Karnataka 2nd PUC Answer Key 2026 Statistics (Model Answers)

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KARNATAKA SCHOOL EXAMINATION AND ASSESSMENT BOARD
2026 II PUC EXAM – 1
SUBJECT: 31 – STATISTICS MODEL ANSWERS MAX. MARKS: 80

Q. NO. ANSWERS MARKS
SECTION – A
I.
1) a) Fertility 1
2) b) 90 1
3) a) 0 1
PQ
4) c) √ n 1
5) d) v = 0 1
II.
6) 100 1
7) Standard normal 1
8) Parameter 1
9) Defective 1
10) Minimum 1
III.
a NRR per woman = 1.1 iv) Population increase 1
b Base year weight v) Laspeyre’s price index number 1
c The mean of a t – distribution vi) 0 1
11)
d Single value proposed as an i) Point estimation 1
estimate
e Physical stock of goods iii) Inventory 1
IV.
12) The capacity of woman to bear children 1
13) War, Flood, etc.. (Any one cause) 1
14) p = ½ = 0.5 1
Maximum size of a test OR Probability of type I error OR Probability of
15) 1
rejecting H0 when it is actually true OR α.
16) Because of non-negativity restriction OR x≥0, y≥0. 1
SECTION - B
V.
17) Prosperity, Recession, Depression, Recovery 1+1
1. The values of the independent variable should have a common
difference. That is the values should be in Arithmetic Progression.
18) 1+1
2. The value of X, for which the value of Y is to be Interpolated must be
one of the values of X.
p = 0.5, q = 0.5
19) Mean = p = 0.5 1
Variance = pq = 0.25 1
1. Poisson distribution has only one parameter λ.
20) 2. x = 0,1, 2, ..........., ∞ is the range of Poisson distribution. 1+1
(Any Two features of Poisson distribution)
Rejecting H0 when it is actually true is called type I error. 1
21)
Accepting H0 when it is actually not true is called type II error. 1

KSEAB MODEL ANSWERS 31 – STATISTICS 2026 II PUC EXAM - 1

2

d̅ 3.4
22) t= 𝑠𝑑 = 2 = 3.4 1+1
√𝑛−1 √5−1
1. Chance cause 1
23)
2. Assignable cause 1
2C3 R 2 × 200 × 5000
24) Q0 = √ = √ = 1000 units 1+1
C1 2
SECTION - C
VI.
p0q0 = 80 400 600 1250 Σp0q0 = 2330 1
p1q0 = 160 480 800 2000 Σp1q0 = 3440 1
∑ p1 q 0 1
CPI = × 100
25) ∑ p0 q 0
3440
CPI = × 100 1
2330
CPI = 147.64 1
(y-1)4 = 0 1
y4 – 4y3 + 6y2 – 4y1 + y0 = 0 1
30 – (4×21) + 6y2 – (4×14) + 8 = 0
y2 = 17 1
26)
y5 – 4y4 + 6y3 – 4y2 + y1 = 0 1
y5 – (4×30) + (6×21) – (4×17) + 14 = 0
y5 = 48 1
n = 3, p = 0.25, q = 0.75
P(X = x) = nCxpxqn-x, x = 0, 1, 2,......, n. 1
27)
i) P(X = 0) = 0.4219 1+1
ii) P(X = 2) = 0.1406 1+1
a = 6, b = 4, n = 3 1
na 3×6 1+1
Mean = = = 1.8
28) a+b 10
nab (a + b − n)
Variance = = 0.56
(a + b)2 (a + b − 1) 1+1
σ = 10.2, n = 25, s2 = 8.2, α = 0.01
H0: σ = 10.2
H1: σ < 10.2 1
29)
𝑛𝑠2
χ2 = 𝜎2 = 16.16 1+1
The critical value at 1% level of significance with 24d.f is 1
χ2 𝑡𝑎𝑏 = 10.9 1
Accept H0, σ = 10.2
Allocation +Solution table 3

