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

More Statistics and Probability Previous-Year Questions — Page 2

Q14 jee_main_2026_24_january_evening Transformation of Mean and Variance

Let

X = x in N : 1 \leq x \leq 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

A. $20$ $20$
B. $80$ $80$
C. $100$ $100$
D. $60$ $60$

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 $X = 1, 2, , 19$ $X = \{1, 2, \dots, 19\}$, we find the mean $x$ $\bar{x}$ and variance $σₓ²$ $\sigma_x^2$.

x = (1 + 2 + + 19)/(19) = ((19 \times 20)/(2))/(19) = 10

Variance(X) = (19² - 1)/(12) = (361 - 1)/(12) = 30

Step 1: Creating System of Equations

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

Using transformations:

y = a x + b 30 = 10a + b (1)

Variance(Y) = a² \cdot Variance(X) 750 = a² \times 30

Step 2: Solving for a and b

From the variance equation:

a² = (750)/(30) = 25 a = \pm 5

Now find the corresponding values of $b$ $b$ 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 $b$ $b$ are $-20$ $-20$ and $80$ $80$.

Sum $= -20 + 80 = 60$ $= -20 + 80 = 60$

Pattern Recognition

When a dataset undergoes linear transformation $Y = aX + b$ $Y = aX + b$, the variance isolates $a²$ $a^2$ entirely decoupled from $b$ $b$. Solve variance first to get $a$ $a$, 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. $5$ $5$
B. $4$ $4$
C. $6$ $6$
D. $7$ $7$

Solution

Related Formula

Mean $x = (Σ xᵢ)/(n)$ $\bar{x} = \frac{\sum x_i}{n}$ Variance $σ² = (Σ xᵢ²)/(n) - ( x)²$ Mean Deviation about Median = $(Σ |xᵢ - M|)/(n)$ $\frac{\sum |x_i - M|}{n}$ where $M$ $M$ is the median.

Core Logic

Let the two missing observations be $a$ $a$ and $b$ $b$. 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

$a² + b² = 58$ $a^2 + b^2 = 58$

We know $a+b = 10 b = 10-a$ $a+b = 10 \implies b = 10-a$.

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: $2, 3, 3, 5, 7, 10, 11, 13, 15, 21$ $2, 3, 3, 5, 7, 10, 11, 13, 15, 21$ Since there are 10 observations, median $M$ $M$ 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 $|xᵢ - M|$ $|x_i - M|$: $|2-8.5|=6.5$ $|2-8.5|=6.5$ $|3-8.5|=5.5$ $|3-8.5|=5.5$ $|3-8.5|=5.5$ $|3-8.5|=5.5$ $|5-8.5|=3.5$ $|5-8.5|=3.5$ $|7-8.5|=1.5$ $|7-8.5|=1.5$ $|10-8.5|=1.5$ $|10-8.5|=1.5$ $|11-8.5|=2.5$ $|11-8.5|=2.5$ $|13-8.5|=4.5$ $|13-8.5|=4.5$ $|15-8.5|=6.5$ $|15-8.5|=6.5$ $|21-8.5|=12.5$ $|21-8.5|=12.5$

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 $a+b$ $a+b$ and $ab$ $ab$.

Step 1: Apply the Mean condition

Given mean $x = 9$ $\bar{x} = 9$ for $N=8$ $N=8$ 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 $σ² = 9.25 = (37)/(4)$ $\sigma^2 = 9.25 = \frac{37}{4}$:

(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 $(a+b)² = a² + b² + 2ab$ $(a+b)^2 = a^2 + b^2 + 2ab$. 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 $a+b+ab$ $a+b+ab$ or $a²+b²$ $a^2+b^2$. These can be calculated using quadratic expansions without solving for $a$ $a$ and $b$ $b$ individually.

Chapter Mix

Class 11 Mathematics: Statistics

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 \geq 3)

is equal to

A. $(7)/(27)$ $\frac{7}{27}$
B. $(4)/(9)$ $\frac{4}{9}$
C. $(8)/(27)$ $\frac{8}{27}$
D. $(1)/(9)$ $\frac{1}{9}$

Solution

Related Formula

Sum of all probabilities in a distribution:

Σx=0\infty P(X = x) = 1

Complementary probability:

P(X \geq 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]

Core Logic

Let's first determine the constant $k$ $k$:

k Σx=0\infty (x+1) 3-x = 1

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

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

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

Step 1: Finding $k$ $k$

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 \cdot ((9)/(4)) = 1 k = (4)/(9)

Step 2: Calculating $P(X \geq 3)$ $P(X \ge 3)$

Calculate initial probability values:

$P(X=0) = k(1)(1) = (4)/(9)$ $P(X=0) = k(1)(1) = \frac{4}{9}$
$P(X=1) = k(2)((1)/(3)) = (2)/(3) \cdot (4)/(9) = (8)/(27)$
$P(X=2) = k(3)((1)/(9)) = (1)/(3) \cdot (4)/(9) = (4)/(27)$

Sum P(X < 3) = (12)/(27) + (8)/(27) + (4)/(27) = (24)/(27) = (8)/(9)

P(X \geq 3) = 1 - (8)/(9) = (1)/(9)

Pattern Recognition

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

Chapter Mix

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

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 2

Solution

Related Formula

Core Logic

Step 1: Creating System of Equations

Step 2: Solving for a and b

Step 3: Finding Sum of b Values

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve for a and b

Step 2: Find the Median

Step 3: Calculate Mean Deviation about Median

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Apply the Mean condition

Step 2: Apply the Variance condition

Step 3: Solve for ab and calculate the target expression

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding k$k$

Step 2: Calculating P(X ≥ 3)$P(X \ge 3)$

Pattern Recognition

Chapter Mix

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

Step 1: Finding $k$ $k$

Step 2: Calculating $P(X \geq 3)$ $P(X \ge 3)$