SSC CGL Statistics – Sampling Theory – Level 3 · Challenge
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N_h S_h = 3,600, 4,800, 3,600 (total 12,000) → 60 × 3,600/12,000 = 18.
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SSC CGL Statistics – Sampling Theory – Level 3 · Challenge – 8 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.
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1.
All samples of size 3 are drawn without replacement from the population 1, 3, 5, 7, 9. The variance of the sampling distribution of the mean is:
Correct Answer: C [4/3]
Explanation:
σ² = 8; (8/3)(5 − 3)/(5 − 1) = 4/3 (listing all 10 sample means 3, 11/3, … , 7 gives the same).
2.
Statement-basedFor simple random sampling with replacement, which of these is/are true?
I. E(x̄) = μ
II. Var(x̄) = σ²/n
III. E(s²) = σ² when s² uses divisor n − 1
I. E(x̄) = μ
II. Var(x̄) = σ²/n
III. E(s²) = σ² when s² uses divisor n − 1
Correct Answer: A [All three]
Explanation:
x̄ is unbiased for μ, Var(x̄) = σ²/n for independent draws, and s² with divisor n − 1 is unbiased for σ².
3.
A population has 3 units with values 2, 6, 10 and samples of size 2 are drawn with replacement. How many possible samples are there, and what is the mean of the sample means?
Correct Answer: C [9 samples; mean 6]
Explanation:
3² = 9 ordered samples; E(x̄) = μ = 6.
4.
For a 95% confidence, ±0.05 survey of a proportion (P unknown) in a village of 1,000 households, the size after the finite-population correction n = n₀/(1 + n₀/N) is:
Correct Answer: B [278]
Explanation:
n₀ = 384.16; n = 384.16/1.38416 = 277.54 → round up to 278.
5.
Which situation calls for cluster sampling rather than stratified sampling?
Correct Answer: D [No list of individuals exists, but a list of villages does]
Explanation:
Cluster sampling needs a frame of groups only; strata need a full frame and are used for distinct groups.
6.
The SD of a large population is 12. With n = 36, the probability that the sample mean differs from μ by more than 3.92 is about:
Correct Answer: A [0.05]
Explanation:
SE = 2; 3.92 = 1.96 SE → two tails 5% (sampling distribution approximately normal).
7.
Arrange in orderArrange in order of the number of possible samples, smallest first (N = 6, n = 3):
(i) WR, order counted (ii) WOR, order ignored (iii) WOR, order counted
(i) WR, order counted (ii) WOR, order ignored (iii) WOR, order counted
Correct Answer: C [(ii), (iii), (i)]
Explanation:
(ii) ⁶C₃ = 20, (iii) ⁶P₃ = 120, (i) 6³ = 216.
8.
A population of 2,400 is split into three strata of sizes 1,200, 800 and 400 with SDs 3, 6 and 9. For n = 60 under Neyman allocation, the first stratum gets:
Correct Answer: B [18]
Explanation:
N_h S_h = 3,600, 4,800, 3,600 (total 12,000) → 60 × 3,600/12,000 = 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)
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