JEE Main Previous Year Statistics Questions and Solutions

Q 11 :

Let $9 = x_{1} < x_{2} < \dots < x_{7}$ be in an A.P. with common difference $d$ . If the standard deviation of $x_{1}, x_{2}, \dots, x_{7}$ is $4$ and the mean is $\bar{x}$ , then $\bar{x} + x_{6}$ is equal to [2023]

$18 (1 + \frac{1}{\sqrt{3}})$
$2 (9 + \frac{8}{\sqrt{7}})$
$34$
$25$

1

2

3

4

(3)

$Mean = \bar{x} = \frac{\sum_{i = 1}^{7} x_{i}}{7} = \frac{\frac{7}{2} [2 a + 6 d]}{7} = a + 3 d = x_{4}$

$Variance = \frac{\sum_{i = 1}^{7} {(x_{i} - \bar{x})}^{2}}{7} = {(4)}^{2} \Rightarrow \frac{\sum_{i = 1}^{7} {(x_{i} - x_{4})}^{2}}{7} = 16$

$\Rightarrow \frac{{(3 d)}^{2} + {(2 d)}^{2} + d^{2} + 0 + d^{2} + {(2 d)}^{2} + {(3 d)}^{2}}{7} = 16$

$\Rightarrow \frac{28 d^{2}}{7} = 16 \Rightarrow d = 2 \Rightarrow \bar{x} = 9 + 3 (2) = 15$

$x_{6} = a + 5 d = 9 + 5 (2) = 19$ $∴$ $\bar{x} + x_{6} = 15 + 19 = 34$

Q 12 :

Let the six numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ be in A.P. and $a_{1} + a_{3} = 10$ . If the mean of these six numbers is $\frac{19}{2}$ and their variance is $σ^{2}$ , then $8 σ^{2}$ is equal to [2023]

105
210
200
220

1

2

3

4

(2)

$Given a_{1}, a_{2}, a_{3}, \dots, a_{6} are in A.P.$

$Sum of these terms in A.P. = \frac{6}{2} [2 a + 5 d] = \frac{19}{2} \times 6$

$\Rightarrow 2 a + 5 d = 19 ...(i)$

$Since a_{1} + a_{3} = 10 \Rightarrow a + a + 2 d = 10$

$\Rightarrow 2 a + 2 d = 10 ...(ii)$

$Solving (i) and (ii), we get a = 2, d = 3$

$So, 2, 5, 8, 11, 14, 17 are the terms.$

$Variance, σ^{2} = \frac{2^{2} + 5^{2} + 8^{2} + 11^{2} + 14^{2} + 17^{2}}{6} - {(\frac{19}{2})}^{2}$

$= \frac{699}{6} - \frac{361}{4} = \frac{466 - 361}{4} = \frac{105}{4}$ $\Rightarrow 8 σ^{2} = \frac{8 \times 105}{4} = 210$

Q 13 :

The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12. If the new mean of the marks is 10.2, then their new variance is equal to: [2023]

3.92
3.96
4.04
4.08

1

2

3

4

(2)

$Mean, μ = 10$

$Variance, σ^{2} = 4$

$Let the number of observations be n$

$Sum of observations \sum x_{i} = 10 n$

$When marks of one student increased from 8 to 12$

$\sum x_{i} = 10 n - 8 + 12 = 10 n + 4$

$New mean = \frac{10 n + 4}{n} = 10.2$ $\Rightarrow n = 20;$ $4 = \frac{\sum x_{i}^{2}}{20} - 10^{2}$

$Incorrect \sum x_{i}^{2} = 104 \times 20 = 2080$ $⇒Correct \sum x_{i}^{2} = 2080 - 8^{2} + 12^{2} = 2160$

$New σ^{2} = \frac{2160}{20} - 10 . 2^{2} = 3.96$

Q 14 :

Let $S$ be the set of all values of $a_{1}$ for which the mean deviation about the mean of 100 consecutive positive integers $a_{1}, a_{2}, a_{3}, \dots, a_{100}$ is $25$ . Then $S$ is [2023]

${99}$
$ϕ$
$N$
${9}$

1

2

3

4

(3)

$Let a_{1} be a natural number.$

$The 100 consecutive positive numbers are a_{1}, a_{1} + 1, a_{1} + 2, a_{1} + 3, \dots, a_{1} + 99$

$Mean (\bar{x}) = \frac{a_{1} + a_{1} + 1 + a_{1} + 2 + \dots + a_{1} + 99}{100} = a_{1} + \frac{99}{2}$

$Mean deviation about the mean = \frac{\sum_{i = 1}^{100} | x_{_{i}}^{2} - \bar{x} |}{100}$

$= \frac{\frac{99}{2} + \frac{97}{2} + \frac{95}{2} + \dots + \frac{99}{2}}{100} = 25$

$Hence, a_{1} is a natural number.$

Q 15 :

