SSC CGL Mock – Paper II Statistics Mock Test 05 – Paper II · Statistics | Online Quiz

1.

In a histogram of a fictional distribution, the heights show frequency per unit of the variable: 0–5: 4, 5–15: 3, 15–30: 2, 30–50: 1.5. What percentage of the observations are below 15?

Correct Answer: D [45.45%]

Explanation:

Frequencies = width × height: 20, 30, 30, 30; N = 110; below 15: 50 → 45.45%.

2.

Diagram-basedThe 'less than' ogive of the weights of some fictional parcels is shown (labels = cumulative frequencies; the first class starts at 10 kg).

Using interpolation, the median weight is

Correct Answer: A [37.27 kg]

Explanation:

N/2 = 35 lies in 30–40: 30 + (35 − 19)/22 × 10 = 37.27 kg.

3.

Direct personal investigation is NOT suitable when

Correct Answer: C [the field of enquiry is very wide]

Explanation:

Personal visits by the investigator are costly and slow for a wide field; they suit small enquiries.

4.

Two pie diagrams show the cost structure of a fictional company. Material takes 72° of a total cost of ₹ 50 lakh in year 1 and 90° of ₹ 48 lakh in year 2. By what percent did the material cost change?

Correct Answer: B [20%]

Explanation:

Year 1: 72/360 × 50 = ₹ 10 lakh; year 2: 90/360 × 48 = ₹ 12 lakh; change = 20% (an increase).

5.

In a continuous distribution with equal classes, the mid-value of the 4th class is 47 and that of the 7th class is 71. What is the lower limit of the first class?

Correct Answer: A [19]

Explanation:

Width = (71 − 47)/3 = 8; 1st mid-value = 47 − 3 × 8 = 23; lower limit = 23 − 4 = 19.

6.

Statement-basedConsider the statements:
I. Figures taken by a researcher from a ministry's published report are secondary data for that researcher.
II. Secondary data are always less reliable than primary data.
Which statement(s) is/are correct?

Correct Answer: A [Only I]

Explanation:

I is correct. II is wrong: published official data may be very reliable; the user must check them.

7.

District-wise rainfall of a state is shown by different shades on an outline map of the state. Such a diagram is a

Correct Answer: D [cartogram]

Explanation:

Data shown on a map (shades, dots) form a cartogram; a pictogram uses picture symbols.

8.

In a histogram, the class 25–40 (frequency 36) is drawn 9 cm tall. The class 40–45 has frequency 15. How tall should its rectangle be?

Correct Answer: B [11.25 cm]

Explanation:

Heights ∝ frequency/width: 36/15 = 2.4 ↔ 9 cm, so 15/5 = 3 ↔ 11.25 cm.

9.

Typing speeds (words a minute) of trainees at a fictional institute: 0–10: 4, 10–20: 8, 20–30: f, 30–40: 12, 40–50: 6. The modal speed is 26 words a minute. The missing frequency f is

Correct Answer: C [20]

Explanation:

20 + (f − 8)/(2f − 8 − 12) × 10 = 26 → (f − 8) = 0.6(2f − 20) → 0.2f = 4 → f = 20.

10.

In a step-deviation calculation, the assumed mean is 200, the class width is 20, Σfd = −48 and N = 60. The mean is

Correct Answer: B [184]

Explanation:

Mean = A + (Σfd/N) × h = 200 + (−48/60) × 20 = 184.

11.

The numbers 1, 2, 3, …, 20 are given weights equal to the numbers themselves (1 gets weight 1, 2 gets weight 2, and so on). The weighted mean is

Correct Answer: C [13.67]

Explanation:

Σw x/Σw = Σx²/Σx = (2n + 1)/3 = 41/3 = 13.67.

12.

Find the median of the fictional distribution with unequal classes 0–10: 8, 10–30: 22, 30–60: 30, 60–100: 20.

Correct Answer: A [40]

Explanation:

N/2 = 40; class 30–60 (cf before 30, width 30): 30 + (40 − 30)/30 × 30 = 40.

13.

Three values have geometric mean 2 and two other values have geometric mean 64. The geometric mean of all five values is

Correct Answer: D [8]

Explanation:

Product = 2³ × 64² = 2³ × 2¹² = 2¹⁵; GM = (2¹⁵) 1/5 = 2³ = 8.

14.

