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

Q13 jee_main_2026_21_jan_morning Mean, Variance and Selection Probability

Let the mean and variance of 7 observations 2, 4, 10, x, 12, 14, y,

x > y

, be 8 and 16 respectively. Two numbers are chosen from

1, 2, 3, x-4, y, 5

one after another without replacement, then the probability, that the smaller number among the two chosen numbers is less than 4, is:

A. $(3)/(5)$ $\frac{3}{5}$
B. $(4)/(5)$ $\frac{4}{5}$
C. $(2)/(5)$ $\frac{2}{5}$
D. $(1)/(3)$ $\frac{1}{3}$

Solution

Related Formula

Mean x = (Σ xᵢ)/(n)

Variance σ² = (Σ xᵢ²)/(n) - ( x)²

Core Logic

Given Mean $= 8$ $= 8$ for 7 observations: $2, 4, 10, x, 12, 14, y$ $2, 4, 10, x, 12, 14, y$.

(2 + 4 + 10 + x + 12 + 14 + y)/(7) = 8

\Rightarrow x + y + 42 = 56 \Rightarrow x + y = 14 (1)

Given Variance $= 16$ $= 16$:

16 = (2² + 4² + 10² + x² + 12² + 14² + y²)/(7) - 8²

16 + 64 = (4 + 16 + 100 + x² + 144 + 196 + y²)/(7)

80 \times 7 = 460 + x² + y² \Rightarrow 560 = 460 + x² + y² \Rightarrow x² + y² = 100 (2)

Step 1: Solve for x and y

Using algebraic identity $(x+y)² = x² + y² + 2xy$ $(x+y)^2 = x^2 + y^2 + 2xy$:

14² = 100 + 2xy \Rightarrow 196 - 100 = 2xy \Rightarrow 2xy = 96 \Rightarrow xy = 48

Since $x+y=14$ $x+y=14$ and $xy=48$ $xy=48$, roots of quadratic $t² - 14t + 48 = 0$ $t^2 - 14t + 48 = 0$ are $8, 6$ $8, 6$. Given $x > y$ $x > y$, we must select $x = 8$ $x = 8$ and $y = 6$ $y = 6$.

Step 2: Construct the set and evaluate Probability

The new set $X$ $X$ is formed by $1, 2, 3, x-4, y, 5$ $\{1, 2, 3, x-4, y, 5\}$. Substituting $x=8$ $x=8$ and $y=6$ $y=6$, we get $1, 2, 3, 4, 6, 5$ $\{1, 2, 3, 4, 6, 5\}$. There are 6 distinct elements: $1, 2, 3, 4, 5, 6$ $\{1, 2, 3, 4, 5, 6\}$.

We choose two numbers without replacement. Total outcomes = $6 \times 5 = 30$ $6 \times 5 = 30$ permutations (or $62 = 15$ $\binom{6}{2} = 15$ combinations). Let's use combinations. Total ways to choose 2 numbers = $62 = 15$ $\binom{6}{2} = 15$.

We need the probability that the smaller number is less than $4$ $4$. $P(smaller < 4) = 1 - P(smaller \geq 4)$ .

Step 3: Final Calculation via Complement

For the smaller number to be $\geq 4$ $\geq 4$, BOTH chosen numbers must be $\geq 4$ $\geq 4$. The available numbers $\geq 4$ $\geq 4$ in the set are $4, 5, 6$ $\{4, 5, 6\}$ (Total 3 numbers). Ways to choose two numbers from these 3 is $32 = 3$ $\binom{3}{2} = 3$.

P(smaller \geq 4) = (3)/(15) = (1)/(5)

P(smaller < 4) = 1 - (1)/(5) = (4)/(5)

Pattern Recognition

When statistical problems ask for "at least one" or "minimum bounding", calculating the complement probability (e.g., both elements strictly exceeding the threshold) cuts combinatorial checks from $3+$ $3+$ cases down to exactly $1$ $1$.

Chapter Mix

Class 11 Maths: Statistics Class 12 Maths: Probability

Q2 jee_main_2026_22_january_evening Mean Deviation about Median

If the mean deviation about the median of the numbers

k, 2k, 3k, , 1000k

is

500

, then

k²

is equal to:

A. 16
B. 4
C. 1
D. 9

Solution

Related Formula

Mean deviation about median is given by:

M.D. = Σi=1ⁿ |Xᵢ - XM|n

Core Logic

For $1000$ $1000$ numbers in A.P., median $XM = (1001k)/(2)$ $X_M = \frac{1001k}{2}$.