Let the mean and variance of 8 numbers $x, y, 10, 12, 6, 12, 4, 8$ be 9 and 9.25 respectively. If $x > y$ , then $3 x - 2 y$ is equal to _____. [2023]

(25)

We have, $x, y, 10, 12, 6, 12, 4, 8$

Mean, $\bar{x} = \frac{x + y + 10 + 12 + 6 + 12 + 4 + 8}{8} = 9$

$\Rightarrow x + y = 20 \dots (i)$

Now, Variance $= \frac{x^{2} + y^{2} + {(10)}^{2} + {(12)}^{2} + {(6)}^{2} + {(12)}^{2} + {(4)}^{2} + {(8)}^{2}}{8} - 81$

$\Rightarrow 9.25 = \frac{x^{2} + y^{2} + 504}{8} - 81$ $\Rightarrow x^{2} + y^{2} = 218 \dots (i i)$

Now, ${(x + y)}^{2} = x^{2} + y^{2} + 2 x y$

$\Rightarrow {(20)}^{2} = 218 + 2 x y$ $\Rightarrow 2 x y = 182$

Now, ${(x - y)}^{2} = 218 - 182 = 36$ $\Rightarrow x - y = \pm 6 \dots (i i i)$

From (i) and (iii), we get $x = 13, y = 7$ $(∵ x > y)$

So, $3 x - 2 y = 39 - 14 = 25$

Q 16 :

Let the mean of the data

$x$ 1 3 5 7 9

Frequency $(f)$ 4 24 28 $α$ 8

be 5. If $m$ and $σ^{2}$ are respectively the mean deviation about the mean and the variance of the data, then $\frac{3 α}{m + σ^{2}}$ is equal to ________ . [2023]

(8)

Given, $\bar{x} = 5$

$\Rightarrow \frac{\sum x_{i} f_{i}}{\sum f_{i}} = \frac{4 + 72 + 140 + 7 α + 72}{64 + α} = 5$

$\Rightarrow 288 + 7 α = 320 + 5 α \Rightarrow 2 α = 32 \Rightarrow α = 16$

Now, $M.D. (\bar{x}) = \frac{\sum f_{i} | x_{i} - \bar{x} |}{\sum f_{i}}$

$= \frac{4 \times 4 + 24 \times 2 + 28 \times 0 + 16 \times 2 + 8 \times 4}{80} = \frac{128}{80} = \frac{8}{5} \Rightarrow m = \frac{8}{5}$

Variance $(σ^{2}) = \frac{\sum f_{i} | x_{i} - \bar{x} |^{2}}{\sum f_{i}}$

$= \frac{4 \times 16 + 24 \times 4 + 28 \times 0 + 16 \times 4 + 8 \times 16}{80} = \frac{352}{80} = \frac{22}{5}$

$∴$ $\frac{3 α}{m + σ^{2}} = \frac{3 \times 16}{\frac{8}{5} + \frac{22}{5}} = \frac{48}{6} = 8$

Q 17 :

The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is __________. [2023]

(269)

Given, Mean $\bar{X} = 50, σ = 12$

$\bar{X} = 50 - \frac{50}{10} = 45$

$\frac{\sum x_{i}^{2}}{10} - 50^{2} = 144$ $[σ^{2} = \frac{\sum x_{i}^{2}}{n} - {(\bar{x})}^{2}]$

$\frac{\sum x_{i}^{2}}{10} = 2500 + 144$

$\frac{\sum x_{i}^{2}}{10} = 2500 + 144 + \frac{400 + 625 - 45^{2} - 50^{2}}{10} = 2294$

$Now correct variance: σ_{T}^{2} = 2294 - {(45)}^{2} = 2294 - 2025 = 269$

Q 18 :

Let $X = {11, 12, 13, \dots, 40, 41}$ and $Y = {61, 62, 63, \dots, 90, 91}$ be the two sets of observations. If $\bar{x}$ and $\bar{y}$ are their respective means and $σ^{2}$ is the variance of all the observations in $X \cup Y$ , then $| \bar{x} + \bar{y} - σ^{2} |$ is equal to _______ . [2023]

(603)

$Mean \bar{x} = \frac{\sum_{i = 11}^{41} i}{31} = \frac{11 + 41}{2} = 26$

$\bar{y} = \frac{\sum_{j = 61}^{91} j}{31} = \frac{61 + 91}{2} = 76$

$Combined mean μ = \frac{31 \times 26 + 31 \times 76}{31 + 31} = \frac{26 + 76}{2} = 51$

$σ^{2} = \frac{1}{62} (\sum_{i = 1}^{31} {(x_{i} - μ)}^{2} + \sum_{i = 1}^{31} {(y_{i} - μ)}^{2}) = 705$

$∴$ $| \bar{x} + \bar{y} - σ^{2} |$ $=$ $| 26 + 76 - 705 |$ $= 603$

Q 19 :

The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted and $a$ and $b$ are respectively mean and variance of remaining 6 observations, then $a + 3 b - 5$ is equal to ______. [2023]

(37)

$Mean = 7, Variance = 16$

Let the observations be $x_{1}, x_{2}, \dots, x_{7}$ .

