SSC CGL Statistics – Random Variables & Probability Distributions – Exam-Level Drill

← Back

SSC CGL Statistics – Random Variables & Probability Distributions – Exam-Level Drill – 13 questions, subject: Statistics (Tier II Paper II). Every question has the correct answer with a one-line explanation; tap "Show Answer". Free practice for TNPSC, UPSC, SSC, Bank and RRB exams – Shanmugam IAS Academy, Coimbatore.

Get daily current affairs & exam alerts – follow us YouTube WhatsApp Channel Telegram Channel Facebook Page Instagram
Share with friends & study groups
Link to this page from your website / blog (HTML code)
1.  

Directions (Q 36–38): The joint probability distribution of X and Y is given below.

Y = 1 Y = 2 Y = 3
X = 1 k 2k k
X = 2 2k k 3k
Data tableThe value of k is:

Correct Answer: A [0.1]

Explanation:

All cells add to 10k = 1 → k = 0.1.

2.  
(Use the directions above – Q 36–38)
Data tableP(Y = 3 | X = 2) is:

Correct Answer: C [0.5]

Explanation:

P(X = 2) = 0.2 + 0.1 + 0.3 = 0.6; 0.3/0.6 = 0.5.

3.  
(Use the directions above – Q 36–38)
Data tableCov(X, Y) is:

Correct Answer: A [0.04]

Explanation:

E(X) = 1.6, E(Y) = 2.1, E(XY) = 3.4 → 3.4 − 3.36 = 0.04.

4.  
A process gives 5% defectives. Using the Poisson approximation, the probability of at most one defective in a sample of 60 is:

Correct Answer: B [0.1991]

Explanation:

m = 3: P(0) + P(1) = e⁻³(1 + 3) = 4e⁻³ ≈ 0.1991.

5.  
Packets filled by a machine weigh N(500 g, σ²). If 6.68% of packets weigh less than 490 g, σ is about:

Correct Answer: D [6.67 g]

Explanation:

Lower tail 0.0668 ↔ z = −1.5 → σ = 10/1.5 ≈ 6.67 g.

6.  
Lifetimes of a bulb are exponential with mean 400 h. A bulb has already worked 300 h. The probability that it works at least 100 h more is:

Correct Answer: A [0.7788]

Explanation:

Memoryless: P(X > 400 | X > 300) = P(X > 100) = e −100/400 ≈ 0.7788.

7.  
The first four moments about the origin are 1, 4, 10 and 46. The distribution is:

Correct Answer: C [Symmetric and mesokurtic (β₂ = 3)]

Explanation:

μ₂ = 3, μ₃ = 10 − 12 + 2 = 0, μ₄ = 46 − 40 + 24 − 3 = 27 → β₁ = 0, β₂ = 27/9 = 3.

8.  
For a binomial distribution with n = 9 and p = 1/3, the third central moment μ₃ is:

Correct Answer: D [2/3]

Explanation:

μ₃ = npq(q − p) = 9 × 1/3 × 2/3 × 1/3 = 2/3 (positive skew since p < 1/2).

9.  
For a Poisson distribution with m = 0.5, the kurtosis coefficient β₂ is:

Correct Answer: B [5]

Explanation:

β₂ = 3 + 1/m = 3 + 2 = 5 (leptokurtic).

10.  
A fair coin is tossed 100 times. Using the normal approximation with continuity correction, P(at least 58 heads) is about:

Correct Answer: C [0.0668]

Explanation:

μ = 50, σ = 5; P(X ≥ 57.5) → z = 1.5 → 1 − 0.9332 = 0.0668.

11.  
For an exponential random variable, P(X > 1) = 1/3. Then P(X > 3) is:

Correct Answer: B [1/27]

Explanation:

P(X > x) = e −λx = (e −λ )ˣ = (1/3)³ = 1/27.

12.  
Reverse questionBy Chebyshev's inequality, the smallest k for which at least 96% of any distribution is guaranteed to lie within μ ± kσ is:

Correct Answer: D [5]

Explanation:

1 − 1/k² = 0.96 → k² = 25 → k = 5.

13.  
X has density f(x) = 2x for 0 ≤ x ≤ 1. Var(X) is:

Correct Answer: A [1/18]

Explanation:

E(X) = ∫2x² dx = 2/3; E(X²) = ∫2x³ dx = 1/2; Var = 1/2 − 4/9 = 1/18.

➤ Full notes, shortcuts and every answer explained – SSC CGL 2026 Statistics (Tier II Paper II) book (₹149): www.shanmugamiasacademy.in/study-materials/ssc-cgl-statistics

📘 SSC CGL 2026 – Statistics (Tier II Paper II)
Here you get the answer with a one-line explanation. 657 original questions with worked answers · notes, formulas, figures and statistical tables (z, t, χ², F) · practice sets + 3 mock tests
Get the book – ₹149 ₹299 →
Score: 0 Answered: 0 / 0