 .

The following is the p.d.f. of continuous r.v.:\(f(x) = \frac{x}{8};\ 0 \lt x \lt 4\)\(= 0\), otherwiseFind (i) expression for c.d.f. of \(X\). (ii) \(F(x)\) at 0.5 and 5

Q.25 The following is the p.d.f. of continuous r.v.:
\(f(x) = \frac{x}{8};\ 0 \lt x \lt 4\)
\(= 0\), otherwise
Find (i) expression for c.d.f. of \(X\). (ii) \(F(x)\) at 0.5 and 5

Doubtly Asked 28 Oct 2026 Board exam · June 2026 hsc 12th Save WhatsApp Report

1 answer

Best answer

Answer: (i) \(F(x) = \frac{x^2}{16}\) for \(0 \lt x \lt 4\) (0 for \(x \le 0\), 1 for \(x \ge 4\)) (ii) \(F(0.5) = \frac{1}{64}\), \(F(5) = 1\)

(i) For \(0 \lt x \lt 4\): \(F(x) = \int_0^x\frac{t}{8}\,dt = \left[\frac{t^2}{16}\right]_0^x = \frac{x^2}{16}\). Also \(F(x) = 0\) for \(x \le 0\) and \(F(x) = 1\) for \(x \ge 4\).

(ii) \(F(0.5) = \frac{(0.5)^2}{16} = \frac{0.25}{16} = \frac{1}{64}\). Since \(5 \ge 4\), \(F(5) = 1\).

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

Was this answer helpful?

The following is the p.d.f. of continuous r.v.:\(f(x) = \frac{x}{8};\ 0 \lt x \lt 4\)\(= 0\), otherwiseFind (i) expression for c.d.f. of \(X\). (ii) \(F(x)\) at 0.5 and 5

1 answer

Your answer