Now, $\frac{x_{1} + x_{2} + \dots + x_{7}}{7} = 8,$ $x_{1} + x_{2} + \dots + x_{7} = 56$

When one observation $14$ is omitted, the mean is

$a = \frac{x_{1} + x_{2} + \dots + x_{7} - 14}{6} \Rightarrow a = \frac{56 - 14}{6} = 7$

$Variance = \frac{\sum_{i = 1}^{7} {(x_{i})}^{2}}{7} - 64 = 16 \Rightarrow \sum_{i = 1}^{7} {(x_{i})}^{2} = 560$

When one observation is omitted, then variance

$b = \frac{\sum_{i = 1}^{6} {(x_{i})}^{2}}{6} - 49 \Rightarrow b = \frac{560 - {(14)}^{2}}{6} - 49 = \frac{364}{6} - 49 = \frac{35}{3}$

So, $a + 3 b - 5 = 7 + 3 \times \frac{35}{3} - 5 = 37$

Q 20 :

If the mean and variance of the frequency distribution

$x_{i}$ 2 4 6 8 10 12 14 16

$f_{i}$ 4 4 $α$ 15 8 $β$ 4 5

are 9 and 15.08 respectively, then the value of $α^{2} + β^{2} - α β$ is ________ . [2023]

(25)

$x_{i}$	$f_{i}$	$x_{i} f_{i}$	$f_{i} x_{i}^{2}$
2	4	8	16
4	4	16	64
6	$α$	$6 α$	$36 α$
8	15	120	960
10	8	80	800
12	$β$	$12 β$	$144 β$
14	$4$	56	784
16	5	80	1280
Total	$40 + α + β$	$360 + 6 α + 12 β$	$3904 + 36 α + 144 β$

$Mean (\bar{x}) = \frac{\sum f_{i} x_{i}}{\sum f_{i}} \Rightarrow \frac{360 + 6 α + 12 β}{40 + α + β} = 9$

$\Rightarrow 360 + 6 α + 12 β = 360 + 9 α + 9 β$

$\Rightarrow 3 α = 3 β \Rightarrow α = β$ ... $(i)$

$Also, variance σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{\sum f_{i}} - {(\frac{\sum f_{i} x_{i}}{\sum f_{i}})}^{2}$

$\Rightarrow \frac{3904 + 36 α + 144 β}{40 + α + β} - {(\bar{x})}^{2} = 15.08$

$\Rightarrow \frac{3904 + 180 α}{40 + 2 α} - {(9)}^{2} = 15.08$ $\Rightarrow α = 5$

From (i), $β = 5$

$∴$ $α^{2} + β^{2} - α β = 25 + 25 - 25 = 25$

Q 21 :

If the mean of the frequency distribution

Class : 0–10 10–20 20–30 30–40 40–50

Frequency : 2 3 $x$ 5 4

is 28, then its variance is _____ . [2023]

(151)

$x_{i}$	$f_{i}$	$x_{i} \cdot f_{i}$
5	2	10
15	3	45
25	$x$	$25 x$
35	5	175
45	4	180
	$n = 14 + x$	$\sum x_{i} f_{i} = 410 + 25 x$

$Mean, \bar{x} = \frac{410 + 25 x}{14 + x}$ $\Rightarrow 28 = \frac{410 + 25 x}{14 + x}$

$\Rightarrow 392 + 28 x = 410 + 25 x \Rightarrow 3 x = 18$

$\Rightarrow x = 6$

$∴$ $n = 20$

$Variance (σ^{2}) = \frac{\sum f_{i} x_{i}^{2}}{n} - {(28)}^{2}$

$= \frac{2 \times 5^{2} + 3 \times 15^{2} + 6 \times 25^{2} + 5 \times 35^{2} + 4 \times 45^{2}}{20} - {(28)}^{2}$

$= \frac{(2 \times 25) + (3 \times 225) + (6 \times 625) + (5 \times 1225) + (4 \times 2025)}{20} - 784$

$= 935 - 784 = 151$

Q 22 :

Let the positive numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $a_{5}$ be in a G.P. Let their mean and variance be $\frac{31}{10}$ and $\frac{m}{n}$ respectively, where $m$ and $n$ are co-prime. If the mean of their reciprocals is $\frac{31}{40}$ and $a_{3} + a_{4} + a_{5} = 14$ , then $m + n$ is equal to _______ . [2023]

(211)

$Let \frac{a}{r^{2}}, \frac{a}{r}, a, a r, a r^{2} be positive numbers in G.P.$

$\Rightarrow \frac{a}{r^{2}} + \frac{a}{r} + a + a r + a r^{2} = 5 \times \frac{31}{10}$ ...(i)

