PYQs / GATE CE / 2025 / Set 2 / Q51 GATE CE 2025 Set 2 — Question 51 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 Random Variables Probability Engineering Mathematics Standard Deviation Statistics & Regression
Engineering Mathematics → Probability → Random Variables
Last updated 5 September 2026
Question Consider a discrete random variable
X X X whose probabilities are given below. The standard deviation of the random variable is
______ (round off to one decimal place).
x i x_i x i 1 2 4 8 P ( X = x i ) P(X = x_i) P ( X = x i ) 0.3 0.1 0.3 0.3
Solution 1. Calculate the mean (expected value) E [ X ] E[X] E [ X ] E [ X ] = ∑ x i P ( x i ) = ( 1 × 0.3 ) + ( 2 × 0.1 ) + ( 4 × 0.3 ) + ( 8 × 0.3 ) E[X] = \sum x_i P(x_i) = (1 \times 0.3) + (2 \times 0.1) + (4 \times 0.3) + (8 \times 0.3) E [ X ] = ∑ x i P ( x i ) = ( 1 × 0.3 ) + ( 2 × 0.1 ) + ( 4 × 0.3 ) + ( 8 × 0.3 ) E [ X ] = 0.3 + 0.2 + 1.2 + 2.4 = 4.1 E[X] = 0.3 + 0.2 + 1.2 + 2.4 = 4.1 E [ X ] = 0.3 + 0.2 + 1.2 + 2.4 = 4.1 2. Calculate E [ X 2 ] E[X^2] E [ X 2 ] E [ X 2 ] = ∑ x i 2 P ( x i ) = ( 1 2 × 0.3 ) + ( 2 2 × 0.1 ) + ( 4 2 × 0.3 ) + ( 8 2 × 0.3 ) E[X^2] = \sum x_i^2 P(x_i) = (1^2 \times 0.3) + (2^2 \times 0.1) + (4^2 \times 0.3) + (8^2 \times 0.3) E [ X 2 ] = ∑ x i 2 P ( x i ) = ( 1 2 × 0.3 ) + ( 2 2 × 0.1 ) + ( 4 2 × 0.3 ) + ( 8 2 × 0.3 ) E [ X 2 ] = ( 1 × 0.3 ) + ( 4 × 0.1 ) + ( 16 × 0.3 ) + ( 64 × 0.3 ) E[X^2] = (1 \times 0.3) + (4 \times 0.1) + (16 \times 0.3) + (64 \times 0.3) E [ X 2 ] = ( 1 × 0.3 ) + ( 4 × 0.1 ) + ( 16 × 0.3 ) + ( 64 × 0.3 ) E [ X 2 ] = 0.3 + 0.4 + 4.8 + 19.2 = 24.7 E[X^2] = 0.3 + 0.4 + 4.8 + 19.2 = 24.7 E [ X 2 ] = 0.3 + 0.4 + 4.8 + 19.2 = 24.7 3. Calculate the variance V a r ( X ) Var(X) V a r ( X ) V a r ( X ) = E [ X 2 ] − ( E [ X ] ) 2 = 24.7 − ( 4.1 ) 2 Var(X) = E[X^2] - (E[X])^2 = 24.7 - (4.1)^2 V a r ( X ) = E [ X 2 ] − ( E [ X ] ) 2 = 24.7 − ( 4.1 ) 2 V a r ( X ) = 24.7 − 16.81 = 7.89 Var(X) = 24.7 - 16.81 = 7.89 V a r ( X ) = 24.7 − 16.81 = 7.89 4. Calculate the standard deviation S D ( X ) SD(X) S D ( X ) S D ( X ) = V a r ( X ) = 7.89 ≈ 2.8089 SD(X) = \sqrt{Var(X)} = \sqrt{7.89} \approx 2.8089 S D ( X ) = V a r ( X ) = 7.89 ≈ 2.8089 Rounding to one decimal place, the standard deviation is 2.8.
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Solution 1. Calculate the mean (expected value) E [ X ] E[X] E [ X ] E [ X ] = ∑ x i P ( x i ) = ( 1 × 0.3 ) + ( 2 × 0.1 ) + ( 4 × 0.3 ) + ( 8 × 0.3 ) E[X] = \sum x_i P(x_i) = (1 \times 0.3) + (2 \times 0.1) + (4 \times 0.3) + (8 \times 0.3) E [ X ] = ∑ x i P ( x i ) = ( 1 × 0.3 ) + ( 2 × 0.1 ) + ( 4 × 0.3 ) + ( 8 × 0.3 ) E [ X ] = 0.3 + 0.2 + 1.2 + 2.4 = 4.1 E[X] = 0.3 + 0.2 + 1.2 + 2.4 = 4.1 E [ X ] = 0.3 + 0.2 + 1.2 + 2.4 = 4.1 2. Calculate E [ X 2 ] E[X^2] E [ X 2 ] E [ X 2 ] = ∑ x i 2 P ( x i ) = ( 1 2 × 0.3 ) + ( 2 2 × 0.1 ) + ( 4 2 × 0.3 ) + ( 8 2 × 0.3 ) E[X^2] = \sum x_i^2 P(x_i) = (1^2 \times 0.3) + (2^2 \times 0.1) + (4^2 \times 0.3) + (8^2 \times 0.3) E [ X 2 ] = ∑ x i 2 P ( x i ) = ( 1 2 × 0.3 ) + ( 2 2 × 0.1 ) + ( 4 2 × 0.3 ) + ( 8 2 × 0.3 ) E [ X 2 ] = ( 1 × 0.3 ) + ( 4 × 0.1 ) + ( 16 × 0.3 ) + ( 64 × 0.3 ) E[X^2] = (1 \times 0.3) + (4 \times 0.1) + (16 \times 0.3) + (64 \times 0.3) E [ X 2 ] = ( 1 × 0.3 ) + ( 4 × 0.1 ) + ( 16 × 0.3 ) + ( 64 × 0.3 ) E [ X 2 ] = 0.3 + 0.4 + 4.8 + 19.2 = 24.7 E[X^2] = 0.3 + 0.4 + 4.8 + 19.2 = 24.7 E [ X 2 ] = 0.3 + 0.4 + 4.8 + 19.2 = 24.7 3. Calculate the variance V a r ( X ) Var(X) V a r ( X ) V a r ( X ) = E [ X 2 ] − ( E [ X ] ) 2 = 24.7 − ( 4.1 ) 2 Var(X) = E[X^2] - (E[X])^2 = 24.7 - (4.1)^2 V a r ( X ) = E [ X 2 ] − ( E [ X ] ) 2 = 24.7 − ( 4.1 ) 2 V a r ( X ) = 24.7 − 16.81 = 7.89 Var(X) = 24.7 - 16.81 = 7.89 V a r ( X ) = 24.7 − 16.81 = 7.89 4. Calculate the standard deviation S D ( X ) SD(X) S D ( X ) S D ( X ) = V a r ( X ) = 7.89 ≈ 2.8089 SD(X) = \sqrt{Var(X)} = \sqrt{7.89} \approx 2.8089 S D ( X ) = V a r ( X ) = 7.89 ≈ 2.8089 Rounding to one decimal place, the standard deviation is 2.8.
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