a numerical function assigning a real number to each outcome in the sample space

A discrete random variable takes

a countable set of values (finite or countably infinite)

A continuous random variable takes

values in an interval (uncountably many)

A valid PMF must satisfy

p(x) ≥ 0 and Σ p(x) = 1 over all x

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 1 Questions 1–10 of 49

Q1 Past Paper · PPSC/FPSC/CSS easy

A random variable is
1. A a numerical function assigning a real number to each outcome in the sample space
2. B always a constant
3. C only a probability
4. D the sample space itself
💡 Explanation:
X(ω) maps outcomes to numbers for algebraic treatment.
Q2 Past Paper · PPSC/FPSC/CSS easy

A discrete random variable takes
1. A only every real number
2. B only zero
3. C values with density only
4. D a countable set of values (finite or countably infinite)
💡 Explanation:
Discrete RVs have jump probabilities at specific points.
Q3 Past Paper · PPSC/FPSC/CSS easy

A continuous random variable takes
1. A only integers
2. B only 0 and 1
3. C a finite list always
4. D values in an interval (uncountably many)
💡 Explanation:
Continuous RVs are modeled with density over intervals.
Q4 Past Paper · PPSC/FPSC/CSS easy

The probability mass function (PMF) p(x) gives
1. A density at x for continuous X
2. B cumulative probability always
3. C P(X = x) for discrete X
4. D variance of X
💡 Explanation:
PMF assigns probabilities to individual discrete values.
Q5 Past Paper · PPSC/FPSC/CSS easy

A valid PMF must satisfy
1. A p(x) can be negative
2. B p(x) ≥ 0 and Σ p(x) = 1 over all x
3. C sum equals 0
4. D p(x) unbounded above
💡 Explanation:
PMF is non-negative and totals to 1.
Q6 Past Paper · PPSC/FPSC/CSS easy

The probability density function (PDF) f(x) for continuous X satisfies
1. A f(x) equals P(X=x)
2. B f(x) ≥ 0 and ∫f(x)dx = 1 over all x
3. C Σf(x)=1
4. D f(x) can be negative
💡 Explanation:
PDF integrates to 1; point probability is zero for continuous X.
Q7 Past Paper · PPSC/FPSC/CSS medium

For continuous X, P(a ≤ X ≤ b) equals
1. A f(a) + f(b)
2. B f(a) × f(b)
3. C Σ f(x)
4. D ∫_a^b f(x) dx
💡 Explanation:
Probability is area under the PDF over the interval.
Q8 Past Paper · PPSC/FPSC/CSS easy

The cumulative distribution function F(x) is defined as
1. A P(X = x) only
2. B P(X ≤ x)
3. C P(X ≥ x) always
4. D E(X)
💡 Explanation:
CDF gives left-tail probability up to x.
Q9 Past Paper · PPSC/FPSC/CSS easy

For any CDF, lim_{x→−∞} F(x) equals
1. A 0
2. B 1
3. C 0.5
4. D undefined
💡 Explanation:
No mass below −∞.
Q10 Past Paper · PPSC/FPSC/CSS easy

For any CDF, lim_{x→+∞} F(x) equals
1. A 1
2. B 0
3. C 0.5
4. D infinity
💡 Explanation:
All probability accumulated by +∞.
Q11 medium

A CDF is always
1. A decreasing
2. B always linear
3. C negative
4. D non-decreasing and right-continuous
💡 Explanation:
F(x2) ≥ F(x1) if x2 ≥ x1.
Q12 medium

For discrete X, F(x) is
1. A a step function with jumps at the values of X
2. B always smooth
3. C always zero
4. D constant
💡 Explanation:
CDF jumps by p(x) at each possible value.
Q13 Past Paper · PPSC/FPSC/CSS easy

The expected value E(X) of discrete X equals
1. A Σ p(x) only
2. B Σ x only
3. C Σ x · p(x)
4. D Σ x² p(x)
💡 Explanation:
Expectation is the probability-weighted average of X.
Q14 Past Paper · PPSC/FPSC/CSS medium

For X with PMF: x=0,p=0.2; x=1,p=0.5; x=2,p=0.3, E(X) equals
1. A 1.0
2. B 1.1
3. C 1.3
4. D 0.8
💡 Explanation:
E(X) = 0×0.2 + 1×0.5 + 2×0.3 = 0.5 + 0.6 = 1.1.
Q15 Past Paper · PPSC/FPSC/CSS medium

For the same distribution, E(X²) equals
1. A 1.7
2. B 1.1
3. C 1.21
4. D 2.0
💡 Explanation:
E(X²) = 0²×0.2 + 1²×0.5 + 2²×0.3 = 0 + 0.5 + 1.2 = 1.7.
Q16 Past Paper · PPSC/FPSC/CSS medium