$and$ $\frac{r^{2}}{a} + \frac{r}{a} + \frac{1}{a} + \frac{1}{a r} + \frac{1}{a r^{2}} = 5 \times \frac{31}{40}$ ...(ii)

$Divide (i) by (ii), we get$

$\frac{a (\frac{1}{r^{2}} + \frac{1}{r} + 1 + r + r^{2})}{\frac{1}{a} (r^{2} + r + 1 + \frac{1}{r} + \frac{1}{r^{2}})} = \frac{5 \times \frac{31}{10}}{5 \times \frac{31}{40}}$

$\Rightarrow a^{2} = 4$

$\Rightarrow a = 2$ $(∵ a cannot be negative)$

$From (i), a (\frac{1}{r^{2}} + \frac{1}{r} + 1 + r + r^{2}) = 5 \times \frac{31}{10}$

$\Rightarrow {(r + \frac{1}{r})}^{2} + (r + \frac{1}{r}) = \frac{31}{4} + 1$

$\Rightarrow 4 t^{2} + 4 t - 35 = 0$ $[Let, r + \frac{1}{r} = t]$

$\Rightarrow t = \frac{5}{2} \Rightarrow r = 2$

$∴$ $Numbers are = \frac{1}{2}, 1, 2, 4, 8$

$∴$ $σ^{2} = \frac{\sum x^{2}}{N} - {(\frac{\sum x}{N})}^{2} = \frac{\frac{1}{4} + 1 + 4 + 16 + 64}{5} - {(\frac{31}{10})}^{2}$

$= \frac{341}{20} - \frac{961}{100} = \frac{186}{25}$ $∴$ $m + n = 186 + 25 = 211$

Q 23 :

If the variance of the frequency distribution

$x_{𝑖}$ 2 3 4 5 6 7 8

Frequency $f_{i}$ 3 6 16 $α$ 9 5 6

is 3, then $α$ is equal to _______. [2023]

(5)

$x_{i}$	$f_{i}$	$d_{i} = x_{i} - 5$	$d_{i}^{2}$	$f_{i} d_{i}^{2}$	$f_{i} d_{i}$
2	3	- 3	9	27	- 9
3	6	- 2	4	24	- 12
4	16	- 1	1	16	- 16
5	$α$	0	0	0	0
6	9	1	1	9	9
7	5	2	4	20	10
8	6	3	9	54	18
Total	$45 + α$			150	0

$σ^{2} = \frac{\sum f_{i} d_{i}^{2}}{\sum f_{i}} - {(\frac{\sum f_{i} d_{i}}{\sum f_{i}})}^{2} = \frac{150}{45 + α} - {(\frac{0}{45 + α})}^{2}$

$\Rightarrow 3 = \frac{150}{45 + α} (∵ variance = 3)$

$\Rightarrow 135 + 3 α = 150 \Rightarrow 3 α = 15 \Rightarrow α = 5$

Q 24 :

Let the mean and variance of 8 numbers $- 10, - 7, - 1, x, y, 9, 2, 16$ be $\frac{7}{2}$ and $\frac{293}{4}$ , respectively. Then the mean of 4 numbers $x, y, x + y + 1,$ $| x - y |$ is: [2026]

9
12
10
11

1

2

3

4

(4)

$Mean = \frac{- 18 + x + y + 2 + 9 + 16}{8} = \frac{7}{2}$

$\Rightarrow \frac{x + y + 9}{8} = \frac{7}{2} \Rightarrow x + y + 9 = 28 . . . . . (1)$

$Variance = \frac{\sum z_{i}^{2}}{8} - μ^{2} = \frac{293}{4}$

$\Rightarrow \frac{10^{2} + 7^{2} + 1^{2} + x^{2} + y^{2} + 2^{2} + 9^{2} + 16^{2}}{8} - {(\frac{7}{2})}^{2} = \frac{293}{4} . . . . (2)$

$Solving (1) & (2) \Rightarrow x = 12, y = 7$

$Mean of (1 + x + y), x, y,$ $| y - x |$ $is$

$\Rightarrow \frac{20 + 12 + 7 + 5}{4} = \frac{44}{4} = 11$

$Option (4)$

Q 25 :

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: [2026]

$\frac{4}{5}$
$\frac{1}{3}$
$\frac{3}{5}$
$\frac{2}{5}$

1

2

3

4

(1)

$Mean (\bar{x}) = 8 (Given)$

$\Rightarrow \frac{2 + 4 + 10 + x + 12 + 14 + y}{7} = 8$

$\Rightarrow x + y = 14 . . . (1)$

$Variance (σ^{2}) = 16 (Given)$

$\Rightarrow 16 = \frac{2^{2} + 4^{2} + 10^{2} + x^{2} + 12^{2} + 14^{2} + y^{2}}{7} - 8^{2}$