M.D. = 2 ((k)/(2) + (3k)/(2) + (5k)/(2) + + 500 terms)1000

M.D. = (2 \cdot (k)/(2) (500)²)/(1000) = (500k)/(2)

Step 1: Evaluation of k and k^2

Given mean deviation is $500$ $500$:

(500k)/(2) = 500 k = 2

Therefore, $k² = 4$ $k^2 = 4$.

Pattern Recognition

Use symmetry of A.P. around median to quickly evaluate absolute deviation sum.

Chapter Mix

Class 11 Maths: Statistics

Q13 jee_main_2026_23_january_morning Mean and Variance

Let the mean and variance of

8

numbers

-10, -7, -1, x, y, 9, 2, 16

be

(7)/(2)

and

(293)/(4)

, respectively. Then the mean of

4

numbers

x, y, x + y + 1

,

|x - y|

is:

A. $11$ $11$
B. $9$ $9$
C. $10$ $10$
D. $12$ $12$

Solution

Related Formula

μ = (Σ xᵢ)/(N)

σ² = (Σ xᵢ²)/(N) - μ²

Core Logic

Calculate $x+y$ $x+y$ using the mean formula:

(-10 - 7 - 1 + x + y + 9 + 2 + 16)/(8) = (7)/(2)

(9 + x + y)/(8) = (7)/(2) \Rightarrow 9 + x + y = 28 \Rightarrow x + y = 19 (1)

Step 1: Calculate sum of squares using Variance

Using the variance formula:

((-10)² + (-7)² + (-1)² + x² + y² + 9² + 2² + 16²)/(8) - ((7)/(2))² = (293)/(4)

(100 + 49 + 1 + x² + y² + 81 + 4 + 256)/(8) - (49)/(4) = (293)/(4)

(491 + x² + y²)/(8) = (293)/(4) + (49)/(4) = (342)/(4) = (684)/(8)

491 + x² + y² = 684 \Rightarrow x² + y² = 193 (2)

Step 2: Solve for x and y

From (1), $y = 19 - x$ $y = 19 - x$. Substitute into (2):

x² + (19 - x)² = 193

x² + 361 - 38x + x² = 193

2x² - 38x + 168 = 0 \Rightarrow x² - 19x + 84 = 0

(x - 12)(x - 7) = 0

Thus, $x = 12$ $x = 12$ and $y = 7$ $y = 7$ (or vice-versa, which doesn't affect absolute differences).

Step 3: Calculate the New Mean

We need the mean of $4$ $4$ numbers: $x, y, x + y + 1, |x - y|$ $x, y, x + y + 1, |x - y|$. Substitute $x = 12, y = 7$ $x = 12, y = 7$: The numbers are $12, 7, (12+7+1), |12-7| = 12, 7, 20, 5$ . New Mean = $(12 + 7 + 20 + 5)/(4) = (44)/(4) = 11$ .

Pattern Recognition

Statistics questions mapping $x+y$ $x+y$ and $x²+y²$ $x^2+y^2$ inherently hide a quadratic symmetric system $(x+y)² - 2xy = x²+y²$ $(x+y)^2 - 2xy = x^2+y^2$. Solving for the roots immediately grants the discrete terms.

Chapter Mix

Class 11 Maths: Statistics

Q18 jee_main_2026_23_january_evening Mean and Variance

If the mean and the variance of the data

Class	4–8	8–12	12–16	16–20
Frequency	3	$λ$ $\lambda$	4	7

are

μ

and 19 respectively, then the value of

λ + μ

is

A. $18$ $18$
B. $21$ $21$
C. $20$ $20$
D. $19$ $19$

Solution

Related Formula

μ = (Σ fᵢ xᵢ)/(Σ fᵢ)

σ² = (Σ fᵢ xᵢ²)/(Σ fᵢ) - μ²

Core Logic

Midpoints ( $xᵢ$ $x_i$): $6, 10, 14, 18$ $6, 10, 14, 18$. Frequencies ( $fᵢ$ $f_i$): $3, λ, 4, 7$ $3, \lambda, 4, 7$. Sum of frequencies $Σ fᵢ = 3 + λ + 4 + 7 = 14 + λ$ .

