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If the p.d.f. of a continuous r.v. \(X\) is\(f(x) = \dfrac{x + 2}{18}\), for \(-2 \lt x \lt 4\)\(\quad = 0\), otherwisethen \(P(|X| \lt 1) =\) _____

Q.1 (viii) If the p.d.f. of a continuous r.v. \(X\) is
\(f(x) = \dfrac{x + 2}{18}\), for \(-2 \lt x \lt 4\)
\(\quad = 0\), otherwise
then \(P(|X| \lt 1) =\) _____

A. \(\dfrac{1}{9}\)

B. \(\dfrac{2}{9}\)

C. \(\dfrac{1}{27}\)

D. \(\dfrac{2}{27}\)

1 answer

Best answer

Answer: B. \(\dfrac{2}{9}\)

Answer: (b) \(\dfrac{2}{9}\)

\(P(|X| \lt 1) = P(-1 \lt X \lt 1) = \displaystyle\int_{-1}^{1}\frac{x + 2}{18}dx = \frac{1}{18}\left[\frac{x^2}{2} + 2x\right]_{-1}^{1} = \frac{1}{18}\left[\left(\tfrac{1}{2} + 2\right) - \left(\tfrac{1}{2} - 2\right)\right] = \frac{4}{18} = \frac{2}{9}\).

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