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Statistics appeared 21 times across 3 years — 2.4% of Mathematics. This question is from Measures of Dispersion.

Year 2026 2025 2024 Total
Questions 7 7 7 21

Let the mean and the standard deviation of the observation 2, 3, 3, 4, 5, 7, a, b be 4 and √(2) respectively. Then the mean deviation about the mode of these observations is :

Solution & Explanation

Core Logic

We have 8 observations: 2, 3, 3, 4, 5, 7, a, b. Given mean x = 4:

(2 + 3 + 3 + 4 + 5 + 7 + a + b)/(8) = 4 24 + a + b = 32 a + b = 8
Step 1: Using the Variance Property

Given standard deviation σ = √(2) Variance σ² = 2. The variance formula is σ² = (Σ xᵢ²)/(n) - ( x)²:

2 = (2² + 3² + 3² + 4² + 5² + 7² + a² + b²)/(8) - 4² 2 = (4 + 9 + 9 + 16 + 25 + 49 + a² + b²)/(8) - 16 = (112 + a² + b²)/(8) - 16 2 + 16 = (112 + a² + b²)/(8) 18 × 8 = 112 + a² + b² 144 = 112 + a² + b² a² + b² = 32
Step 2: Solving for a and b and finding the Mode

We know (a+b)² = a² + b² + 2ab 8² = 32 + 2ab 64 = 32 + 2ab 2ab = 32 ab = 16. Solving a+b=8 and ab=16 gives a=4 and b=4.

The complete data set is 2, 3, 3, 4, 4, 4, 5, 7. The value appearing with highest frequency is 4 (occurs 3 times), so Mode = 4.

Step 3: Calculating Mean Deviation about Mode

Mean Deviation about Mode is:

M.D. = Σ |xᵢ - Mode|n = (|2-4| + |3-4| + |3-4| + |4-4| + |4-4| + |4-4| + |5-4| + |7-4|)/(8) = (2 + 1 + 1 + 0 + 0 + 0 + 1 + 3)/(8) = (8)/(8) = 1
Pattern Recognition

When a+b=2√(ab) (here 8 = 2√(16)), the roots are guaranteed to be equal, meaning a=b. Recognizing this condition instantly avoids full quadratic polynomial substitution steps.

Chapter Mix

Class 11 Mathematics: Statistics

Reference Study Guides

More Statistics Previous-Year Questions — Page 4

Q4 jee_main_2024_29_january_evening Mean and Variance
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
  • A. (4)/(5)
  • B. (77)/(12)
  • C. (5)/(4)
  • D. (105)/(4)

Solution

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

Q14 jee_main_2024_27_jan_morning Standard Deviation
Let a₁, a₂, ,a₁₀ be 10 observations such that Σk=1¹⁰ak=50 and Σk
  • A. 5
  • B. √(5)
  • C. 10
  • D. √(115)

Solution

Related Formula
σ = √((Σ aᵢ²)/(n) - ((Σ aᵢ)/(n))²) (Σi=1ⁿ aᵢ)² = Σi=1ⁿ aᵢ² + 2 Σk < j ak aⱼ
Core Logic

Given: Σ aᵢ = 50 Σk < j ak aⱼ = 1100 Number of observations, n = 10.

To find the standard deviation, we need the sum of squares, Σ aᵢ². We use the algebraic identity for the square of a sum of n terms.

Step 1: Finding the Sum of Squares

Substitute the known values into the identity:

(Σ aᵢ)² = Σ aᵢ² + 2 Σk < j ak aⱼ (50)² = Σ aᵢ² + 2(1100) 2500 = Σ aᵢ² + 2200 Σ aᵢ² = 2500 - 2200 = 300
Step 2: Calculating Standard Deviation

Now, apply the variance formula:

σ² = (Σ aᵢ²)/(n) - ((Σ aᵢ)/(n))² σ² = (300)/(10) - ((50)/(10))² σ² = 30 - (5)² σ² = 30 - 25 = 5

The standard deviation σ is the square root of variance:

σ = √(5)
Pattern Recognition

Whenever you see pairwise products Σ aᵢ aⱼ in a statistics problem, immediately bridge it to Σ aᵢ² using the multinomial expansion identity. The variance formula handles the rest organically.