Calculate Mean ( $μ$ $\mu$):

μ = (6(3) + 10(λ) + 14(4) + 18(7))/(14 + λ)

μ = (18 + 10λ + 56 + 126)/(14 + λ) = (10λ + 200)/(λ + 14) = 10 + (60)/(λ + 14)

Since data is standard, $μ$ $\mu$ and $λ$ $\lambda$ typically hold integer values, hinting $λ + 14$ $\lambda + 14$ must be a factor of 60. Therefore $λ$ $\lambda$ could be $1, 6, 16$ $1, 6, 16$.

Step 1: Applying Variance

Variance $σ² = 19$ $\sigma^2 = 19$.

σ² = (Σ fᵢ xᵢ²)/(14 + λ) - μ²

Σ fᵢ xᵢ² = 3(36) + λ(100) + 4(196) + 7(324) = 108 + 100λ + 784 + 2268 = 100λ + 3160

Substitute into variance:

19 = (100λ + 3160)/(λ + 14) - ((10λ + 200)/(λ + 14))²

Step 2: Testing Candidates

Let's test the potential integer $λ = 6$ $\lambda = 6$ (a factor of 60 is $14+6=20$ $14+6=20$): If $λ = 6$ $\lambda = 6$:

μ = 10 + (60)/(20) = 10 + 3 = 13

Let's verify variance for $λ = 6$ $\lambda = 6$:

σ² = (100(6) + 3160)/(20) - 13² = (3760)/(20) - 169 = 188 - 169 = 19

This perfectly matches the given variance.

Thus, $λ = 6$ $\lambda = 6$ and $μ = 13$ $\mu = 13$.

λ + μ = 6 + 13 = 19

Pattern Recognition

Instead of solving a horrible cubic/quadratic equation for the variance parameter, use the integer division trait of the mean to test valid candidates.

Chapter Mix

Class 11 Maths: Statistics

Q18 jee_main_2026_24_january_morning Variance and Mean Replacement

The mean and variance of a data of

10

observations are

10

and

2

, respectively. If an observation

α

in this data is replaced by

β

, then the mean and variance become

10.1

and

1.99

, respectively. Then

α + β

equals.

A. $10$ $10$
B. $15$ $15$
C. $5$ $5$
D. $20$ $20$

Solution

Related Formula

Mean x = (Σ xᵢ)/(n)

Variance σ² = (Σ xᵢ²)/(n) - x²

Core Logic

Let the 9 unchanged numbers be $x₁, x₂, , x₉$ $x_1, x_2, \dots, x_9$. Case 1 (with $α$ $\alpha$):

Σi=1⁹ xᵢ + α = 10 \times 10 = 100 \Rightarrow Σi=1⁹ xᵢ = 100 - α

Σi=1⁹ xᵢ² + α²10 - (10)² = 2 \Rightarrow Σi=1⁹ xᵢ² + α² = 1020 \Rightarrow Σi=1⁹ xᵢ² = 1020 - α²

Step 1: Applying Replacement Conditions

Case 2 (with $β$ $\beta$): Mean = 10.1

Σi=1⁹ xᵢ + β10 = 10.1 \Rightarrow 100 - α + β = 101 \Rightarrow β - α = 1

Variance = 1.99

Σi=1⁹ xᵢ² + β²10 - (10.1)² = 1.99

Σi=1⁹ xᵢ² + β² = 10(1.99 + 102.01) = 10(104) = 1040

Step 2: Solving for Alpha and Beta

Substitute $Σi=1⁹ xᵢ² = 1020 - α²$ $\sum_{i=1}^9 x_i^2 = 1020 - \alpha^2$:

1020 - α² + β² = 1040 \Rightarrow β² - α² = 20

(β - α)(β + α) = 20

Since $β - α = 1$ $\beta - \alpha = 1$:

(1)(α + β) = 20 \Rightarrow α + β = 20

(Solving gives $α = 19/2, β = 21/2$ $\alpha = 19/2, \beta = 21/2$).

Pattern Recognition

When a single term is replaced, immediately establish the difference of sums and difference of sum-of-squares. The identity $β² - α²$ $\beta^2 - \alpha^2$ gracefully factors to use the established $β - α$ $\beta - \alpha$ value.

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

Solution

Related Formula

Core Logic

Step 1: Solve for x and y

Step 2: Construct the set and evaluate Probability

Step 3: Final Calculation via Complement

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluation of k and k^2

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate sum of squares using Variance

Step 2: Solve for x and y

Step 3: Calculate the New Mean

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Applying Variance

Step 2: Testing Candidates

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Applying Replacement Conditions

Step 2: Solving for Alpha and Beta

Pattern Recognition

Chapter Mix

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