A cyclist rides for equal lengths of time at 12, 15 and 18 km/h. Her average speed is

Correct Answer: C [15 km/h]

Explanation:

Equal TIMES → arithmetic mean: (12 + 15 + 18)/3 = 15 km/h (HM is for equal distances).

15.

Assertion–ReasonAssertion (A): For two unequal positive numbers, AM × HM = GM².
Reason (R): For any set of positive numbers, AM ≥ GM ≥ HM.

Correct Answer: B [Both A and R are true but R is not the correct explanation of A]

Explanation:

Both are true, but A follows from AM = (a + b)/2, HM = 2ab/(a + b), not from the inequality R.

16.

The mean of 50 observations is 38. Each of the first 20 observations is increased by 5 and each of the remaining observations is decreased by 2. The new mean is

Correct Answer: A [38.8]

Explanation:

Change in total = 20×5 − 30×2 = 40; new mean = 38 + 40/50 = 38.8.

17.

Monthly incomes (₹ thousand) of 60 households in a fictional ward: 0–20: 5, 20–40: 15, 40–60: 25, 60–80: 10, 80–100: 5. Above what income do the richest 10% of households lie (P₉₀)?

Correct Answer: B [78]

Explanation:

90% of 60 = 54; class 60–80 (cf before 45): 60 + (54 − 45)/10 × 20 = 78.

18.

Using the (n + 1) rule for positions, the interquartile range of the fictional data 12, 15, 17, 20, 22, 24, 25, 28, 30, 33, 35, 38, 41, 45 is

Correct Answer: D [16.5]

Explanation:

Q₁ at 3.75th value = 17 + 0.75 × 3 = 19.25; Q₃ at 11.25th = 35 + 0.25 × 3 = 35.75; IQR = 16.5.

19.

Two groups of equal size have means 20 and 30 and variances 16 and 36. The variance of the two groups taken together is

Correct Answer: C [51]

Explanation:

σ² = (16 + 36)/2 + ((30 − 20)/2)² = 26 + 25 = 51.

20.

The variance of the first 10 odd numbers 1, 3, 5, …, 19 is

Correct Answer: B [33]

Explanation:

Odd numbers = 2k − 1, so variance = 4 × (n² − 1)/12 = (n² − 1)/3 = 33.

21.

Ten fictional observations have mean 15 and SD 4. If the observation 21 is removed, the SD of the remaining nine is

Correct Answer: A [3.65]

Explanation:

Σx = 150 − 21 = 129; Σx² = 10(16 + 225) − 441 = 1969; σ² = 1969/9 − (129/9)² = 13.3333; σ ≈ 3.65.

22.

Marks of 40 candidates in a fictional aptitude test: 5–15: 4, 15–25: 10, 25–35: 16, 35–45: 7, 45–55: 3. Their standard deviation (divide by N) is

Correct Answer: D [10.53]

Explanation:

x̄ = 28.75; σ² = 110.94; σ ≈ 10.53.

23.

Assertion–ReasonAssertion (A): The coefficient of variation is not a useful measure when the mean is close to zero.
Reason (R): The coefficient of variation is the standard deviation divided by the mean, × 100.

Correct Answer: A [Both A and R are true and R is the correct explanation of A]

Explanation:

With a mean near zero, σ/x̄ becomes huge and unstable – this follows from the formula in R.

24.

Using the (n + 1) rule, the coefficient of quartile deviation of the fictional data 41, 45, 48, 52, 55, 57, 60, 64, 66, 70, 73 is

Correct Answer: B [0.16]

Explanation:

Q₁ = 3rd value = 48; Q₃ = 9th value = 66; (Q₃ − Q₁)/(Q₃ + Q₁) = 18/114 ≈ 0.16.

25.

Two-quantity comparisonFor the values 4, 4, 10, 10 (divide by n):
Quantity I: standard deviation
Quantity II: half the range

Correct Answer: C [Quantity I = Quantity II]

Explanation:

Mean 7; every deviation is ±3, so σ = 3; half the range = 6/2 = 3.

26.

The variance of the first n natural numbers is 10. The value of n is

Correct Answer: A [11]

Explanation:

(n² − 1)/12 = 10 → n² = 121 → n = 11.

27.

X is the morning sale and Y the evening sale of a fictional bakery, with Var(X) = 25, Var(Y) = 16 and Cov(X, Y) = 12. The variance of the difference X − Y is

Correct Answer: D [17]

Explanation:

Var(X − Y) = Var X + Var Y − 2 Cov = 25 + 16 − 24 = 17.

