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

Let the five observations be x1, x2, x3, x4, x5. Given total mean: Given mean of first four observations: Substituting this back, we find the fifth observation: Using the variance of the 5 observations: 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. Class 11 Mathematics: Statistics

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

Let the five observations be x1, x2, x3, x4, x5. Given total mean: Given mean of first four observations: Substituting this back, we find the fifth observation: Using the variance of the 5 observations: 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. Class 11 Mathematics: Statistics

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

Let the five observations be x1, x2, x3, x4, x5. Given total mean: Given mean of first four observations: Substituting this back, we find the fifth observation: Using the variance of the 5 observations: 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. Class 11 Mathematics: Statistics

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

Let the five observations be x1, x2, x3, x4, x5. Given total mean: Given mean of first four observations: Substituting this back, we find the fifth observation: Using the variance of the 5 observations: 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. Class 11 Mathematics: Statistics

If the mean and variance of five observations are (24/5) and (194/25) respectively and the mean of first four…

Year	2026	2025	2024	Total
Questions	7	7	7	21

More Statistics Previous-Year Questions — Page 4

Q14 jee_main_2024_27_jan_morning Standard Deviation

Let

a₁, a₂, ,a₁₀

be 10 observations such that

Σk=1¹⁰ak=50

and

Σk$\sum_{\forall k. Then the standard deviation of aᵢ$a_{i}$ is equal to :

A. $5$ $5$
B. $\sqrt(5)$ $\sqrt{5}$
C. $10$ $10$
D. $\sqrt(115)$ $\sqrt{115}$

Solution

Related Formula

σ = \sqrt((Σ aᵢ²)/(n) - ((Σ aᵢ)/(n))²)

(Σi=1ⁿ aᵢ)² = Σi=1ⁿ aᵢ² + 2 Σk < j ak aⱼ

Core Logic

Given: $Σ aᵢ = 50$ $\sum a_i = 50$ $Σk < j ak aⱼ = 1100$ $\sum_{k < j} a_k a_j = 1100$ Number of observations, $n = 10$ $n = 10$.

To find the standard deviation, we need the sum of squares, $Σ aᵢ²$ $\sum a_i^2$. We use the algebraic identity for the square of a sum of $n$ $n$ 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 $σ$ $\sigma$ is the square root of variance:

σ = \sqrt(5)

Pattern Recognition

Whenever you see pairwise products $Σ aᵢ aⱼ$ $\sum a_i a_j$ in a statistics problem, immediately bridge it to $Σ aᵢ²$ $\sum a_i^2$ 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 $n=10$ $n=10$.

Sum of known observations:

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

The mean $x = 56$ $\bar{x} = 56$:

(448 + α + β)/(10) = 56

448 + α + β = 560

α + β = 112

Step 1: Use Variance Equation

The variance $σ² = 66.2$ $\sigma^2 = 66.2$. 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

$= 25678$ $= 25678$

Insert this into the variance equation:

(25678 + α² + β²)/(10) - 3136 = 66.2

Step 2: Solve for Squares Sum

Isolate $α² + β²$ $\alpha^2 + \beta^2$:

(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 $α²+β²$ $\alpha^2+\beta^2$ given mean and variance, you do not need to solve the complex polynomial system to find the individual values of $α$ $\alpha$ and $β$ $\beta$. The variance equation isolates $α²+β²$ $\alpha^2+\beta^2$ 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

$xᵢ$ $x_i$	0	1	5	6	10	12	17
$fᵢ$ $f_i$	3	2	3	2	6	3	3

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:

$xᵢ$ $x_i$	$fᵢ$ $f_i$	$fᵢ xᵢ$ $f_i x_i$	$fᵢ xᵢ²$ $f_i x_i^2$
0	3	0	0
1	2	2	2
5	3	15	75
6	2	12	72
10	6	60	600
12	3	36	432
17	3	51	867
	$Σ fᵢ = 22$ $\Sigma f_i = 22$	$Σ fᵢ xᵢ = 176$ $\Sigma f_i x_i = 176$	$Σ fᵢ xᵢ² = 2048$ $\Sigma f_i x_i^2 = 2048$

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 $(Σ fᵢ xᵢ²)/(N) - μ²$ $\frac{\Sigma f_i x_i^2}{N} - \mu^2$ 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.

Class	0-4	4-8	8-12	12-16	16-20
Frequency	3	9	10	8	6

Then

20M

is equal to:

A. $416$ $416$
B. $104$ $104$
C. $52$ $52$
D. $208$ $208$

Solution

Related Formula

M = l + ( ((N)/(2) - cf)/(f) ) \times h

Core Logic

Constructing the Cumulative Frequency (CF) table:

Class	Frequency	Cumulative frequency
0-4	3	3
4-8	9	12
8-12	10	22
12-16	8	30
16-20	6	36

Total frequency $N = 36$ $N = 36$. Therefore, $(N)/(2) = 18$ $\frac{N}{2} = 18$.

Step 1: Identifying median class

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

Step 2: Calculating median

M = 8 + ( (18 - 12)/(10) ) \times 4

M = 8 + (6)/(10) \times 4

M = 8 + 2.4 = 10.4

We need to find $20M$ $20M$:

20M = 20 \times 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

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

Solution & Explanation

Related Formula

Core Logic

Step 1: Finding the Sum of Squares

Step 2: Variance of First Four Observations

Pattern Recognition

Chapter Mix

Reference Study Guides

More Statistics Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Finding the Sum of Squares

Step 2: Calculating Standard Deviation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Use Variance Equation

Step 2: Solve for Squares Sum

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating the Mean

Step 2: Calculating the Variance

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Identifying median class

Step 2: Calculating median

Pattern Recognition

Chapter Mix

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