Class 11 Maths — Paper 2 📊

Arrange in order: 3, 3, 5, 5, 5, 7, 9
Mean = (3+3+5+5+5+7+9)/7 = 37/7 ≈ 5.28
Median = middle value = 4th value = 5
Mode = most frequent = 5 (appears 3 times)

Quartiles divide arranged data into 4 equal parts.
Q1 = lower quartile (25th percentile)
Q2 = median (50th percentile)
Q3 = upper quartile (75th percentile)

For 2,4,6,8,10,12,14 (n=7):
Q1 = value at (n+1)/4 = 2nd position = 4
Q3 = value at 3(n+1)/4 = 6th position = 12

IQR = Q3 − Q1
It measures the spread of the middle 50% of data. It is not affected by extreme values.
For above data: IQR = 12 − 4 = 8

Sample space (S): Set of all possible outcomes of an experiment. Example: Tossing a coin → S = {H, T}

Event: Any subset of sample space.

Probability P(E) = n(E)/n(S) = (favourable outcomes)/(total outcomes). Range: 0 ≤ P(E) ≤ 1

S = {1,2,3,4,5,6}, n(S) = 6
Even numbers = {2,4,6} → P(even) = 3/6 = 1/2
Numbers > 4 = {5,6} → P(>4) = 2/6 = 1/3

S = {HH, HT, TH, TT}, n(S) = 4
At least one head = {HH, HT, TH} → n = 3
P(at least one head) = 3/4 = 0.75

P(A∪B) = P(A) + P(B) − P(A∩B)
If A and B are mutually exclusive (P(A∩B)=0):
P(A∪B) = P(A) + P(B)

a = 2, d = 3
nth term: aₙ = a + (n−1)d = 2 + (n−1)3 = 3n−1
Sum of n terms: Sₙ = n/2[2a+(n−1)d] = n/2[4+3(n−1)] = n(3n+1)/2

a=1, d=2, n=20
S₂₀ = 20/2[2(1)+(20−1)2] = 10[2+38] = 10×40 = 400

a=3, r=2
8th term: a₈ = ar⁷ = 3×2⁷ = 3×128 = 384
S₁₀ = a(rⁿ−1)/(r−1) = 3(2¹⁰−1)/(2−1) = 3×1023 = 3069

a=1, r=1/2 (|r|<1, so sum exists)
S∞ = a/(1−r) = 1/(1−1/2) = 1/(1/2) = 2

SI = PRT/100 = 5000×8×3/100 = ₹1200
Amount = P + SI = 5000 + 1200 = ₹6200

A = P(1+R/100)ⁿ = 10000(1+10/100)² = 10000×(1.1)² = 10000×1.21 = ₹12100
CI = A − P = 12100 − 10000 = ₹2100

For same P, R, T:
CI > SI (except for T=1 year when they are equal)
Difference = P × (R/100)² for 2 years
= 10000 × (10/100)² = 10000 × 0.01 = ₹100

Linear Programming (LPP): method to find maximum or minimum value of a linear function subject to linear constraints.

Objective function: The function to be maximized or minimized. Example: Z = 3x + 5y

Constraints: The restrictions in the form of linear inequalities. Example: x+y ≤ 10, x≥0, y≥0

Corner points of feasible region: (0,0), (4,0), (0,4)
Z at (0,0) = 0
Z at (4,0) = 12
Z at (0,4) = 16
Maximum Z = 16 at (0,4)

The feasible region is the set of all points satisfying all the constraints simultaneously. It is the common region of all constraint inequalities. The optimal solution (maximum or minimum) always occurs at a corner (vertex) of the feasible region.

Permutation: arrangement of objects where ORDER matters. ⁿPᵣ = n!/(n−r)!

Combination: selection of objects where ORDER does NOT matter. ⁿCᵣ = n!/r!(n−r)!

Example: Selecting 2 from {A,B,C}: Combinations = {AB,AC,BC} = 3. Permutations = {AB,BA,AC,CA,BC,CB} = 6.

⁵P₃ = 5!/(5−3)! = 5!/2! = 120/2 = 60
⁵C₃ = 5!/3!×2! = 120/6×2 = 120/12 = 10

This is arrangement (order matters) → permutation.
5! = 5×4×3×2×1 = 120 ways

Order doesn't matter → combination.
⁸C₃ = 8!/3!×5! = (8×7×6)/(3×2×1) = 336/6 = 56 ways

Mean = (2+4+4+4+5+5+7+9)/8 = 40/8 = 5
Deviations from mean: −3,−1,−1,−1,0,0,2,4
Squared deviations: 9,1,1,1,0,0,4,16
Variance σ² = (9+1+1+1+0+0+4+16)/8 = 32/8 = 4
SD σ = √4 = 2

CV = (σ/mean) × 100
It is used to compare variability of two different data sets with different units or means. Lower CV = more consistent/uniform data. Higher CV = more variable data.

Mean = (6+7+10+12+13+4+8)/7 = 60/7 ≈ 8.57
|deviations| = 2.57, 1.57, 1.43, 3.43, 4.43, 4.57, 0.57
MD = sum/7 = 18.57/7 ≈ 2.65