PYQs / GATE CE / 2015 / Set 2 / Q36 GATE CE 2015 Set 2 — Question 36 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. NAT +2 / -0 Medium PMF, PDF & CDF Probability Engineering Mathematics
Engineering Mathematics → Probability → PMF, PDF & CDF
Last updated 5 September 2026
Question The probability density function of a random variable,
x x x is
f ( x ) = x 4 ( 4 − x 2 ) for 0 ≤ x ≤ 2 f(x) = \frac{x}{4}(4 - x^2) \text{ for } 0 \leq x \leq 2 f ( x ) = 4 x ( 4 − x 2 ) for 0 ≤ x ≤ 2 = 0 otherwise = 0 \text{ otherwise} = 0 otherwise The mean,
μ x \mu_x μ x of the random variable is
__________ .
Correct answer 1.06 to 1.07
Solution The mean
μ x \mu_x μ x (expected value
E [ x ] E[x] E [ x ] ) of a continuous random variable is given by:
μ x = ∫ − ∞ ∞ x f ( x ) d x \mu_x = \int_{-\infty}^{\infty} x f(x) dx μ x = ∫ − ∞ ∞ x f ( x ) d x Given the PDF:
f ( x ) = x 4 ( 4 − x 2 ) f(x) = \frac{x}{4}(4 - x^2) f ( x ) = 4 x ( 4 − x 2 ) for
0 ≤ x ≤ 2 0 \leq x \leq 2 0 ≤ x ≤ 2 Calculating the integral:
μ x = ∫ 0 2 x ⋅ [ x 4 ( 4 − x 2 ) ] d x \mu_x = \int_{0}^{2} x \cdot \left[ \frac{x}{4}(4 - x^2) \right] dx μ x = ∫ 0 2 x ⋅ [ 4 x ( 4 − x 2 ) ] d x μ x = ∫ 0 2 1 4 ( 4 x 2 − x 4 ) d x \mu_x = \int_{0}^{2} \frac{1}{4}(4x^2 - x^4) dx μ x = ∫ 0 2 4 1 ( 4 x 2 − x 4 ) d x μ x = 1 4 [ 4 x 3 3 − x 5 5 ] 0 2 \mu_x = \frac{1}{4} \left[ \frac{4x^3}{3} - \frac{x^5}{5} \right]_0^2 μ x = 4 1 [ 3 4 x 3 − 5 x 5 ] 0 2 μ x = 1 4 [ ( 4 ( 2 ) 3 3 − ( 2 ) 5 5 ) − 0 ] \mu_x = \frac{1}{4} \left[ \left( \frac{4(2)^3}{3} - \frac{(2)^5}{5} \right) - 0 \right] μ x = 4 1 [ ( 3 4 ( 2 ) 3 − 5 ( 2 ) 5 ) − 0 ] μ x = 1 4 [ 32 3 − 32 5 ] \mu_x = \frac{1}{4} \left[ \frac{32}{3} - \frac{32}{5} \right] μ x = 4 1 [ 3 32 − 5 32 ] μ x = 32 4 [ 1 3 − 1 5 ] = 8 [ 5 − 3 15 ] \mu_x = \frac{32}{4} \left[ \frac{1}{3} - \frac{1}{5} \right] = 8 \left[ \frac{5 - 3}{15} \right] μ x = 4 32 [ 3 1 − 5 1 ] = 8 [ 15 5 − 3 ] μ x = 8 × 2 15 = 16 15 ≈ 1.0667 \mu_x = 8 \times \frac{2}{15} = \frac{16}{15} \approx 1.0667 μ x = 8 × 15 2 = 15 16 ≈ 1.0667 The answer range is 1.06 to 1.07.
Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker. Review answer and solution without JavaScript Interactive answer checking needs JavaScript. The published solution is available below.
Correct answer 1.06 to 1.07
Solution The mean
μ x \mu_x μ x (expected value
E [ x ] E[x] E [ x ] ) of a continuous random variable is given by:
μ x = ∫ − ∞ ∞ x f ( x ) d x \mu_x = \int_{-\infty}^{\infty} x f(x) dx μ x = ∫ − ∞ ∞ x f ( x ) d x Given the PDF:
f ( x ) = x 4 ( 4 − x 2 ) f(x) = \frac{x}{4}(4 - x^2) f ( x ) = 4 x ( 4 − x 2 ) for
0 ≤ x ≤ 2 0 \leq x \leq 2 0 ≤ x ≤ 2 Calculating the integral:
μ x = ∫ 0 2 x ⋅ [ x 4 ( 4 − x 2 ) ] d x \mu_x = \int_{0}^{2} x \cdot \left[ \frac{x}{4}(4 - x^2) \right] dx μ x = ∫ 0 2 x ⋅ [ 4 x ( 4 − x 2 ) ] d x μ x = ∫ 0 2 1 4 ( 4 x 2 − x 4 ) d x \mu_x = \int_{0}^{2} \frac{1}{4}(4x^2 - x^4) dx μ x = ∫ 0 2 4 1 ( 4 x 2 − x 4 ) d x μ x = 1 4 [ 4 x 3 3 − x 5 5 ] 0 2 \mu_x = \frac{1}{4} \left[ \frac{4x^3}{3} - \frac{x^5}{5} \right]_0^2 μ x = 4 1 [ 3 4 x 3 − 5 x 5 ] 0 2 μ x = 1 4 [ ( 4 ( 2 ) 3 3 − ( 2 ) 5 5 ) − 0 ] \mu_x = \frac{1}{4} \left[ \left( \frac{4(2)^3}{3} - \frac{(2)^5}{5} \right) - 0 \right] μ x = 4 1 [ ( 3 4 ( 2 ) 3 − 5 ( 2 ) 5 ) − 0 ] μ x = 1 4 [ 32 3 − 32 5 ] \mu_x = \frac{1}{4} \left[ \frac{32}{3} - \frac{32}{5} \right] μ x = 4 1 [ 3 32 − 5 32 ] μ x = 32 4 [ 1 3 − 1 5 ] = 8 [ 5 − 3 15 ] \mu_x = \frac{32}{4} \left[ \frac{1}{3} - \frac{1}{5} \right] = 8 \left[ \frac{5 - 3}{15} \right] μ x = 4 32 [ 3 1 − 5 1 ] = 8 [ 15 5 − 3 ] μ x = 8 × 2 15 = 16 15 ≈ 1.0667 \mu_x = 8 \times \frac{2}{15} = \frac{16}{15} \approx 1.0667 μ x = 8 × 15 2 = 15 16 ≈ 1.0667 The answer range is 1.06 to 1.07.
Understand the concept, then try another question Revisit Engineering Mathematics with concept notes, common mistakes and an original worked example before your next attempt.
More questions on Probability ← Q35 Full paper Q37 → 