28.

For a variable x, μ₂ = 4, μ₃ = −6 and μ₄ = 40. If y = (x − 10)/2, the coefficient β₂ of y is

Correct Answer: C [2.5]

Explanation:

β₂ is unchanged by change of origin and scale: β₂ = 40/4² = 2.5 (μ₄(y) = 40/16, μ₂(y) = 1).

29.

For the number of goals per match in a fictional football league, the first three moments about 3 goals are 1, 10 and 40. The moment measure of skewness γ₁ = μ₃/σ³ is about

Correct Answer: A [0.44]

Explanation:

μ₂ = 10 − 1 = 9 (σ = 3); μ₃ = 40 − 30 + 2 = 12; γ₁ = 12/27 ≈ 0.44.

30.

For a fictional income distribution, P₁₀ = 18, P₅₀ = 40 and P₉₀ = 70 (₹ thousand). Kelly's coefficient of skewness (P₉₀ + P₁₀ − 2P₅₀)/(P₉₀ − P₁₀) is

Correct Answer: C [0.15]

Explanation:

(70 + 18 − 80)/(70 − 18) = 8/52 ≈ 0.15.

31.

Karl Pearson's coefficient of skewness 3(mean − median)/σ of a fictional distribution is −0.5. Its mean is 40 and median 42. The coefficient of variation is

Correct Answer: B [30%]

Explanation:

σ = 3(40 − 42)/(−0.5) = 12; CV = 12/40 × 100 = 30%.

32.

Reverse questionFor any distribution μ₄ ≥ μ₂², i.e. β₂ ≥ 1. Which pair of values of (μ₂, μ₄) is therefore impossible?

Correct Answer: D [μ₂ = 4, μ₄ = 12]

Explanation:

β₂ = 12/16 = 0.75 < 1 is impossible; the others give β₂ = 1, 3 and 5.

33.

Statement-basedConsider the statements:
I. Central moments are unaffected by a change of origin.
II. The coefficients β₁ and β₂ are unaffected by a change of origin and of scale.
Which statement(s) is/are correct?

Correct Answer: C [Both I and II]

Explanation:

μ_r uses x − x̄ (no origin effect); β₁ and β₂ are ratios in which the scale factor cancels.

34.

For a fictional distribution, Bowley's coefficient of skewness is 0.25, the interquartile range is 40 and the median is 50. The lower quartile Q₁ is

Correct Answer: C [35]

Explanation:

Q₃ + Q₁ = 2Md + Sk(Q₃ − Q₁) = 100 + 10 = 110 and Q₃ − Q₁ = 40 → Q₁ = 35.

35.

Reverse questionWhich set of correlation coefficients among three variables is NOT possible?

Correct Answer: B [r₁₂ = 0.8, r₁₃ = −0.4, r₂₃ = 0.7]

Explanation:

For it R²₁.₂₃ = (0.64 + 0.16 + 0.448)/0.51 > 1 (equivalently 1 − Σr² + 2r₁₂r₁₃r₂₃ = −0.738 < 0).

36.

Blood pressure (X₁), age (X₂) and hours of exercise a week (X₃) of fictional patients give r₁₂ = 0.5, r₁₃ = −0.6 and r₂₃ = −0.4. The multiple correlation of X₁ on X₂ and X₃ is about

Correct Answer: A [0.66]

Explanation:

R² = (0.25 + 0.36 − 2 × 0.5 × (−0.6) × (−0.4))/(1 − 0.16) = 0.37/0.84 ≈ 0.4405; R ≈ 0.66.

37.

Output (X₁), labour (X₂) and capital (X₃) of fictional units: r₁₂ = 0.6, r₁₃ = 0.5, r₂₃ = 0.4, σ₁ = 10, σ₂ = 8. The partial regression coefficient b₁₂.₃ = (r₁₂ − r₁₃r₂₃)/(1 − r₂₃²) × σ₁/σ₂ is about

Correct Answer: D [0.60]

Explanation:

(0.6 − 0.2)/0.84 × 1.25 ≈ 0.60.

38.