30) IBFS: x11 = 75, x12 = 225, x22 = 100, x23 = 400, x33 = 100, x34 = 500 1

Total transportation cost = Rs. 21200 1
Year : 1 2 3 4 5
P - Sn : 3500 4500 5500 6500 7500 1
31) ΣCi : 500 1100 1900 3000 4500 1
Tn : 4000 5600 7400 9500 12000 1
A(n) : 4000 2800 2466.67 2375 2400 1

KSEAB MODEL ANSWERS 31 – STATISTICS 2026 II PUC EXAM - 1

3
The machine is replaced at the end of 4th year. 1

VII.
X: Weekly wages of workers
X ~ N(μ = 3000, σ2 = 5002)
x − μ x − 3000
Z= = ~ N(0, 1)
σ 500 1
i) P(X<3500)
32) Z=1
1+1
P(Z<1) = area (1 to -∞) = 1 - 0.1587 = 0.8413

ii) P(2500 < X < 3500)
Z1 = -1, Z2 = 1
P(-1 < Z < 1)
1+1
= 0.6826
H0 : μ1 = μ2
H1 : μ1 ≠ μ2 1
𝑥̅1 − 𝑥̅2
𝑍= ~ 𝑍 𝑡𝑒𝑠𝑡
𝑠12 𝑠22 1
33) √ +
𝑛
1 𝑛 2
= 2.2 1
Ztab = (-2.58, 2.58) 1
Accept H0, μ1 = μ2 1
λ| = 4
LCL = λ| − 3√λ| = −2. (taken as 0) 1+1
34) 1
CL = λ| = 4
1+1
UCL = λ| + 3√λ| = 10
x+y = 350 (0,350) (350,0)
2x+y = 600 (0,600) (300,0)
2 lines + graph (Feasible region) 3

35) Corner points Z = 2x+3y
(0, 350) 1050
(0, 600) 1800 (Max) 1
(250, 100) 800
Optimal solution is x = 0, y = 600, max Z = 1800 1
SECTION - D
VIII.
ASDR =
Dx
× 1000 or 1 calculation of ASDR should be shown 1
Px

1
Locality A, ASDR: 17, 6, 13, 43 1+1
ΣP = 30000, ΣPA = 450000
36) ∑ PA
STDR[A] = = 15 1+1
∑P 1
Locality B, ASDR: 20, 5, 10, 40 1
ΣPB = 420000 1
STDR(B) = 14 1
Locality B is more healthy.
Items p0 q0 p1 q1 p0q0 p1q0 p0q1 p1q1
37) A 10 15 14 25 150 210 250 350
B 14 20 20 30 280 400 420 600

KSEAB MODEL ANSWERS 31 – STATISTICS 2026 II PUC EXAM - 1

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C 17 30 15 45 510 450 765 675
D 20 25 30 32 500 750 640 960
1440 1810 2075 2585

Σp0q0 = 1440 1
Σp1q0 = 1810 1
Σp0q1 = 2075 1
Σp1q1 = 2585 1

TRT P01 × P10 = 1 + Verification 1+2

FRT P01 × Q01 = V01 + Verification 1+2
38) a) 3 yearly moving sum : - 59 57 53 49 54 57 60 - 2
3 yearly moving average: - 19.67 19 17.67 16.33 18 19 20 - 2
Position of the trend values 1

b) x : -5 -3 -1 1 3 5 Σx = 0
x2: 25 9 1 1 9 25 Σ x2 = 70 2
xy: -80 -60 -19 14 54 105 Σxy = 14
Σy = 630

∑y 1
a= = 18
n

∑ xy
b= = 0.2
∑ x2 1

Y = 18+0.2x 1

SECTION – E
35) Linear Programming Problem solving steps in graphical method 5

KSEAB MODEL ANSWERS 31 – STATISTICS 2026 II PUC EXAM - 1