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

Year 2026 2025 2024 Total
Questions 7 7 7 21

For a statistical data x₁, x₂, …, x₁₀ of 10 values, a student obtained the mean as 5.5 and Σi=1¹⁰ xᵢ² = 371 . He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively. The variance of the corrected data is :

Solution & Explanation

Related Formula

The standard statistical variance equation for a sample size n is defined as:

σ² = (Σ xᵢ²)/(n) - ( x)²
Core Logic

Calculate the initial incorrect \sum of observations using the given incorrect mean:

xold = 5.5 = Σ xold10 Σ xold = 55

The given incorrect \sum of squares is:

Σ xold² = 371
Step 1: Compute Corrected Sum of Observations

Adjust the linear \sum by subtracting the incorrect inputs and adding the true values:

Σ xnew = 55 - (4 + 5) + (6 + 8) = 55 - 9 + 14 = 60

Calculate the new corrected mean value:

xnew = (60)/(10) = 6
Step 2: Compute Corrected Sum of Squares

Adjust the \sum of squares by swapping the squared entries:

Σ xnew² = 371 - (4² + 5²) + (6² + 8²) Σ xnew² = 371 - (16 + 25) + (36 + 64) Σ xnew² = 371 - 41 + 100 = 430
Step 3: Calculate Corrected Variance

Substitute the corrected values into the standard variance formula:

σnew² = Σ xnew²10 - ( xnew)² σnew² = (430)/(10) - (6)² = 43 - 36 = 7
Pattern Recognition

When updating statistical aggregates like mean and variance after data correction, always compute the corrected linear \sum and \sum of squares separately before recombining them into the variance formula.

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_2025_24_jan_morning

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