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

Mean of N observations: Variance of N observations: Given mean is 5 for 5 observations: Given variance is 10: Using standard algebraic identity (a+b)2 = a2 + b2 + 2ab: Solving a+b=14 and ab=40: Since a > b, we obtain a = 10 and b = 4. 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: Variance of new set: 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. Class 11 Mathematics: Statistics and Probability

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

Mean of N observations: Variance of N observations: Given mean is 5 for 5 observations: Given variance is 10: Using standard algebraic identity (a+b)2 = a2 + b2 + 2ab: Solving a+b=14 and ab=40: Since a > b, we obtain a = 10 and b = 4. 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: Variance of new set: Note that adding a changing factor like

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

Mean of N observations: Variance of N observations: Given mean is 5 for 5 observations: Given variance is 10: Using standard algebraic identity (a+b)2 = a2 + b2 + 2ab: Solving a+b=14 and ab=40: Since a > b, we obtain a = 10 and b = 4. 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: Variance of new set: 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. Class 11 Mathematics: Statistics and Probability

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

Mean of N observations: Variance of N observations: Given mean is 5 for 5 observations: Given variance is 10: Using standard algebraic identity (a+b)2 = a2 + b2 + 2ab: Solving a+b=14 and ab=40: Since a > b, we obtain a = 10 and b = 4. 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: Variance of new set: Note that adding a changing factor like

Let the Mean and Variance of five observations x1 = 1, x2 = 3, x3 = a, x4 = 7 and x5 = b, a

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. $(4)/(5)$ $\frac{4}{5}$
B. $(77)/(12)$ $\frac{77}{12}$
C. $(5)/(4)$ $\frac{5}{4}$
D. $(105)/(4)$ $\frac{105}{4}$

Solution

Related Formula

Mean X = (Σ xᵢ)/(n)

Variance σ² = (Σ xᵢ²)/(n) - ( X)²

Core Logic

Let the five observations be $x₁, x₂, x₃, x₄, x₅$ $x_1, x_2, x_3, x_4, x_5$. 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 $x₅$ $x_5$, 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$\sum_{\forall k. Then the standard deviation of aᵢ$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

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:

$xᵢ$ $x_i$	$fᵢ$ $f_i$	$fᵢ xᵢ$ $f_i x_i$	$fᵢ xᵢ²$ $f_i x_i^2$
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

Let the Mean and Variance of five observations x₁ = 1, x₂ = 3, x₃ = a, x₄ = 7$x_1 = 1, x_2 = 3, x_3 = a, x_4 = 7$ and x₅ = b, a > b,$x_5 = b, a > b,$ be 5$5$ and 10$10$ respectively. Then the Variance of the observations n + xₙ, n = 1, 2, 5$n + x_n, n = 1, 2, \dots \dots 5$ is

Solution & Explanation

Related Formula

Core Logic

Step 1: Finding a$a$ and b$b$

Step 2: Constructing new set and finding variance

Pattern Recognition

Chapter Mix

Reference Study Guides

More Statistics and Probability Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Finding the Sum of Squares

Step 2: Variance of First Four Observations

Pattern Recognition

Chapter Mix

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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Step 1: Finding $a$ $a$ and $b$ $b$