For the same distribution, Var(X) equals
1. A 1.7
2. B 0.6
3. C 0.49
4. D 1.1
💡 Explanation:
Var(X) = E(X²) − [E(X)]² = 1.7 − 1.21 = 0.49.
Q17 Past Paper · PPSC/FPSC/CSS easy

Variance Var(X) is defined as
1. A E(X)² only
2. B E[(X − μ)²] = E(X²) − [E(X)]²
3. C E(X) only
4. D standard deviation squared always incorrectly as E(X)
💡 Explanation:
Variance measures squared deviation from the mean.
Q18 Past Paper · PPSC/FPSC/CSS easy

The standard deviation σ equals
1. A Var(X) squared
2. B the positive square root of Var(X)
3. C E(X) − μ
4. D P(X ≤ μ)
💡 Explanation:
σ = √Var(X) restores original units.
Q19 Past Paper · PPSC/FPSC/CSS easy

Linearity of expectation gives E(aX + b) as
1. A aE(X) × b
2. B E(X) + a + b
3. C a²E(X)
4. D aE(X) + b
💡 Explanation:
Expectation is linear regardless of independence.
Q20 Past Paper · PPSC/FPSC/CSS medium

If X and Y are independent, E(XY) equals
1. A E(X) + E(Y)
2. B E(X) / E(Y)
3. C E(X)E(Y)
4. D E(X²)
💡 Explanation:
Independence factorizes expectation of product.
Q21 Past Paper · PPSC/FPSC/CSS medium

Var(aX + b) equals
1. A a²Var(X)
2. B aVar(X) + b
3. C aVar(X)
4. D Var(X) + b²
💡 Explanation:
Adding constant b does not affect spread; scaling squares in variance.
Q22 Past Paper · PPSC/FPSC/CSS medium

If X and Y are independent, Var(X + Y) equals
1. A Var(X) × Var(Y)
2. B Var(X) + Var(Y)
3. C Var(X) − Var(Y)
4. D [Var(X) + Var(Y)]²
💡 Explanation:
Variances add for independent sums.
Q23 Past Paper · PPSC/FPSC/CSS easy

The PMF of a Bernoulli(p) trial is
1. A P(X=0)=p always
2. B uniform on 0,1,2
3. C P(X=1)=1 always
4. D P(X=1)=p, P(X=0)=1−p
💡 Explanation:
Bernoulli models one success/failure with parameter p.
Q24 Past Paper · PPSC/FPSC/CSS easy

For Bernoulli(p), E(X) equals
1. A 1 − p
2. B p(1 − p)
3. C p
4. D p²
💡 Explanation:
E(X) = 0×(1−p) + 1×p = p.
Q25 Past Paper · PPSC/FPSC/CSS medium

For Bernoulli(p), Var(X) equals
1. A p²
2. B p(1 − p)
3. C 1 − p
4. D p
💡 Explanation:
Var = p − p² = p(1−p).
Q26 easy

The mode of a distribution is
1. A the value of x with highest probability (discrete) or highest density (continuous)
2. B always the mean
3. C always the median
4. D E(X²)
💡 Explanation:
Mode is the most likely value.
Q27 medium

The median m satisfies
1. A P(X ≤ m) ≥ 0.5 and P(X ≥ m) ≥ 0.5
2. B P(X = m) = 1 always
3. C m = E(X) always
4. D m = mode always
💡 Explanation:
Median splits the distribution in half.
Q28 medium

The relationship between PDF and CDF is
1. A F(x) = f(x) × f(x)
2. B f(x) = F(x) − 1
3. C f(x) = dF(x)/dx where derivative exists
4. D no relationship
💡 Explanation:
Differentiating CDF yields PDF for continuous X.
Q29 hard

For discrete X, recovering PMF from CDF: p(x) equals
1. A F(x) × 2
2. B F(x) − F(x−) (jump at x)
3. C 1 − F(x) always
4. D Σ F(x)
💡 Explanation:
Jump size in CDF equals point mass.
Q30 hard

Chebyshev's inequality bounds
1. A P(X = μ) exactly
2. B P(|X − μ| ≥ kσ) ≤ 1/k² for k > 0
3. C only normal distributions
4. D Var(X) = 0
💡 Explanation:
Chebyshev holds for any distribution with finite variance.
Q31 Past Paper · PPSC/FPSC/CSS easy

A fair die score X. E(X) equals
1. A 3.5
2. B 3
3. C 4
4. D 3.6
💡 Explanation:
E(X) = (1+2+3+4+5+6)/6 = 21/6 = 3.5.
Q32 hard

For fair die X, Var(X) equals
1. A 35/12
2. B 7/12
3. C 2.5
4. D 3.5
💡 Explanation:
E(X²)=(1+4+9+16+25+36)/6=91/6; Var=91/6−12.25=35/12≈2.917.
Q33 Past Paper · PPSC/FPSC/CSS medium