Scores of seven fictional applicants in a written test: 72, 65, 65, 80, 58, 58, 90; in an interview: 70, 68, 74, 74, 60, 62, 85. Spearman's rank correlation with the correction for ties is about

Correct Answer: D [0.87]

Explanation:

Ranks (highest = 1) – test: 3, 4.5, 4.5, 2, 6.5, 6.5, 1; interview: 4, 5, 2.5, 2.5, 7, 6, 1; Σd² = 6; three tied pairs add 3 × 0.5 = 1.5; ρ = 1 − 6(7.5)/336 ≈ 0.87.

39.

X and Y have equal standard deviations. The correlation coefficient between U = X + Y and V = X − Y is

Correct Answer: B [0]

Explanation:

Cov(U, V) = Var X − Var Y = 0 when σₓ = σᵧ, so r(U, V) = 0 whatever r(X, Y) is.

40.

A sample of 25 pairs gives r = 0.4. Using the probable error PE = 0.6745(1 − r²)/√n and the rule 'r is significant if r > 6 PE', the conclusion is

Correct Answer: C [PE ≈ 0.113; r is not significant]

Explanation:

PE = 0.6745 × 0.84/5 ≈ 0.113; 6 PE ≈ 0.680 > 0.4 → not significant.

41.

For fictional towns, X₁ = crime rate, X₂ = police strength and X₃ = population density have r₁₂ = −0.5, r₁₃ = 0.6, r₂₃ = −0.6. Allowing for density, the correlation between crime and police strength (r₁₂.₃) is

Correct Answer: A [−0.22]

Explanation:

(−0.5 − (0.6)(−0.6))/√(0.64 × 0.64) = (−0.5 + 0.36)/0.64 = −0.22.

42.

X and Y are independent with σₓ = 3 and σᵧ = 4. The correlation coefficient between X and X + Y is

Correct Answer: B [0.6]

Explanation:

Cov(X, X + Y) = Var X = 9; σ(X + Y) = √(9 + 16) = 5; r = 9/(3 × 5) = 0.6.

43.

In a fictional town of 800 adults, 300 read a newspaper (A), 400 watch the news on TV (B) and 200 do both. Yule's Q between A and B is

Correct Answer: A [0.5]

Explanation:

(AB) = 200, (Aβ) = 100, (αB) = 200, (αβ) = 300; Q = (60000 − 20000)/(60000 + 20000) = 0.5.

44.

The partial correlation coefficient r₁₂.₃ is zero when

Correct Answer: C [r₁₂ = r₁₃ r₂₃]

Explanation:

Its numerator is r₁₂ − r₁₃r₂₃; it vanishes exactly when r₁₂ = r₁₃r₂₃ (r₁₂ = 0 alone is not enough).

45.

The multiple correlation coefficient R₁.₂₃ always lies between

Correct Answer: D [0 and 1]

Explanation:

R is the correlation between X₁ and its best linear estimate from X₂, X₃; it is never negative.

46.

Three fictional urns contain (2 white, 4 black), (3 white, 2 black) and (4 white, 1 black) balls. An urn is chosen at random and two balls drawn from it together are both white. The probability that the third urn was chosen is

Correct Answer: D [18/29]

Explanation:

Likelihoods C(w, 2)/C(n, 2): 1/15, 3/10, 3/5; Bayes: (3/5)/(29/30) = 18/29.

47.

Two fair coins are tossed. A: first coin shows head; B: second coin shows head; C: both coins show the same face. Which statement is correct?

Correct Answer: B [A, B, C are pairwise independent but not mutually independent]

Explanation:

Each pair: P = 1/4 = (1/2)(1/2); but P(A∩B∩C) = 1/4 ≠ 1/8.

48.

The top card of a well-shuffled pack is removed and put aside face down, unseen. The probability that the next card is a king is

Correct Answer: A [1/13]

Explanation:

4/52 × 3/51 + 48/52 × 4/51 = 1/13 (by symmetry, every position is equally likely to hold a king).

49.

A and B throw a fair die in turns, A first. Whoever first throws a six wins. The probability that A wins is

Correct Answer: C [6/11]

Explanation:

A wins on throws 1, 3, 5, …: (1/6)[1 + (5/6)² + …] = (1/6)/(1 − 25/36) = 6/11.

50.

A token is drawn from a box of tokens numbered 1 to 300 at a fictional exhibition. The chance that its number has 3, 5 or 7 as a factor is

Correct Answer: A [0.54]

Explanation:

100 + 60 + 42 − (20 + 14 + 8) + 2 = 162 → 162/300 = 0.54.