$\Rightarrow x^{2} + y^{2} = 100 . . . (2)$

$∵$ ${(x + y)}^{2} = x^{2} + y^{2} + 2 x y$

$\Rightarrow x y = 48 (sum is 14, product is 48)$

$Since problem states x > y$

$∴$ $x = 8 and y = 6$

$Now set X = {1, 2, 3, 4, 6, 5}$

Now we choose two numbers one after another without replacement

Total outcomes $= 6 \times 5 = 30$

We want the probability that the smaller number among the two is less than 4

$P (smaller < 4) = 1 - P (smaller \geq 4)$

$= 1 - \frac{6}{30} = \frac{4}{5}$

Q 26 :

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: [2026]

5
15
10
20

1

2

3

4

(4)

$Let first 10 numbers are x_{1}, x_{2}, \dots, x_{9}, α$

$\Rightarrow α + \sum_{i = 1}^{9} x_{i} = 100 \Rightarrow \sum_{i = 1}^{9} x_{i} = 100 - α$

$Variance = (\frac{\sum x_{i}^{2}}{n}) - {(\frac{\sum x_{i}}{n})}^{2}$

$\Rightarrow \frac{\sum x_{i}^{2}}{n} = 98$

$\Rightarrow x_{1}^{2} + x_{2}^{2} + \dots + x_{9}^{2} + α^{2} = 1020 \Rightarrow \sum x_{i}^{2} = 1020 - α^{2}$

$In second case, let numbers are x_{1}, x_{2}, \dots, x_{9}, β$

$100 - α + β = 101 \Rightarrow α - β + 1 = 0$

$\frac{\sum x_{i}^{2} + β^{2}}{10} - {(10.1)}^{2} = 1.99$

$\Rightarrow β^{2} - α^{2} = 20$

$α = \frac{19}{2}, β = \frac{21}{2}$

$α + β = \frac{19 + 21}{2} = 20$

Q 27 :

Let $X = {x \in ℕ : 1 \leq x \leq 19}$ and for some $a, b \in ℝ, Y = {a x + b : x \in X} .$ If the mean and variance of the elements of Y are 30 and 750, respectively, then the sum of all possible values of $b$ is [2026]

100
20
80
60

1

2

3

4

(4)

$\sum y_{i} = a \sum x_{i} + \sum b$

$= a (1 + 2 + \dots + 19) + 19 b$

$\frac{\sum y_{i}}{19} = \frac{a \cdot 19 \cdot 20}{2 \cdot 19} + b$

$30 = 10 a + b . . . (1)$

$Variance of X = \frac{\sum x_{i}^{2}}{19} - {(\frac{\sum x_{i}}{19})}^{2}$

$= \frac{19 \cdot 20 \cdot 39}{19 \cdot 6} - {(10)}^{2} = 30$

$Variance of Y = a^{2} (variance of X)$

$750 = a^{2} \cdot 30$

$a^{2} = 25 \Rightarrow a = \pm 5$

$if a = + 5 \Rightarrow b = 30 - 50 = - 20 from (1)$

$if a = - 5 \Rightarrow b = 30 + 50 = 80 from (1)$

$sum of values of b = 80 - 20 = 60$

$option (4)$

Q 28 :

If the mean and the variance of the data

Class

4–8

8–12

12–16

16–20

Frequency

3

$λ$

4

7

are $μ and 19$ respectively, then the value of $λ + μ$ is. [2026]

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

18
20
21
19

1

2

3

4

(4)

$μ = \frac{\sum f_{i} x_{i}}{\sum f_{i}} = \frac{18 + 10 λ + 56 + 126}{14 + λ}$

$= \frac{200 + 10 λ}{λ + 14} = 10 + (\frac{60}{λ + 14})$

$λ + 14 is a multiple of 60 \Rightarrow λ = 1, 6 or 16$

$σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{λ + 14} - μ^{2}$

$σ^{2} = \frac{6^{2} (3) + 10^{2} (λ) + 14^{2} (4) + 18^{2} (7)}{λ + 14} - μ^{2}$

$for λ = 6, μ = 10 + 3 = 13$

$λ + μ = 19$

Q 29 :

A random variable X takes values 0, 1, 2, 3 with probabilities $\frac{2 a + 1}{30}, \frac{8 a - 1}{30}, \frac{4 a + 1}{30}, b$ respectively,

where $a, b \in R$ . Let $μ and σ$ respectively be the mean and standard deviation of X such that $σ^{2} + μ^{2} = 2 .$

Then $\frac{a}{b}$ is equal to : [2026]

3
30
60
12

1

2

3

4

(3)

$x$	0	1	2	3
$p (x)$	$\frac{2 a + 1}{30}$	$\frac{8 a - 1}{30}$	$\frac{4 a + 1}{30}$	$b$