If Y = 2X + 3 and E(X)=4, Var(X)=5, then E(Y) equals
1. A 8
2. B 13
3. C 11
4. D 10
💡 Explanation:
E(Y) = 2×4 + 3 = 11.
Q34 Past Paper · PPSC/FPSC/CSS medium

With the same Y = 2X + 3, Var(Y) equals
1. A 10
2. B 5
3. C 25
4. D 20
💡 Explanation:
Var(Y) = 2²×5 = 20.
Q35 medium

Joint PMF p(x,y) gives
1. A P(X = x and Y = y)
2. B P(X = x) only
3. C E(XY) always
4. D Cov(X,Y)
💡 Explanation:
Joint PMF specifies simultaneous probabilities.
Q36 medium

Marginal PMF p_X(x) is found by
1. A summing p(x,y) over all y
2. B multiplying over y
3. C setting y = x always
4. D integrating CDF only
💡 Explanation:
Marginalization sums (or integrates) out other variables.
Q37 Past Paper · PPSC/FPSC/CSS medium

X and Y are independent iff
1. A E(X) = E(Y)
2. B p(x,y) = p_X(x) p_Y(y) for all x,y
3. C Var(X) = Var(Y)
4. D ρ = 1
💡 Explanation:
Independence factorizes the joint distribution.
Q38 Past Paper · PPSC/FPSC/CSS medium

Cov(X,Y) = E(XY) − E(X)E(Y). If independent, Cov(X,Y) equals
1. A 1
2. B Var(X)
3. C 0
4. D E(X)E(Y)
💡 Explanation:
Independence implies zero covariance (converse not always true).
Q39 medium

Correlation ρ(X,Y) equals
1. A Cov(X,Y)²
2. B σ_X + σ_Y
3. C Cov(X,Y) / (σ_X σ_Y)
4. D E(XY) only
💡 Explanation:
Correlation is standardized covariance in [−1,1].
Q40 hard

A function g(X) is a random variable. E[g(X)] for discrete X equals
1. A g(E(X)) always
2. B Σ g(x) p(x)
3. C Σ g(x) only without weights
4. D E(X)^g
💡 Explanation:
LOTUS: expectation of function uses same weights p(x).
Q41 hard

Jensen's inequality for convex g states E[g(X)] is
1. A always equal to g(E(X))
2. B ≥ g(E(X)) for convex g
3. C ≤ g(E(X)) for convex g
4. D unrelated to convexity
💡 Explanation:
Convex functions: average of g ≥ g of average.
Q42 hard

The moment generating function M(t) = E(e^{tX}) helps find moments because
1. A M(t) equals Var(X)
2. B M(t) is always 1
3. C E(X^n) equals the nth derivative of M(t) at t=0
4. D derivatives give only median
💡 Explanation:
Differentiate MGF and evaluate at 0 for moments.
Q43 medium

Percentile p_k satisfies
1. A P(X = p_k) = k
2. B p_k = E(X)
3. C p_k = mode
4. D P(X ≤ p_k) = k/100 for continuous X
💡 Explanation:
k-th percentile is value with k% below it.
Q44 easy

Interquartile range (IQR) equals
1. A Q1 + Q3
2. B median − mean
3. C σ²
4. D Q3 − Q1
💡 Explanation:
IQR spans middle 50% of data/distribution.
Q45 Past Paper · PPSC/FPSC/CSS easy

For X uniform on {1,2,3,4}, E(X) equals
1. A 2.5
2. B 2
3. C 3
4. D 2.25
💡 Explanation:
E(X) = (1+2+3+4)/4 = 10/4 = 2.5.
Q46 hard

A RV with E(X)=μ and Var(X)=0 must satisfy
1. A P(X=0)=1
2. B uniform distribution
3. C P(X=μ)=0
4. D P(X = μ) = 1
💡 Explanation:
Zero variance means no spread: degenerate at μ.
Q47 medium

The law of large numbers states that
1. A sample mean equals μ for any n
2. B Var(X) goes to infinity
3. C the sample mean converges to μ as n increases
4. D PMF changes with n
💡 Explanation:
LLN: X̄ₙ → μ in probability under iid finite μ.
Q48 easy

A probability histogram for discrete data uses
1. A smooth curves only
2. B negative heights
3. C areas unrelated to probability
4. D bar heights equal to probabilities (or relative frequencies)
💡 Explanation:
Bar area/height represents probability mass at each value.
Q49 medium

The coefficient of variation CV equals
1. A μ / σ
2. B σ² / μ
3. C μ × σ
4. D σ / μ (standard deviation divided by mean)
💡 Explanation:
CV measures relative variability, useful for comparing scales.

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