51.

In a four-option test, a fictional student knows the answer to a question with probability 3/4 and otherwise guesses. Given that the answer is correct, the probability that the student knew it is

Correct Answer: D [12/13]

Explanation:

Bayes: (3/4)/(3/4 + 1/4 × 1/4) = (12/16)/(13/16) = 12/13.

52.

P(A) = 0.6, P(B) = 0.5 and P(A | B) = 0.4. The value of P(B | A′) (A′ = not A) is

Correct Answer: C [0.75]

Explanation:

P(A ∩ B) = 0.4 × 0.5 = 0.2; P(A′ ∩ B) = 0.5 − 0.2 = 0.3; P(A′) = 0.4 → 0.3/0.4 = 0.75.

53.

Five people P, Q, R, S, T sit in a row in random order. The probability that P and Q sit next to each other is

Correct Answer: B [2/5]

Explanation:

Treat PQ as one block: 2 × 4!/5!

54.

Three letters are put at random into three addressed envelopes, one letter in each. The probability that no letter goes into its correct envelope is

Correct Answer: C [1/3]

Explanation:

Of the 3! = 6 ways, only 2 (the two cyclic shifts) leave every letter misplaced → 1/3.

55.

Assertion–ReasonAssertion (A): If P(A | B) > P(A), then P(B | A) > P(B) (all probabilities positive).
Reason (R): P(A | B) P(B) = P(B | A) P(A) = P(A ∩ B).

Correct Answer: A [Both A and R are true and R is the correct explanation of A]

Explanation:

From R, P(B | A)/P(B) = P(A | B)/P(A) > 1 – R proves A.

56.

For a binomial variable with n = 6, P(X = 1) = P(X = 2). The value of p is

Correct Answer: D [2/7]

Explanation:

6pq⁵ = 15p²q⁴ → 6q = 15p → 6 = 21p → p = 2/7.

57.

Calls per hour at a fictional night desk follow a Poisson law; an hour with 3 calls is three times as likely as an hour with 4 calls. The standard deviation of the hourly number of calls is

Correct Answer: A [1.15]

Explanation:

P(4)/P(3) = m/4 = 1/3 → m = 4/3; SD = √(4/3) ≈ 1.15.

58.

In a fictional normal population, 31% of the items are below 45 and 8% are above 64. Taking the corresponding z-values as −0.5 and 1.4, the mean and SD are

Correct Answer: B [mean 50, SD 10]

Explanation:

μ − 0.5σ = 45 and μ + 1.4σ = 64 → 1.9σ = 19 → σ = 10, μ = 50.

59.

X and Y show whether each of two fictional machines is busy (1) or idle (0) at a random instant, with P(0, 0) = 0.2, P(0, 1) = 0.1, P(1, 0) = 0.1, P(1, 1) = 0.6. The variance of the number of busy machines, Var(X + Y), is

Correct Answer: D [0.64]

Explanation:

Var X = Var Y = 0.21; Cov = 0.6 − 0.49 = 0.11; Var(X + Y) = 0.42 + 2 × 0.11 = 0.64.

60.

A fictional variable X has mean 50 and SD 5; nothing else is known about its distribution. By Chebyshev's inequality, P(40 < X < 60) is at least

Correct Answer: A [3/4]

Explanation:

60 − 50 = 10 = 2σ; P(|X − μ| < 2σ) ≥ 1 − 1/4 = 3/4.

61.

The time (hours) to serve a fictional customer is exponential with λ = 0.5 per hour. Using e^−0.5 = 0.6065 and e^−1.5 = 0.2231, the probability that a service takes between 1 and 3 hours is

Correct Answer: B [0.3834]

Explanation:

P(1 < X < 3) = e −0.5 − e −1.5 = 0.3834.

62.

0.4% of fictional tyres are faulty. In a lot of 500, using the Poisson approximation and e⁻² = 0.1353, the probability of at most two faulty tyres is

Correct Answer: C [0.6765]

Explanation:

m = 500 × 0.004 = 2; e⁻²(1 + 2 + 2²/2) = 0.1353 × 5 = 0.6765.

63.

For a normal variable and Φ(2.5) = 0.9938, the probability that it lies more than 2.5 standard deviations away from its mean (on either side) is

Correct Answer: C [0.0124]

Explanation:

Answer: 0.0124. 2 × (1 − 0.9938) = 0.0124.