$σ^{2} = \sum x_{i}^{2} p (x_{i}) - μ^{2}$

$σ^{2} + μ^{2} = \sum x_{i}^{2} p (x_{i})$

$= 0 + 1 (\frac{8 a - 1}{30}) + 4 (\frac{4 a + 1}{30}) + 9 b$

$\Rightarrow \frac{24 a + 270 b + 3}{30} = 2$

$24 a + 270 b = 57$

$8 a + 90 b = 19 . . . (1)$

Also,

$\sum p (i) = 1$

$\frac{2 a + 1}{30} + \frac{8 a - 1}{30} + \frac{4 a + 1}{30} + b = 1$

$14 a + 30 b = 29 . . . (2)$

Solving (1) and (2),

$a = 2, b = \frac{1}{30}, \frac{a}{b} = 60$

Q 30 :

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 [2026]

6
7
4
5

1

2

3

4

(4)

$\frac{2 + 3 + 5 + 10 + 11 + 13 + 15 + 21 + a + b}{10} = 9$

$\frac{80 + a + b}{10} = 9 \Rightarrow a + b = 10 . . . (1)$

$\frac{\sum x_{i}^{2}}{10} - {(\frac{\sum x_{i}}{10})}^{2} = 34.2$

$\frac{2^{2} + 3^{2} + 5^{2} + 10^{2} + 11^{2} + 13^{2} + 15^{2} + 21^{2} + a^{2} + b^{2}}{10} - {(9)}^{2} = 34.2$

$\frac{1094 + a^{2} + b^{2}}{10} - 81 = 34.2$

$1094 + a^{2} + b^{2} - 810 = 342$

$a^{2} + b^{2} = 58 . . . (2)$

$a = 7, b = 3 or a = 3, b = 7$

$Numbers = 2, 3, 5, 7, 10, 11, 13, 15, 21$

$Mean = \frac{7 + 10}{2} = 8.5$

$M.D. = \frac{6.5 + 5.5 + 5.5 + 3.5 + 1.5 + 1.5 + 2.5 + 4.5 + 6.5 + 12.5}{10}$

$= \frac{50}{5}$

$= 5$

Q 31 :

The mean and standard deviation of the marks of 200 candidates were found to be 40 and 15 respectively. Later, it was discovered that one student's marks 40 was wrongly read as 50. The correct mean and standard deviation, respectively are

14.98, 39.95
39.95, 14.98
39.95, 224.5
39.00, 14.00

1

2

3

4

(2)

$Corrected \sum x = 40 \times 200 - 50 + 40 = 7990$

$∴$ $Corrected \bar{x} = \frac{7990}{200} = 39.95$

$Incorrect \sum x^{2} = n [σ^{2} + {\bar{x}}^{2}] = 200 [15^{2} + 40^{2}] = 365000$

$Correct \sum x^{2} = 365000 - 2500 + 1600 = 364100$

$∴$ $Corrected σ = \sqrt{\frac{364100}{200} - {(39.95)}^{2}} = \sqrt{1820.5 - 1596} = \sqrt{224.5} = 14.98$

Q 32 :

An online exam is attempted by 40 candidates out of which 15 are boys. The average marks obtained by boys is 10 with variance 2. The variance of marks obtained by 25 girls is also 2 and the average marks of all 40 candidates is 12.5. If $μ$ is average marks of the 25 girls & $σ^{2}$ is variance of marks of all 40 candidates then $20 σ^{2} - 8 μ$ = ______

(3)

$μ = \frac{40 \times 12.5 - 10 \times 15}{25} = 14$

$σ^{2} = 2 + \frac{15 \times 25}{40 \times 40} {(4)}^{2} = 5.75$

Q 33 :

Let $\bar{x}$ , M and $σ^{2}$ be respectively the mean, mode and variance of $n$ observations $x_{1}, x_{2}, \dots, x_{n}$ and $d_{i} = - x_{i} - a, i = 1, 2, \dots, n$ where $a$ is any number.

Statement 1: Variance of $d_{1}, d_{2}, \dots, d_{n}$ is $σ^{2}$ .

Statement 2: Mean and mode of $d_{1}, d_{2}, \dots, d_{n}$ are $- \bar{x} - a$ and $- M - a$ respectively.

Statement 1 and Statement 2 are both false
Statement 1 and Statement 2 are both true
Statement 1 is true and Statement 2 is false
Statement 1 is false and Statement 2 is true

1

2

3

4

(2)

$S_{1} \to V (a x + b) = a^{2} V (x)$

$S_{2} \to mean (a x \pm b) = a mean (X) \pm b$

$Mode (x_{1}, x_{2}, \dots, x_{n}) = K$

$Mode (a x_{1} \pm α, a x_{2} \pm α, \dots, a x_{n} \pm α) = a k \pm α$

$Central tendencies depend on shift of origin & scaling$

$Variance depends on scaling but not shift of origin$

Q 34 :

$\sum_{i = 1}^{10} (x_{i} - 5) = 5$ and $\sum_{i = 1}^{10} {(x_{i} - 5)}^{2} = 125$ , variance of $x_{1}, x_{2}, \dots, x_{10}$ will be

$\frac{9}{4}$
$\frac{25}{4}$
$\frac{49}{4}$
$\frac{81}{4}$

1

2

3

4

(3)

