Random Variables and Mathematical Expectation MCQs 2026

49 questions with detailed answers · 29 from past papers · 5 quiz batches available

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Page 1 of 1Questions 1–10 of 49
  1. Q1Past Paper · PPSC/FPSC/CSSeasy

    A random variable is

    1. Aa numerical function assigning a real number to each outcome in the sample space
    2. Balways a constant
    3. Conly a probability
    4. Dthe sample space itself
    💡 Explanation:

    X(ω) maps outcomes to numbers for algebraic treatment.

  2. Q2Past Paper · PPSC/FPSC/CSSeasy

    A discrete random variable takes

    1. Aonly every real number
    2. Bonly zero
    3. Cvalues with density only
    4. Da countable set of values (finite or countably infinite)
    💡 Explanation:

    Discrete RVs have jump probabilities at specific points.

  3. Q3Past Paper · PPSC/FPSC/CSSeasy

    A continuous random variable takes

    1. Aonly integers
    2. Bonly 0 and 1
    3. Ca finite list always
    4. Dvalues in an interval (uncountably many)
    💡 Explanation:

    Continuous RVs are modeled with density over intervals.

  4. Q4Past Paper · PPSC/FPSC/CSSeasy

    The probability mass function (PMF) p(x) gives

    1. Adensity at x for continuous X
    2. Bcumulative probability always
    3. CP(X = x) for discrete X
    4. Dvariance of X
    💡 Explanation:

    PMF assigns probabilities to individual discrete values.

  5. Q5Past Paper · PPSC/FPSC/CSSeasy

    A valid PMF must satisfy

    1. Ap(x) can be negative
    2. Bp(x) ≥ 0 and Σ p(x) = 1 over all x
    3. Csum equals 0
    4. Dp(x) unbounded above
    💡 Explanation:

    PMF is non-negative and totals to 1.

  6. Q6Past Paper · PPSC/FPSC/CSSeasy

    The probability density function (PDF) f(x) for continuous X satisfies

    1. Af(x) equals P(X=x)
    2. Bf(x) ≥ 0 and ∫f(x)dx = 1 over all x
    3. CΣf(x)=1
    4. Df(x) can be negative
    💡 Explanation:

    PDF integrates to 1; point probability is zero for continuous X.

  7. Q7Past Paper · PPSC/FPSC/CSSmedium

    For continuous X, P(a ≤ X ≤ b) equals

    1. Af(a) + f(b)
    2. Bf(a) × f(b)
    3. CΣ f(x)
    4. D∫_a^b f(x) dx
    💡 Explanation:

    Probability is area under the PDF over the interval.

  8. Q8Past Paper · PPSC/FPSC/CSSeasy

    The cumulative distribution function F(x) is defined as

    1. AP(X = x) only
    2. BP(X ≤ x)
    3. CP(X ≥ x) always
    4. DE(X)
    💡 Explanation:

    CDF gives left-tail probability up to x.

  9. Q9Past Paper · PPSC/FPSC/CSSeasy

    For any CDF, lim_{x→−∞} F(x) equals

    1. A0
    2. B1
    3. C0.5
    4. Dundefined
    💡 Explanation:

    No mass below −∞.

  10. Q10Past Paper · PPSC/FPSC/CSSeasy

    For any CDF, lim_{x→+∞} F(x) equals

    1. A1
    2. B0
    3. C0.5
    4. Dinfinity
    💡 Explanation:

    All probability accumulated by +∞.

  11. Q11medium

    A CDF is always

    1. Adecreasing
    2. Balways linear
    3. Cnegative
    4. Dnon-decreasing and right-continuous
    💡 Explanation:

    F(x2) ≥ F(x1) if x2 ≥ x1.

  12. Q12medium

    For discrete X, F(x) is

    1. Aa step function with jumps at the values of X
    2. Balways smooth
    3. Calways zero
    4. Dconstant
    💡 Explanation:

    CDF jumps by p(x) at each possible value.

