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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 2

Q14 jee_main_2026_24_january_evening Transformation of Mean and Variance
Let X = x in N : 1 ≤ x ≤ 19 and for some a, b in R, Y = ax + b : x in X. If the mean and variance of the elements of Y are 30 and 750, respectively, then the sum of all possible values of b is
  • A. 20
  • B. 80
  • C. 100
  • D. 60

Solution

Related Formula
If yᵢ = axᵢ + b, then y = a x + b Variance(Y) = a² Variance(X) Variance of first n natural numbers: (n² - 1)/(12)
Core Logic

For X = 1, 2, , 19, we find the mean x and variance σₓ².

x = (1 + 2 + + 19)/(19) = ((19 × 20)/(2))/(19) = 10 Variance(X) = (19² - 1)/(12) = (361 - 1)/(12) = 30
Step 1: Creating System of Equations

Given Mean of Y = 30 and Variance of Y = 750.

Using transformations:

y = a x + b 30 = 10a + b (1) Variance(Y) = a² · Variance(X) 750 = a² × 30
Step 2: Solving for a and b

From the variance equation:

a² = (750)/(30) = 25 a = ± 5

Now find the corresponding values of b from equation (1): If a = 5 b = 30 - 10(5) = 30 - 50 = -20 If a = -5 b = 30 - 10(-5) = 30 + 50 = 80

Step 3: Finding Sum of b Values

The possible values for b are -20 and 80.

Sum = -20 + 80 = 60

Pattern Recognition

When a dataset undergoes linear transformation Y = aX + b, the variance isolates a² entirely decoupled from b. Solve variance first to get a, then back-substitute into the mean equation.

Chapter Mix

Class 11 Maths: Statistics

Q15 jee_main_2026_28_january_morning Mean Deviation and Variance
The mean and variance of 10 observations are 9 and 34.2, respectively. If 8 of these observations are 2, 3, 5, 10, 11, 13, 15, 21, then the mean deviation about the median of all the 10 observations is
  • A. 5
  • B. 4
  • C. 6
  • D. 7

Solution

Related Formula

Mean x = (Σ xᵢ)/(n) Variance σ² = (Σ xᵢ²)/(n) - ( x)² Mean Deviation about Median = (Σ |xᵢ - M|)/(n) where M is the median.

Core Logic

Let the two missing observations be a and b. From the given mean:

(2 + 3 + 5 + 10 + 11 + 13 + 15 + 21 + a + b)/(10) = 9 (80 + a + b)/(10) = 9 a + b = 10

From the given variance:

(Σ xᵢ²)/(10) - 9² = 34.2 (2² + 3² + 5² + 10² + 11² + 13² + 15² + 21² + a² + b²)/(10) = 34.2 + 81 = 115.2
Step 1: Solve for a and b

Sum of squares of knowns:

4 + 9 + 25 + 100 + 121 + 169 + 225 + 441 = 1094 1094 + a² + b² = 1152

a² + b² = 58

We know a+b = 10 b = 10-a.

a² + (10-a)² = 58 2a² - 20a + 100 = 58 2a² - 20a + 42 = 0 a² - 10a + 21 = 0 (a-3)(a-7) = 0

Thus, the missing observations are 3 and 7.

Step 2: Find the Median

Arrange all 10 observations in ascending order: 2, 3, 3, 5, 7, 10, 11, 13, 15, 21 Since there are 10 observations, median M is the average of the 5th and 6th terms:

M = (7 + 10)/(2) = 8.5
Step 3: Calculate Mean Deviation about Median

Find absolute deviations |xᵢ - M|: |2-8.5|=6.5 |3-8.5|=5.5 |3-8.5|=5.5 |5-8.5|=3.5 |7-8.5|=1.5 |10-8.5|=1.5 |11-8.5|=2.5 |13-8.5|=4.5 |15-8.5|=6.5 |21-8.5|=12.5

Sum of absolute deviations:

6.5 + 5.5 + 5.5 + 3.5 + 1.5 + 1.5 + 2.5 + 4.5 + 6.5 + 12.5 = 50 Mean Deviation = (50)/(10) = 5
Chapter Mix

Class 11 Mathematics: Statistics

Q61 jee_main_2025_02_april_evening Mean and Variance
If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then a + b + ab is equal to :
  • A. 105
  • B. 103
  • C. 100
  • D. 106

Solution

Related Formula
Mean: x = (Σ xᵢ)/(N) Variance: σ² = (Σ xᵢ²)/(N) - x²
Core Logic

We set up algebraic equations using the definitions of mean and variance to determine the values of a+b and ab.

Step 1: Apply the Mean condition

Given mean x = 9 for N=8 observations:

(6 + 4 + a + 8 + b + 12 + 10 + 13)/(8) = 9 53 + a + b = 72 a + b = 19 --- (1)
Step 2: Apply the Variance condition

Given variance σ² = 9.25 = (37)/(4):

(36 + 16 + a² + 64 + b² + 144 + 100 + 169)/(8) - 81 = (37)/(4) (529 + a² + b²)/(8) = 81 + 9.25 = 90.25 = (361)/(4) 529 + a² + b² = 722 a² + b² = 193 --- (2)
Step 3: Solve for ab and calculate the target expression

We know (a+b)² = a² + b² + 2ab. Substitute equations (1) and (2):

19² = 193 + 2ab 361 = 193 + 2ab 2ab = 168 ab = 84

Now, calculate the target value:

a + b + ab = 19 + 84 = 103
Pattern Recognition

Direct symmetric evaluation: Statistics problems in JEE with missing observations often ask for symmetric combinations of variables like a+b+ab or a²+b². These can be calculated using quadratic expansions without solving for a and b individually.