64.

A fictional random variable has density f(x) = 2(1 − x) for 0 ≤ x ≤ 1. Its variance is

Correct Answer: B [1/18]

Explanation:

E(X) = 1/3; E(X²) = 2(1/3 − 1/4) = 1/6; Var = 1/6 − 1/9 = 1/18.

65.

Assertion–ReasonAssertion (A): A binomial distribution with p = 0.5 is symmetrical.
Reason (R): When p = q, P(X = k) = P(X = n − k) for every k.

Correct Answer: A [Both A and R are true and R is the correct explanation of A]

Explanation:

C(n, k) = C(n, n − k) and pᵏqⁿ⁻ᵏ = qᵏpⁿ⁻ᵏ when p = q – R gives the symmetry in A.

66.

Goals scored by two fictional teams in a match are independent Poisson variables with means 1.2 and 0.8. Using e⁻² = 0.1353, the probability that the match ends 0–0 is

Correct Answer: A [0.1353]

Explanation:

X + Y is Poisson with mean 2; P(X + Y = 0) = e⁻² = 0.1353.

67.

Four fictional workers did 1, 3, 5 and 7 hours of overtime. All 6 samples of two workers (without replacement) are listed. The variance of the 6 sample means is

Correct Answer: D [5/3]

Explanation:

σ² = 5; Var(x̄) = (σ²/n)(N − n)/(N − 1) = (5/2)(2/3) = 5/3.

68.

A sample of 5 is drawn by simple random sampling without replacement from 50 fictional shops. The probability that a particular shop is included is

Correct Answer: B [1/10]

Explanation:

Answer: 1/10. P = n/N = 5/50 = 1/10.

69.

A fictional town has two strata, urban (60% of households) and rural (40%). Sample means of monthly spending (₹ hundred) are 50 (urban) and 70 (rural). The stratified estimate of the town's mean is

Correct Answer: D [58]

Explanation:

Answer: 58. 0.6 × 50 + 0.4 × 70 = 58.

70.

The central limit theorem states that, for a large sample size n, the distribution of the sample mean

Correct Answer: A [is approximately normal whatever the shape of the population]

Explanation:

x̄ is approximately N(μ, σ²/n) for large n, for any population with finite variance.

71.

The coefficient of variation of farm sizes in a fictional state is 40%. To estimate the mean within ±5% of its value with z = 2, the sample size needed is

Correct Answer: C [256]

Explanation:

n = (z × CV/e)² = (2 × 40/5)² = 16² = 256.

72.

Statement-basedConsider the statements:
I. Sampling error tends to decrease as the sample size increases.
II. Non-sampling errors necessarily decrease as the sample size increases.
Which statement(s) is/are correct?

Correct Answer: A [Only I]

Explanation:

Non-sampling errors may even grow with a bigger survey (more field staff, more processing).

73.

Villages are chosen so that a village with twice the population has twice the chance of selection. This is

Correct Answer: D [probability proportional to size (PPS) sampling]

Explanation:

Selection probability tied to a size measure – PPS sampling.

74.

H₀: μ = 50 is tested against H₁: μ > 50 at 5% (z = 1.645) with σ = 10 and n = 25 (fictional data). If the true mean is 55, the power of the test is about (Φ(0.855) = 0.8037)

Correct Answer: B [0.8037]

Explanation:

Reject if x̄ > 50 + 1.645 × 2 = 53.29; power = P(Z > (53.29 − 55)/2) = P(Z > −0.855) = 0.8037.

75.

Fictional typing speeds: method A – 21, 24, 19, 23, 18, 21; method B – 25, 27, 24, 28, 26. Using the pooled t-test (df = 9) at the 1% level (two-sided, t₀.₀₀₅,₉ = 3.250), the conclusion is

Correct Answer: A [t ≈ 4.13; methods differ significantly]

Explanation:

x̄_A = 21, x̄_B = 26; s_p² = 4; t = 5/√(s_p²(1/6 + 1/5)) ≈ 4.13 > 3.250.

76.

A fictional plant cross gives four classes in the numbers 290, 102, 94, 26 (total 512). Theory says 9 : 3 : 3 : 1. With χ²₀.₀₅,₃ = 7.815, the test gives

Correct Answer: C [χ² ≈ 1.56; the data agree with the 9 : 3 : 3 : 1 ratio]

Explanation:

Expected 288, 96, 96, 32; χ² ≈ 1.56 < 7.815.

