Karnataka 2nd PUC Question Paper 2025 Answer Key Statistics

Government of Karnataka
Karnataka Secondary Education Examination Board

Question Paper
ANSWER KEY

1
KARNATAKA SCHOOL EXAMINATION AND ASSESSMENT BOARD
II PUC EXAMINATION – 1, MARCH 2025
Subject: 31-STATISTICS MODEL ANSWERS MAX. MARKS: 80
Q. NO. SECTION – A MARKS

I. 1. d) Fecundity 1
2. b) 100 1
3. a) Mean = variance 1
4. b) Accepting H0 when it is not true. 1
5. c) Σai = Σbj 1

II. 6. Retail 1
7. Asymptotic 1
8. Size 1
9. Defect 1
10. Degenerate 1
III. 11.
a) iii) Σ Annual ASFRs 1
b) vi) Factor Reversal Test 1
c) v) x = 0,1 1
d) i) H0: P1 = P2 1
e) iv) Inventory model – II 1

IV. 12. Size of the cohort. 1
13. War, Floods, Strikes, Lockouts, Earthquakes etc. (Any one) 1
14. Mean = 0 1
15. Statistical constant of the population. 1
16. Maximum of the row minimums. 1

Q. NO. V. SECTION – B MARKS

17. 110 1
120 1
18. (i) There are no sudden jumps in the values of dependent variable from one
period to another. 1
(ii) There is a sort of uniformity in the rise or fall of the values of the dependent
variable. 1
(iii) There will be no consecutive missing values in the series. (Any two)
19. S.D. (X) = √pq 1
= 0.4 1
20. Median = 9.34 1
Mode = 8 1
21. Point estimation 1
Interval estimation 1
22. t = Sd
̅
d
⁄
1
√n−1
d.f. = n-1 1
23. UCL = λ′ + 3√λ′ 1
= 4 + 3√4 = 10 1
24. 2C3 R
Q0 = √ 1
C1
= 500 units 1

IIPUC EXAM-1, MARCH 2025 KSEAB 31-STATISTICS MODEL ANSWERS

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Q. NO. VI. SECTION – C MARKS
25. p1 4600
P = p x 100 Or P = 4000 x 100 = 115 1
0
P: 115 120 126 150 120 1
WP: 2300 1200 1260 3000 2400 ΣWP: 10,160 1
ΣWP 10160
CLI(FBM) = = = 127 1+1
ΣW 80
26. 10 14 - 27 46 70 -
y0 y1 y2 y3 y4 y5 y6 1
(y – 1)5 = 0 or Δ5 y0 = 0 ⇒ y5 – 5y4 + 10y3 – 10y2 + 5y1 – y0 = 0 1
y2 = 17 1
y6 – 5y5 + 10y4 – 10y3 + 5y2 – y1 = 0 1
y6 = 89 1
27. n = 4, p = 0.4, q = 0.6, p(x) = 4Cx (0.4)x (0.6)4−x ; x = 0,1, 2, 3, 4. 1
(a) p(2) = 4C2 (0.4)2 (0.6)4−2 1
= 0.3456 1
(b) P(X ≥1) = 1 – p(0) = 1 – 4C0 (0.4)0 (0.6)4−0 1
= 0.8704 1
na
28. Mean, E(X) = 1
a+b
5x6
= 6+4 = 3 1
nab(a+b−n)
Variance, V(X) = (a+b)2 1
(a+b−1)
5x6x4(6+4−5) 600
= (6+4)2 = = 0.6667 1+1
(6+4−1) 900
29. H0: μ = μ0 (55) and H1: μ ≠ μ0 (55) 1
x̅−μ
Z= s 0 = 2 1+1
⁄
√n
k = ±1.96 1
∴ H0 is rejected. 1
30. A1 dominates A2 and A3. B2 dominates B1, B3 and B4. A1 dominates A4. 2
Best strategy for Player A is A1, Best strategy for Player B is B2, 1
Value of the game, v = 0. 1
Game is fair. 1
31. ΣCi : 1000 3000 6100 10600 16600 1
P – Sn: 10000 14000 17000 20000 22000 1
T(n) : 11000 17000 23100 30600 38600 1
A(n) : 11000 8500 7700 7650 7720 1
∴ The optimal replacement period, n = 4 years. 1
Q. NO. VII. MARKS
32. μ = 600, σ = 50, Z =
X − 600
is a S.N.V. 1
50
P(550 ≤ X ≤ 650) = P(–1 ≤ Z ≤ 1) 1+1
= 0.8413 – 0.1587 = 0.6826 1
Among 400 workers, 400 x 0.6826 = 273.04 ≅ 273 workers. 1
33. H0: Accidents occur uniformly throughout the week, and
H1: Accidents do not occur uniformly throughout the week. 1
(O−E)2
Table, finding (O-E), E
values 1+1
(O−E)2
χ2 = ∑ E =8 1
6 d.f., k2 = 16.8, ∴ H0 is Accepted. 1
34. C.L. = R̅ = ΣR = 5 1
k
L.C.L. = D3R̅=0 1+1
U.C.L. = D4R̅ = 2.285 x 5 = 11.41 1+1
35. Co-ordinates: (0,2), (4,0) and (0,4), (6,0). 1
Drawing two lines. 1
Identification of F.R. and its corner points: A(0,4), B(6,0), C(0,2) and D(4,0) 1
Objective function values: ZA = 20, ZB = 24, ZC = 10 and ZD = 16 1
Maximum value of Z = 24 and optimal solution: x = 6 & y = 0 1

IIPUC EXAM-1, MARCH 2025 KSEAB 31-STATISTICS MODEL ANSWERS

3
Q. NO. VIII. SECTION – D MARKS

36. (a) WSFR =
Number of female births in a specified age group in a year
x 1000
Total number of females in that particular age group in a year
252
Or WSFR15-19 = 14000 x 1000 = 18 1
WSFR: 18 65 70 42 24 14 3 ΣWSFR: 236 2
GRR = i x ΣWSFR = 5 x 236 = 1180 1+1

Number of deaths in a specific age group in a year
(b) ASDR = x 1000
Total population in that age group in a year
143
Or ASDR0-21 = 11000 x 1000 = 13 1

ASDR(A): 13 05 13 30 1
PA: 130000 75000 195000 300000 ΣPA = 7,00,000 1
ΣPA 700000
STDRA = ΣP = 50000 = 14 1+1

37. p1 q0 : 504 672 270 420 Σ p1 q0 = 1866 1
p0 q 0 : 384 504 324 315 Σ p0 q0 = 1527 1
p1 q1 : 630 840 450 600 Σ p1 q1 = 2520 1
p0 q1 : 480 630 540 450 Σ p0 q1 = 2100 1
Σp q
P01(L) = Σp1 q0 x 100 = 122.2 1+1
0 0
Σp1 q1
P01(P) = Σp q x 100 = 120 1+1
0 1
P01(F) = √P01 (L) x P01 (P) = 121.095 1+1
38. Table, n = 5, Σx = 0, Σy = 130, Σx2 = 10, Σx3 = 0, Σx4 = 34, Σxy = -10, Σx2y = 274 5
(When x: -2, -1, 0, 1, 2)

By substituting and solving the normal equations,
a = 24, b = –1 and c = 1 1+1+1
The quadratic trend equation is:
y = 24 – x + x2 1
ŷ2024 = 30 1

Q. NO. SECTION – E MARKS
(for visually challenged students only)

35. Procedure of solving the linear programming problem graphically. 5

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IIPUC EXAM-1, MARCH 2025 KSEAB 31-STATISTICS MODEL ANSWERS

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