  13. Q13Past Paper · PPSC/FPSC/CSSeasy

    The expected value E(X) of discrete X equals

    1. AΣ p(x) only
    2. BΣ x only
    3. CΣ x · p(x)
    4. DΣ x² p(x)
    💡 Explanation:

    Expectation is the probability-weighted average of X.

  14. Q14Past Paper · PPSC/FPSC/CSSmedium

    For X with PMF: x=0,p=0.2; x=1,p=0.5; x=2,p=0.3, E(X) equals

    1. A1.0
    2. B1.1
    3. C1.3
    4. D0.8
    💡 Explanation:

    E(X) = 0×0.2 + 1×0.5 + 2×0.3 = 0.5 + 0.6 = 1.1.

  15. Q15Past Paper · PPSC/FPSC/CSSmedium

    For the same distribution, E(X²) equals

    1. A1.7
    2. B1.1
    3. C1.21
    4. D2.0
    💡 Explanation:

    E(X²) = 0²×0.2 + 1²×0.5 + 2²×0.3 = 0 + 0.5 + 1.2 = 1.7.

  16. Q16Past Paper · PPSC/FPSC/CSSmedium

    For the same distribution, Var(X) equals

    1. A1.7
    2. B0.6
    3. C0.49
    4. D1.1
    💡 Explanation:

    Var(X) = E(X²) − [E(X)]² = 1.7 − 1.21 = 0.49.

  17. Q17Past Paper · PPSC/FPSC/CSSeasy

    Variance Var(X) is defined as

    1. AE(X)² only
    2. BE[(X − μ)²] = E(X²) − [E(X)]²
    3. CE(X) only
    4. Dstandard deviation squared always incorrectly as E(X)
    💡 Explanation:

    Variance measures squared deviation from the mean.

  18. Q18Past Paper · PPSC/FPSC/CSSeasy

    The standard deviation σ equals

    1. AVar(X) squared
    2. Bthe positive square root of Var(X)
    3. CE(X) − μ
    4. DP(X ≤ μ)
    💡 Explanation:

    σ = √Var(X) restores original units.

  19. Q19Past Paper · PPSC/FPSC/CSSeasy

    Linearity of expectation gives E(aX + b) as

    1. AaE(X) × b
    2. BE(X) + a + b
    3. Ca²E(X)
    4. DaE(X) + b
    💡 Explanation:

    Expectation is linear regardless of independence.

  20. Q20Past Paper · PPSC/FPSC/CSSmedium

    If X and Y are independent, E(XY) equals

    1. AE(X) + E(Y)
    2. BE(X) / E(Y)
    3. CE(X)E(Y)
    4. DE(X²)
    💡 Explanation:

    Independence factorizes expectation of product.

  21. Q21Past Paper · PPSC/FPSC/CSSmedium

    Var(aX + b) equals

    1. Aa²Var(X)
    2. BaVar(X) + b
    3. CaVar(X)
    4. DVar(X) + b²
    💡 Explanation:

    Adding constant b does not affect spread; scaling squares in variance.

  22. Q22Past Paper · PPSC/FPSC/CSSmedium

    If X and Y are independent, Var(X + Y) equals

    1. AVar(X) × Var(Y)
    2. BVar(X) + Var(Y)
    3. CVar(X) − Var(Y)
    4. D[Var(X) + Var(Y)]²
    💡 Explanation:

    Variances add for independent sums.

  23. Q23Past Paper · PPSC/FPSC/CSSeasy

    The PMF of a Bernoulli(p) trial is

    1. AP(X=0)=p always
    2. Buniform on 0,1,2
    3. CP(X=1)=1 always
    4. DP(X=1)=p, P(X=0)=1−p
    💡 Explanation:

    Bernoulli models one success/failure with parameter p.