77.

Fictional samples: n₁ = 100, x̄₁ = 72, s₁ = 10; n₂ = 144, x̄₂ = 68, s₂ = 12. With z = 1.96, the 95% confidence interval for μ₁ − μ₂ is

Correct Answer: B [1.23 to 6.77]

Explanation:

SE = √(100/100 + 144/144) = √2 ≈ 1.414; 4 ± 1.96 × 1.414.

78.

An estimator T of θ has variance 4 and bias 1.5. Its mean squared error is

Correct Answer: A [6.25]

Explanation:

MSE = Var + bias² = 4 + 2.25 = 6.25.

79.

A sample 3.2, 7.9, 5.4, 9.1, 6.6 comes from a uniform distribution on (0, θ). The maximum likelihood estimate of θ is

Correct Answer: D [9.1]

Explanation:

L(θ) = θ⁻ⁿ for θ ≥ max xᵢ, decreasing in θ → θ̂ = max xᵢ = 9.1 (the moment estimate 2x̄ is different).

80.

Of 50 fictional vaccinated persons 35 stayed healthy; of 50 unvaccinated, 25 stayed healthy. With χ²₀.₀₅,₁ = 3.841 and no correction, the χ² test gives

Correct Answer: C [χ² ≈ 4.17; significant at 5%]

Explanation:

Expected 30, 20, 30, 20; χ² = 25/30 + 25/20 + 25/30 + 25/20 ≈ 4.17 > 3.841.

81.

A fictional study reports the 95% confidence interval for a mean as 12.4 to 15.6. At the 5% level (two-sided), the hypothesis H₀: μ = 16 is

Correct Answer: A [rejected, since 16 lies outside the interval]

Explanation:

A two-sided 5% test rejects exactly those μ₀ outside the 95% confidence interval.

82.

Two fictional labs measure the same quantity: lab A (n = 13) has s² = 40, lab B (n = 16) has s² = 12. With F₀.₀₅(12, 15) = 2.48, the F statistic is (and it is significant)

Correct Answer: B [3.33]

Explanation:

Answer: 3.33. F = 40/12 ≈ 3.33 > 2.48.

83.

A sample of 20 fictional tablets gives s² = 9. H₀: σ² = 16 is tested two-sided at 5% with χ²₀.₉₇₅,₁₉ = 8.907 and χ²₀.₀₂₅,₁₉ = 32.852. The result is

Correct Answer: D [χ² ≈ 10.69; do not reject H₀]

Explanation:

χ² = 19 × 9/16 ≈ 10.69, which lies between 8.907 and 32.852.

84.

Fictional yields of three fertilisers (rows) in four seasons (columns), one plot each: 15, 18, 18, 21 / 12, 13, 17, 18 / 20, 22, 25, 29. With F₀.₀₅(2, 6) = 5.14 and F₀.₀₅(3, 6) = 4.76, the two-way ANOVA shows that

Correct Answer: C [both fertilisers and seasons differ significantly]

Explanation:

SSR = 168, SSC = 82, SSE = 8; F(rows) = 63, F(columns) = 20.5.

85.

In a fictional one-way ANOVA of 4 treatments with 24 observations, MSB = 20 and F = 2.5. The total sum of squares is

Correct Answer: B [220]

Explanation:

MSE = 20/2.5 = 8; SSB = 20 × 3 = 60; SSE = 8 × 20 = 160; SST = 220.

86.

Fictional pulse rates after three exercises: E₁ – 23, 25, 27, 25; E₂ – 30, 28, 32, 34; E₃ – 26, 24, 25, 25. At the 1% level (F₀.₀₁(2, 9) = 8.02), one-way ANOVA gives

Correct Answer: A [F ≈ 14.4; significant at 1%]

Explanation:

Means 25, 31, 25; SSB = 96, SSW = 30; F = (96/2)/(30/9) ≈ 14.4.

87.

A randomised block design has 3 treatments in 6 blocks (one plot per treatment in each block). The error degrees of freedom are

Correct Answer: D [10]

Explanation:

Answer: 10. (b − 1)(t − 1) = 5 × 2 = 10.

88.