Chapter Mix

Class 11 Maths: Statistics

Q27 jee_main_2024_29_jan_morning Mean and Variance
If the mean and variance of the data 65, 68, 58, 44, 48, 45, 60, α, β, 60 where α gt β are 56 and 66.2 respectively, then α²+β² is equal to
Numerical Answer. Answer: 6344 to 6344

Solution

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

We are given 10 observations: 65, 68, 58, 44, 48, 45, 60, α, β, 60. Total n=10.

Sum of known observations:

S = 65 + 68 + 58 + 44 + 48 + 45 + 60 + 60 = 448

The mean x = 56:

(448 + α + β)/(10) = 56 448 + α + β = 560 α + β = 112
Step 1: Use Variance Equation

The variance σ² = 66.2. Using the computational formula for variance:

(Σ xᵢ²)/(10) - (56)² = 66.2

Calculate the sum of squares of known observations:

Σ xknown² = 65² + 68² + 58² + 44² + 48² + 45² + 60² + 60² = 4225 + 4624 + 3364 + 1936 + 2304 + 2025 + 3600 + 3600

= 25678

Insert this into the variance equation:

(25678 + α² + β²)/(10) - 3136 = 66.2
Step 2: Solve for Squares Sum

Isolate α² + β²:

(25678 + α² + β²)/(10) = 3136 + 66.2 (25678 + α² + β²)/(10) = 3202.2 25678 + α² + β² = 32022 α² + β² = 32022 - 25678 α² + β² = 6344
Pattern Recognition

If a question asks solely for α²+β² given mean and variance, you do not need to solve the complex polynomial system to find the individual values of α and β. The variance equation isolates α²+β² automatically as a single chunk.

Chapter Mix

Class 11 Mathematics: Statistics

Q30 jee_main_2024_30_january_evening Variance
The variance σ² of the data
xᵢ0156101217
fᵢ3232633
is
Numerical Answer. Answer: 29 to 29

Solution

Related Formula
Mean ( x) = (Σ fᵢ xᵢ)/(Σ fᵢ) Variance (σ²) = (1)/(N) Σ fᵢ xᵢ² - ( x)²
Core Logic

Let's build the summation table for calculating mean and variance:

xᵢfᵢfᵢ xᵢfᵢ xᵢ²
0300
1222
531575
621272
10660600
12336432
17351867
Σ fᵢ = 22Σ fᵢ xᵢ = 176Σ fᵢ xᵢ² = 2048

Step 1: Calculating the Mean
x = (Σ fᵢ xᵢ)/(Σ fᵢ) = (176)/(22) = 8
Step 2: Calculating the Variance
σ² = (1)/(N) Σ fᵢ xᵢ² - ( x)² σ² = (1)/(22)(2048) - 8² σ² = 93.0909 - 64 σ² = 29.0909

Rounding to the nearest integer as indicated by the official answer key gives 29.

Pattern Recognition

For grouped discrete data, the computational formula (Σ fᵢ xᵢ²)/(N) - μ² minimizes subtraction errors compared to tracking raw deviations point-by-point.

Chapter Mix

Class 11 Maths: Statistics

Q16 jee_main_2024_30_jan_morning Measures of Central Tendency
Let M denote the median of the following frequency distribution.
Class0-44-88-1212-1616-20
Frequency391086
Then 20M is equal to:
  • A. 416
  • B. 104
  • C. 52
  • D. 208

Solution

Related Formula
M = l + ( ((N)/(2) - cf)/(f) ) × h
Core Logic

Constructing the Cumulative Frequency (CF) table:

ClassFrequencyCumulative frequency
0-433
4-8912
8-121022
12-16830
16-20636

Total frequency N = 36. Therefore, (N)/(2) = 18.

Step 1: Identifying median class

Since 18 lies in the cumulative frequency interval > 12 and ≤ 22, the median class is 8-12. Here, l = 8 (lower limit), cf = 12 (CF of previous class), f = 10 (frequency of current class), h = 4 (class size).

Step 2: Calculating median
M = 8 + ( (18 - 12)/(10) ) × 4 M = 8 + (6)/(10) × 4 M = 8 + 2.4 = 10.4

We need to find 20M:

20M = 20 × 10.4 = 208
Pattern Recognition

Finding the cumulative frequency sequence safely identifies the median class. Plugging into the standard linear interpolation formula yields the exact median.

Chapter Mix

Class 11 Maths: Statistics

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