Chapter Mix

Class 11 Mathematics: Statistics

Q54 jee_main_2025_03_april_evening Measures of Dispersion
Let the Mean and Variance of five observations x₁ = 1, x₂ = 3, x₃ = a, x₄ = 7 and x₅ = b, a > b, be 5 and 10 respectively. Then the Variance of the observations n + xₙ, n = 1, 2, 5 is
  • A. 17
  • B. 16.4
  • C. 17.4
  • D. 16

Solution

Related Formula

Mean of N observations:

x = (Σ xᵢ)/(N)

Variance of N observations:

σ² = (Σ xᵢ²)/(N) - ( x)²
Core Logic

Given mean is 5 for 5 observations:

(1 + 3 + a + 7 + b)/(5) = 5 a + b + 11 = 25 a + b = 14 --- (1)

Given variance is 10:

(1² + 3² + a² + 7² + b²)/(5) - 5² = 10 (59 + a² + b²)/(5) = 35 a² + b² = 175 - 59 = 116 --- (2)
Step 1: Finding a and b

Using standard algebraic identity (a+b)² = a² + b² + 2ab:

14² = 116 + 2ab 196 = 116 + 2ab ab = 40

Solving a+b=14 and ab=40:

a(14-a) = 40 a² - 14a + 40 = 0 (a-10)(a-4) = 0

Since a > b, we obtain a = 10 and b = 4.

Step 2: Constructing new set and finding variance

The original set is x₁ = 1, x₂ = 3, x₃ = 10, x₄ = 7, x₅ = 4. We construct the new set yₙ = n + xₙ:

  • y₁ = 1 + 1 = 2
  • y₂ = 2 + 3 = 5
  • y₃ = 3 + 10 = 13
  • y₄ = 4 + 7 = 11
  • y₅ = 5 + 4 = 9
  • Mean of new set:

y = (2 + 5 + 13 + 11 + 9)/(5) = (40)/(5) = 8

Variance of new set:

σnew² = (2² + 5² + 13² + 11² + 9²)/(5) - 8² σnew² = (4 + 25 + 169 + 121 + 81)/(5) - 64 = (400)/(5) - 64 = 80 - 64 = 16
Pattern Recognition

Note that adding a changing factor like +n is different from adding a constant C to each observation (which leaves variance unchanged). In this case, calculate individual xₙ variables directly first before applying transformations.

Chapter Mix

Class 11 Mathematics: Statistics and Probability

Q63 jee_main_2025_03_april_evening Probability Distributions
If the probability that the random variable X takes the value x is given by P(X = x) = k(x + 1) 3-x, x = 0, 1, 2, 3, where k is a constant, then P(X ≥ 3) is equal to
  • A. (7)/(27)
  • B. (4)/(9)
  • C. (8)/(27)
  • D. (1)/(9)

Solution

Related Formula

Sum of all probabilities in a distribution:

Σx=0∞ P(X = x) = 1

Complementary probability:

P(X ≥ 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]
Core Logic

Let's first determine the constant k:

k Σx=0∞ (x+1) 3-x = 1

This is an Arithmetico-Geometric Progression (AGP). Let S = Σx=0∞ (x+1)((1)/(3))^x:

S = 1 + (2)/(3) + (3)/(9) + (4)/(27) + --- (1) (1)/(3)S = (1)/(3) + (2)/(9) + (3)/(27) + --- (2)
Step 1: Finding k

Subtracting (2) from (1):

S(1 - (1)/(3)) = 1 + (1)/(3) + (1)/(9) + (1)/(27) + (2)/(3)S = (1)/(1 - 1/3) = (3)/(2) S = (9)/(4)

Substitute back:

k · ((9)/(4)) = 1 k = (4)/(9)
Step 2: Calculating P(X ≥ 3)

Calculate initial probability values:

  • P(X=0) = k(1)(1) = (4)/(9)
  • P(X=1) = k(2)((1)/(3)) = (2)/(3) · (4)/(9) = (8)/(27)
  • P(X=2) = k(3)((1)/(9)) = (1)/(3) · (4)/(9) = (4)/(27)
Sum P(X < 3) = (12)/(27) + (8)/(27) + (4)/(27) = (24)/(27) = (8)/(9) P(X ≥ 3) = 1 - (8)/(9) = (1)/(9)
Pattern Recognition

The infinite AGP sum with factor (x+1)r^x always converges to (1)/((1-r)²). Here r = 1/3, so the sum is (1)/((2/3)²) = 9/4. This mental check saves doing the full subtraction sequence.

Chapter Mix

Class 11 Mathematics: Statistics and Probability Class 11 Mathematics: Sequences and Series

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