SSC CGL Statistics – Mock Test M2 – Probability, Sampling, Inference & ANOVA (Chapters 6–10)

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SSC CGL Statistics – Mock Test M2 – Probability, Sampling, Inference & ANOVA (Chapters 6–10) – 25 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.  
What is the chance that a card picked at random from a full pack of 52 is a red face card?

Correct Answer: A [3/26]

Explanation:

6 red face cards (J, Q, K of hearts and diamonds) → 6/52 = 3/26.

2.  
Three fair coins are tossed. The probability of getting at most two heads is:

Correct Answer: D [7/8]

Explanation:

Answer: 7/8. Only HHH is excluded → 7/8.

3.  
P(A) = 0.5, P(B) = 0.7 and P(B ∣ A) = 0.6. P(A ∪ B) is:

Correct Answer: C [0.9]

Explanation:

P(A ∩ B) = 0.5 × 0.6 = 0.3; 0.5 + 0.7 − 0.3 = 0.9.

4.  
Urn I has 2 white and 3 black balls; Urn II has 4 white and 1 black. Urn I is chosen with probability 1/3 and Urn II with probability 2/3, and a ball drawn is white. The probability that it came from Urn I is:

Correct Answer: B [1/5]

Explanation:

(1/3 × 2/5)/(1/3 × 2/5 + 2/3 × 4/5) = (2/15)/(10/15) = 1/5.

5.  
A and B are independent with P(A) = 0.3 and P(B) = 0.2. The probability that neither occurs is:

Correct Answer: C [0.56]

Explanation:

Answer: 0.56. 0.7 × 0.8 = 0.56.

6.  
X takes the values 0, 1 and 2 with probabilities k, 2k and 3k. E(X) is:

Correct Answer: A [4/3]

Explanation:

6k = 1 → k = 1/6; E(X) = 0 + 2/6 + 6/6 = 8/6 = 4/3.

7.  
Var(X) = 5. The variance of 2X + 3 is:

Correct Answer: B [20]

Explanation:

Var(aX + b) = a²Var(X) = 4 × 5 = 20.

8.  
A binomial distribution has mean 8 and variance 6. The number of trials n is:

Correct Answer: D [32]

Explanation:

q = 6/8 = 0.75 → p = 0.25 → n = 8/0.25 = 32.

9.  
Four fair coins are tossed together. What is the chance of getting exactly three heads?

Correct Answer: A [1/4]

Explanation:

Answer: 1/4. C(4, 3)/2⁴ = 4/16 = 1/4.

10.  
For a Poisson variable X, P(X = 2) = P(X = 3). Then P(X = 0) is (e⁻³ ≈ 0.0498):

Correct Answer: C [0.0498]

Explanation:

m² e⁻ᵐ/2 = m³ e⁻ᵐ/6 → m = 3; P(0) = e⁻³ ≈ 0.0498.

11.  
A Poisson distribution has mean 9. Its standard deviation is:

Correct Answer: B [3]

Explanation:

Answer: 3. Variance = mean = 9 → SD = 3.

12.  
X is normal with mean 50 and SD 10. Given that the area under the standard normal curve between 0 and 2 is 0.4772, P(X > 70) is:

Correct Answer: D [0.0228]

Explanation:

z = (70 − 50)/10 = 2; P = 0.5 − 0.4772 = 0.0228.

13.  
X is normal with mean 60 and SD 20. Given P(0 < Z < 0.6745) = 0.25, the upper quartile of X is (to two decimals):

Correct Answer: C [73.49]

Explanation:

Q₃ = μ + 0.6745σ = 60 + 13.49 = 73.49.

14.  
X is exponential with mean 4. P(X > 10 ∣ X > 2) equals:

Correct Answer: D [e⁻²]

Explanation:

Memoryless property: P(X > 10 ∣ X > 2) = P(X > 8) = e⁻⁸ᐟ⁴ = e⁻².

15.  
Reverse questionrandom samples of 25 from a large population give a standard error of the mean of 3. The population SD is:

Correct Answer: B [15]

Explanation:

Answer: 15. SE = σ/√n → σ = 3 × √25 = 15.

16.  
How many different samples of size 2 can be drawn without replacement (order ignored) from a population of 6 units?

Correct Answer: A [15]

Explanation:

Answer: 15. C(6, 2) = 15.

17.  
Three strata have 600, 300 and 100 units. A stratified sample of 50 is taken with proportional allocation. The sample size from the second stratum is:

Correct Answer: B [15]

Explanation:

Answer: 15. 50 × 300/1000 = 15.

18.  
Match the followingMatch the sampling method with its description:
P. Systematic Q. Stratified R. Cluster S. Quota
1. whole groups are selected at random 2. every k-th unit after a random start 3. random samples from each homogeneous subgroup 4. interviewer fills fixed numbers in each category (non-random)

Correct Answer: C [P-2, Q-3, R-1, S-4]

Explanation:

Systematic – every k-th; stratified – within strata; cluster – whole groups; quota – non-probability.

19.  
A sample of 100 has mean 80; the population SD is 15. The 95% confidence interval for the population mean (z = 1.96) is:

Correct Answer: A [77.06 to 82.94]

Explanation:

80 ± 1.96 × 15/10 = 80 ± 2.94.

20.  
H₀: μ = 50 against H₁: μ ≠ 50. A sample of 64 gives x̄ = 52; σ = 8. At the 5% level (z = 1.96):

Correct Answer: D [z = 2; reject H₀]

Explanation:

z = (52 − 50)/(8/8) = 2 > 1.96 → reject.

21.  
Two independent small samples of sizes 10 and 12 are compared with the pooled t test. The degrees of freedom are:

Correct Answer: A [20]

Explanation:

Answer: 20. n₁ + n₂ − 2 = 20.

22.  
A 4 × 5 contingency table is tested for independence of the two attributes by χ². How many degrees of freedom does the test have?

Correct Answer: B [12]

Explanation:

Answer: 12. (4 − 1)(5 − 1) = 12.

23.  
Assertion–Reason
A: Lowering the level of significance from 5% to 1% (same sample) makes a Type II error more likely.
R: A smaller rejection region rejects H₀ less often, including when H₀ is false.

Correct Answer: C [Both A and R are true and R explains A]

Explanation:

Shrinking the critical region lowers α but raises β.

24.  
One-way ANOVA with 4 groups and 24 observations gives SS between = 60 and SS within = 100. The F ratio is:

Correct Answer: D [4]

Explanation:

MSB = 60/3 = 20; MSE = 100/20 = 5; F = 4.

25.  
Six fertilisers are compared on 4 blocks, each fertiliser once in every block. How many degrees of freedom does the error sum of squares have?

Correct Answer: D [15]

Explanation:

Answer: 15. (6 − 1)(4 − 1) = 15.

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