  24. Q24Past Paper · PPSC/FPSC/CSSeasy

    For Bernoulli(p), E(X) equals

    1. A1 − p
    2. Bp(1 − p)
    3. Cp
    4. Dp²
    💡 Explanation:

    E(X) = 0×(1−p) + 1×p = p.

  25. Q25Past Paper · PPSC/FPSC/CSSmedium

    For Bernoulli(p), Var(X) equals

    1. Ap²
    2. Bp(1 − p)
    3. C1 − p
    4. Dp
    💡 Explanation:

    Var = p − p² = p(1−p).

  26. Q26easy

    The mode of a distribution is

    1. Athe value of x with highest probability (discrete) or highest density (continuous)
    2. Balways the mean
    3. Calways the median
    4. DE(X²)
    💡 Explanation:

    Mode is the most likely value.

  27. Q27medium

    The median m satisfies

    1. AP(X ≤ m) ≥ 0.5 and P(X ≥ m) ≥ 0.5
    2. BP(X = m) = 1 always
    3. Cm = E(X) always
    4. Dm = mode always
    💡 Explanation:

    Median splits the distribution in half.

  28. Q28medium

    The relationship between PDF and CDF is

    1. AF(x) = f(x) × f(x)
    2. Bf(x) = F(x) − 1
    3. Cf(x) = dF(x)/dx where derivative exists
    4. Dno relationship
    💡 Explanation:

    Differentiating CDF yields PDF for continuous X.

  29. Q29hard

    For discrete X, recovering PMF from CDF: p(x) equals

    1. AF(x) × 2
    2. BF(x) − F(x−) (jump at x)
    3. C1 − F(x) always
    4. DΣ F(x)
    💡 Explanation:

    Jump size in CDF equals point mass.

  30. Q30hard

    Chebyshev's inequality bounds

    1. AP(X = μ) exactly
    2. BP(|X − μ| ≥ kσ) ≤ 1/k² for k > 0
    3. Conly normal distributions
    4. DVar(X) = 0
    💡 Explanation:

    Chebyshev holds for any distribution with finite variance.

  31. Q31Past Paper · PPSC/FPSC/CSSeasy

    A fair die score X. E(X) equals

    1. A3.5
    2. B3
    3. C4
    4. D3.6
    💡 Explanation:

    E(X) = (1+2+3+4+5+6)/6 = 21/6 = 3.5.

  32. Q32hard

    For fair die X, Var(X) equals

    1. A35/12
    2. B7/12
    3. C2.5
    4. D3.5
    💡 Explanation:

    E(X²)=(1+4+9+16+25+36)/6=91/6; Var=91/6−12.25=35/12≈2.917.

  33. Q33Past Paper · PPSC/FPSC/CSSmedium

    If Y = 2X + 3 and E(X)=4, Var(X)=5, then E(Y) equals

    1. A8
    2. B13
    3. C11
    4. D10
    💡 Explanation:

    E(Y) = 2×4 + 3 = 11.

  34. Q34Past Paper · PPSC/FPSC/CSSmedium

    With the same Y = 2X + 3, Var(Y) equals

    1. A10
    2. B5
    3. C25
    4. D20
    💡 Explanation:

    Var(Y) = 2²×5 = 20.

  35. Q35medium

    Joint PMF p(x,y) gives

    1. AP(X = x and Y = y)
    2. BP(X = x) only
    3. CE(XY) always
    4. DCov(X,Y)
    💡 Explanation:

    Joint PMF specifies simultaneous probabilities.

  36. Q36medium

    Marginal PMF p_X(x) is found by

    1. Asumming p(x,y) over all y
    2. Bmultiplying over y
    3. Csetting y = x always
    4. Dintegrating CDF only
    💡 Explanation:

    Marginalization sums (or integrates) out other variables.