$Let x_{i} - 5 = d_{i} σ_{x}^{2}, σ_{d}^{2} = \frac{\sum d_{i}^{2}}{n} - {(\frac{\sum d_{i}}{n})}^{2}$

$= \frac{125}{10} - {(\frac{5}{10})}^{2}$

$= \frac{25}{2} - \frac{1}{4}$

$= \frac{49}{4}$

Q 35 :

The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is

$\sqrt{3.86}$
$1.8$
$\sqrt{3.96}$
$1.94$

1

2

3

4

(3)

$\frac{\sum x_{i}}{20} = 10, \sum x_{i} = 10 \times 20 = 200$

$If 8 replaced by 12 then \sum x_{i} = 200 - 8 + 12 = 204$

$Now, correct mean (\bar{x}) = \frac{\sum x_{i}}{20} = \frac{204}{20} = 10.2$

$Standard deviation = 2$

$So, variance = {(S . D)}^{2} = 2^{2} = 4$

$By definition,$ $\Rightarrow \frac{\sum x_{i}^{2}}{20} - {(\frac{\sum x_{i}}{20})}^{2} = 4 \Rightarrow \frac{\sum x_{i}^{2}}{20} - {(10)}^{2} = 4$

$\Rightarrow \sum x_{i}^{2} = 2080$

$\Rightarrow V_{a r} = \frac{\sum x_{i}^{2}}{20} - {(\frac{\sum x_{i}}{20})}^{2} = \frac{2160}{20} - {(10.2)}^{2}$

Q 36 :

The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and $σ^{2}$ respectively. If the variance of all the 30 numbers in the two sets is 13, then 3 $σ^{2}$ is equal to

33
36
30
27

1

2

3

4

(3)

$\sum x_{i} = 15 \times 12 and \frac{\sum x_{i}^{2}}{15} - 12^{2} = 14, \sum y_{i} = 15 \times 14 and \frac{\sum y_{i}^{2}}{15} - 14^{2} = σ$

$Now 13 = \frac{(14 + 144) \times 15 + (σ^{2} + 196) \times 15}{30} - 13^{2} \Rightarrow 3 σ^{2} = 30$

Q 37 :

If the standard deviation of $a_{1}, a_{2}, a_{3}, \dots, a_{2025}$ is $λ$ , then the value of standard deviation of $2 a_{1} - 1, 2 a_{2} - 1, 2 a_{3} - 1, \dots, 2 a_{2025} - 1$ is

$3 λ$
$2 λ$
$λ - 1$
$2 λ + 1$

1

2

3

4

(2)

As standard deviation is independent of change of origin

$∴$ It remains same $∴$ $S . D = 2 λ$

Topic Question Set

Q 11 :

Let $9 = x_{1} < x_{2} < \dots < x_{7}$ be in an A.P. with common difference $d$ . If the standard deviation of $x_{1}, x_{2}, \dots, x_{7}$ is $4$ and the mean is $\bar{x}$ , then $\bar{x} + x_{6}$ is equal to [2023]

Q 12 :

Let the six numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ be in A.P. and $a_{1} + a_{3} = 10$ . If the mean of these six numbers is $\frac{19}{2}$ and their variance is $σ^{2}$ , then $8 σ^{2}$ is equal to [2023]

Q 13 :

The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12. If the new mean of the marks is 10.2, then their new variance is equal to: [2023]

Q 14 :

Let $S$ be the set of all values of $a_{1}$ for which the mean deviation about the mean of 100 consecutive positive integers $a_{1}, a_{2}, a_{3}, \dots, a_{100}$ is $25$ . Then $S$ is [2023]

Q 15 :

Let the mean and variance of 8 numbers $x, y, 10, 12, 6, 12, 4, 8$ be 9 and 9.25 respectively. If $x > y$ , then $3 x - 2 y$ is equal to _____. [2023]

Q 16 :

Let the mean of the data

$x$ 1 3 5 7 9

Frequency $(f)$ 4 24 28 $α$ 8

be 5. If $m$ and $σ^{2}$ are respectively the mean deviation about the mean and the variance of the data, then $\frac{3 α}{m + σ^{2}}$ is equal to ________ . [2023]

Q 17 :

The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is __________. [2023]

Q 19 :

The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted and $a$ and $b$ are respectively mean and variance of remaining 6 observations, then $a + 3 b - 5$ is equal to ______. [2023]

Q 20 :

If the mean and variance of the frequency distribution

$x_{i}$ 2 4 6 8 10 12 14 16

$f_{i}$ 4 4 $α$ 15 8 $β$ 4 5

are 9 and 15.08 respectively, then the value of $α^{2} + β^{2} - α β$ is ________ . [2023]

Q 21 :

If the mean of the frequency distribution

Class : 0–10 10–20 20–30 30–40 40–50

Frequency : 2 3 $x$ 5 4

is 28, then its variance is _____ . [2023]