In the one-way model xᵢⱼ = μ + αᵢ + eᵢⱼ (equal group sizes), the treatment effects αᵢ satisfy

Correct Answer: C [Σαᵢ = 0, and H₀ says every αᵢ = 0]

Explanation:

αᵢ = μᵢ − μ are deviations from the overall mean, so they add to zero; H₀: α₁ = … = αₖ = 0.

89.

Fictional exports (₹ crore) for 2017–2023 are 52, 55, 61, 60, 66, 70, 72. Fitting a straight-line trend by least squares (origin 2020), the trend estimate for 2025 is about

Correct Answer: D [79.25]

Explanation:

a = 62.286, b = Σxy/Σx² = 95/28 = 3.393; 2025 → x = 5: 79.25.

90.

Quarterly sales of a fictional brand (Q1 of year 1 to Q3 of year 2): 30, 42, 38, 26, 34, 46, 40. The ratio to the centred 4-quarterly moving average for Q3 of year 1 (value 38) is about

Correct Answer: C [110.14]

Explanation:

Moving averages 34, 35; centred 34.5; 38/34.5 × 100 ≈ 110.14.

91.

Seasonal indices worked out for a fictional series are 102, 98, 110, 94 for quarters 1–4. After adjusting them to total 400, the index of quarter 3 is about

Correct Answer: A [108.91]

Explanation:

Total = 404; multiply by 400/404: 110 × 400/404 ≈ 108.91.

92.

A fictional series grows by roughly the same percentage every year. The most suitable trend curve is

Correct Answer: B [the exponential curve y = abˣ (log y linear in x)]

Explanation:

Constant growth RATE → log y = log a + x log b is linear: exponential trend.

93.

A fictional firm's annual trend is Y = 600 + 48x (Y = annual sales, origin 1 July 2020, x in years). Converting to a monthly trend (origin 1 July 2020, x in months), the trend value for the middle of January 2021 (x = 6.5) is about

Correct Answer: D [52.17]

Explanation:

Monthly: y = 50 + 0.3333x; at x = 6.5: 52.17.

94.

In the multiplicative model, after the trend T and seasonal S are removed, the quantity Y/(T × S) contains

Correct Answer: C [the cyclical and irregular components (C × I)]

Explanation:

Y/(T·S) = C·I; smoothing it further separates cycles from random variation (residual method).

95.

Fictional data: prices ₹ 10, 20, 5 (year 0) and ₹ 12, 25, 6 (year 1); quantities 30, 10, 40 (year 0) and 25, 12, 50 (year 1). Laspeyres' index of year 1 on base 0 (as a ratio) multiplied by Laspeyres' index of year 0 on base 1 (as a ratio) is about

Correct Answer: A [0.9984]

Explanation:

L₀₁ = 1.2143, L₁₀ = Σp₀q₁/Σp₁q₁ = 0.8222; product ≈ 0.9984 ≠ 1 – Laspeyres fails the time reversal test.

96.

Base prices of three fictional items are ₹ 40, 25, 10 with base quantities 5, 8, 20. Current prices of the first two are ₹ 50, 30. If Laspeyres' price index is 125, the current price of the third item is

Correct Answer: B [₹ 13]

Explanation:

Σp₀q₀ = 600; Σp₁q₀ = 1.25 × 600 = 750 = 250 + 240 + 20p → p = 13.

97.

A fictional index (2012 = 100) is 140 in 2016 and 175 in 2019. From 2019 a new index (2019 = 100) is used, which reads 120 in 2022. On the base 2016 = 100, the 2022 figure is

Correct Answer: C [150]

Explanation:

2019 on base 2016: 175/140 × 100 = 125; 2022: 125 × 120/100 = 150.

98.

A fictional industry's money-wage index rose to 180 and the CPI to 150 (same base = 100). The real wage index is

Correct Answer: B [120]

Explanation:

Answer: 120. 180/150 × 100 = 120.

99.

Group weights of a fictional consumer price index: food 50, fuel 20, clothing 10, miscellaneous 20; group indices 180, 150, 120 and 200. The overall index is

Correct Answer: D [172]

Explanation:

(9000 + 3000 + 1200 + 4000)/100 = 172.

100.

While choosing the base year for a fictional price index, the statistician should prefer a year that is

Correct Answer: A [normal – free from wars, droughts, booms and slumps]

Explanation:

An abnormal base distorts every index computed on it; a normal, recent year is chosen.

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