  37. Q37Past Paper · PPSC/FPSC/CSSmedium

    X and Y are independent iff

    1. AE(X) = E(Y)
    2. Bp(x,y) = p_X(x) p_Y(y) for all x,y
    3. CVar(X) = Var(Y)
    4. Dρ = 1
    💡 Explanation:

    Independence factorizes the joint distribution.

  38. Q38Past Paper · PPSC/FPSC/CSSmedium

    Cov(X,Y) = E(XY) − E(X)E(Y). If independent, Cov(X,Y) equals

    1. A1
    2. BVar(X)
    3. C0
    4. DE(X)E(Y)
    💡 Explanation:

    Independence implies zero covariance (converse not always true).

  39. Q39medium

    Correlation ρ(X,Y) equals

    1. ACov(X,Y)²
    2. Bσ_X + σ_Y
    3. CCov(X,Y) / (σ_X σ_Y)
    4. DE(XY) only
    💡 Explanation:

    Correlation is standardized covariance in [−1,1].

  40. Q40hard

    A function g(X) is a random variable. E[g(X)] for discrete X equals

    1. Ag(E(X)) always
    2. BΣ g(x) p(x)
    3. CΣ g(x) only without weights
    4. DE(X)^g
    💡 Explanation:

    LOTUS: expectation of function uses same weights p(x).

  41. Q41hard

    Jensen's inequality for convex g states E[g(X)] is

    1. Aalways equal to g(E(X))
    2. B≥ g(E(X)) for convex g
    3. C≤ g(E(X)) for convex g
    4. Dunrelated to convexity
    💡 Explanation:

    Convex functions: average of g ≥ g of average.

  42. Q42hard

    The moment generating function M(t) = E(e^{tX}) helps find moments because

    1. AM(t) equals Var(X)
    2. BM(t) is always 1
    3. CE(X^n) equals the nth derivative of M(t) at t=0
    4. Dderivatives give only median
    💡 Explanation:

    Differentiate MGF and evaluate at 0 for moments.

  43. Q43medium

    Percentile p_k satisfies

    1. AP(X = p_k) = k
    2. Bp_k = E(X)
    3. Cp_k = mode
    4. DP(X ≤ p_k) = k/100 for continuous X
    💡 Explanation:

    k-th percentile is value with k% below it.

  44. Q44easy

    Interquartile range (IQR) equals

    1. AQ1 + Q3
    2. Bmedian − mean
    3. Cσ²
    4. DQ3 − Q1
    💡 Explanation:

    IQR spans middle 50% of data/distribution.

  45. Q45Past Paper · PPSC/FPSC/CSSeasy

    For X uniform on {1,2,3,4}, E(X) equals

    1. A2.5
    2. B2
    3. C3
    4. D2.25
    💡 Explanation:

    E(X) = (1+2+3+4)/4 = 10/4 = 2.5.

  46. Q46hard

    A RV with E(X)=μ and Var(X)=0 must satisfy

    1. AP(X=0)=1
    2. Buniform distribution
    3. CP(X=μ)=0
    4. DP(X = μ) = 1
    💡 Explanation:

    Zero variance means no spread: degenerate at μ.

  47. Q47medium

    The law of large numbers states that

    1. Asample mean equals μ for any n
    2. BVar(X) goes to infinity
    3. Cthe sample mean converges to μ as n increases
    4. DPMF changes with n
    💡 Explanation:

    LLN: X̄ₙ → μ in probability under iid finite μ.

  48. Q48easy

    A probability histogram for discrete data uses

    1. Asmooth curves only
    2. Bnegative heights
    3. Careas unrelated to probability
    4. Dbar heights equal to probabilities (or relative frequencies)
    💡 Explanation:

    Bar area/height represents probability mass at each value.

  49. Q49medium

    The coefficient of variation CV equals

    1. Aμ / σ
    2. Bσ² / μ
    3. Cμ × σ
    4. Dσ / μ (standard deviation divided by mean)
    💡 Explanation:

    CV measures relative variability, useful for comparing scales.