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Statistics appeared 21 times across 3 years — 2.4% of Mathematics. This question is from Mean and Variance.

Year 2026 2025 2024 Total
Questions 7 7 7 21

If the mean and variance of five observations are (24)/(5) and (194)/(25) respectively and the mean of first four observations is (7)/(2), then the variance of the first four observations is equal to

Solution & Explanation

Related Formula
Mean X = (Σ xᵢ)/(n) Variance σ² = (Σ xᵢ²)/(n) - ( X)²
Core Logic

Let the five observations be x₁, x₂, x₃, x₄, x₅. Given total mean:

(x₁ + x₂ + x₃ + x₄ + x₅)/(5) = (24)/(5) x₁ + x₂ + x₃ + x₄ + x₅ = 24

Given mean of first four observations:

(x₁ + x₂ + x₃ + x₄)/(4) = (7)/(2) x₁ + x₂ + x₃ + x₄ = 14

Substituting this back, we find the fifth observation:

14 + x₅ = 24 x₅ = 10
Step 1: Finding the Sum of Squares

Using the variance of the 5 observations:

σ² = (194)/(25) = (x₁² + x₂² + x₃² + x₄² + x₅²)/(5) - ((24)/(5))² (194)/(25) = (x₁² + x₂² + x₃² + x₄² + 100)/(5) - (576)/(25) (194 + 576)/(25) = (x₁² + x₂² + x₃² + x₄² + 100)/(5) (770)/(5) = x₁² + x₂² + x₃² + x₄² + 100 154 = x₁² + x₂² + x₃² + x₄² + 100 x₁² + x₂² + x₃² + x₄² = 54
Step 2: Variance of First Four Observations
Variance₄ = Σi=1⁴ xᵢ²4 - ( Σi=1⁴ xᵢ4)² Variance₄ = (54)/(4) - ((7)/(2))² = (54)/(4) - (49)/(4) = (5)/(4)
Pattern Recognition

Isolate the missing elements sequentially. Use the sum of elements first to find x₅, then use the sum of squares equation to find the squared sum of the subset.

Chapter Mix

Class 11 Mathematics: Statistics

Reference Study Guides

More Statistics Previous-Year Questions — Page 5

Q12 jee_main_2024_31_jan_evening Mean and Variance
Let the mean and the variance of 6 observation a, b, 68, 44, 48, 60 be 55 and 194, respectively if a > b, then a + 3b is
  • A. 200
  • B. 190
  • C. 180
  • D. 210

Solution

Related Formula
Mean x = (Σ xᵢ)/(n) Variance σ² = Σ (xᵢ - x)²n
Core Logic

Mean is 55:

(a + b + 68 + 44 + 48 + 60)/(6) = 55 220 + a + b = 330 a + b = 110

Variance is 194:

((a-55)² + (b-55)² + (68-55)² + (44-55)² + (48-55)² + (60-55)²)/(6) = 194 (a-55)² + (b-55)² + 13² + (-11)² + (-7)² + 5² = 1164 (a-55)² + (b-55)² + 169 + 121 + 49 + 25 = 1164 (a-55)² + (b-55)² = 800

Expand the squares using a+b=110: a² + b² - 110(a+b) + 2(3025) = 800 a² + b² - 110(110) + 6050 = 800 a² + b² = 6850 Using (a+b)² = 12100 a²+b²+2ab = 12100 2ab = 12100 - 6850 = 5250. (a-b)² = a²+b² - 2ab = 6850 - 5250 = 1600 a-b = 40 (since a>b).

Solving a+b=110 and a-b=40:

a = 75, b = 35

Finally, evaluate a + 3b:

a + 3b = 75 + 3(35) = 180
Chapter Mix

Class 11 Maths: Statistics

More Statistics Questions — jee_main_2024_29_january_evening

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