Q 23 :

If the variance of the frequency distribution

$x_{𝑖}$ 2 3 4 5 6 7 8

Frequency $f_{i}$ 3 6 16 $α$ 9 5 6

is 3, then $α$ is equal to _______. [2023]

Q 24 :

Let the mean and variance of 8 numbers $- 10, - 7, - 1, x, y, 9, 2, 16$ be $\frac{7}{2}$ and $\frac{293}{4}$ , respectively. Then the mean of 4 numbers $x, y, x + y + 1,$ $| x - y |$ is: [2026]

Q 26 :

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: [2026]

Q 27 :

Let $X = {x \in ℕ : 1 \leq x \leq 19}$ and for some $a, b \in ℝ, Y = {a x + b : x \in X} .$ If the mean and variance of the elements of Y are 30 and 750, respectively, then the sum of all possible values of $b$ is [2026]

Q 28 :

If the mean and the variance of the data

Class

4–8

8–12

12–16

16–20

Frequency

3

$λ$

4

7

are $μ and 19$ respectively, then the value of $λ + μ$ is. [2026]

Q 30 :

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 [2026]

Q 31 :

The mean and standard deviation of the marks of 200 candidates were found to be 40 and 15 respectively. Later, it was discovered that one student's marks 40 was wrongly read as 50. The correct mean and standard deviation, respectively are

Q 34 :

$\sum_{i = 1}^{10} (x_{i} - 5) = 5$ and $\sum_{i = 1}^{10} {(x_{i} - 5)}^{2} = 125$ , variance of $x_{1}, x_{2}, \dots, x_{10}$ will be

Q 35 :

The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is

Q 36 :

The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and $σ^{2}$ respectively. If the variance of all the 30 numbers in the two sets is 13, then 3 $σ^{2}$ is equal to

Q 37 :

If the standard deviation of $a_{1}, a_{2}, a_{3}, \dots, a_{2025}$ is $λ$ , then the value of standard deviation of $2 a_{1} - 1, 2 a_{2} - 1, 2 a_{3} - 1, \dots, 2 a_{2025} - 1$ is

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Class :	0–10	10–20	20–30	30–40	40–50
Frequency :	2	3	$x$	5	4

Topic Question Set

Q 11 : Let 9=x1<x2<⋯<x7 be in an A.P. with common difference d. If the standard deviation of x1,x2,…,x7 is 4 and the mean is x¯, then x¯+x6 is equal to [2023]

Q 12 : Let the six numbers a1,a2,a3,a4,a5,a6 be in A.P. and a1+a3=10. If the mean of these six numbers is 192 and their variance is σ2, then 8σ2 is equal to [2023]

Q 13 : The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12. If the new mean of the marks is 10.2, then their new variance is equal to: [2023]

Q 14 : Let S be the set of all values of a1 for which the mean deviation about the mean of 100 consecutive positive integers a1,a2,a3,…,a100 is 25. Then S is [2023]

Q 15 : Let the mean and variance of 8 numbers x,y,10,12,6,12,4,8 be 9 and 9.25 respectively. If x>y, then 3x−2y is equal to _____. [2023]

Q 16 : Let the mean of the data x 1 3 5 7 9 Frequency (f) 4 24 28 α 8 be 5. If m and σ2 are respectively the mean deviation about the mean and the variance of the data, then 3αm+σ2 is equal to ________ . [2023]

Q 17 : The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is __________. [2023]

Q 18 : Let X={11,12,13,…,40,41} and Y={61,62,63,…,90,91} be the two sets of observations. If x¯ and y¯ are their respective means and σ2 is the variance of all the observations in X∪Y, then |x¯+y¯-σ2| is equal to _______ . [2023]

Q 19 : The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted and a and b are respectively mean and variance of remaining 6 observations, then a+3b−5 is equal to ______. [2023]

Q 20 : If the mean and variance of the frequency distribution xi 2 4 6 8 10 12 14 16 fi 4 4 α 15 8 β 4 5 are 9 and 15.08 respectively, then the value of α2+β2-αβ is ________ . [2023]

Q 21 : If the mean of the frequency distribution Class : 0–10 10–20 20–30 30–40 40–50 Frequency : 2 3 x 5 4 is 28, then its variance is _____ . [2023]

Q 22 : Let the positive numbers a1,a2,a3,a4 and a5 be in a G.P. Let their mean and variance be 3110 and mn respectively, where m and n are co-prime. If the mean of their reciprocals is 3140 and a3+a4+a5=14, then m+n is equal to _______ . [2023]

Q 23 : If the variance of the frequency distribution x𝑖 2 3 4 5 6 7 8 Frequency fi 3 6 16 α 9 5 6 is 3, then α is equal to _______. [2023]

Q 24 : Let the mean and variance of 8 numbers -10, -7, -1, x, y, 9, 2, 16 be 72 and 2934, respectively. Then the mean of 4 numbers x, y, x+y+1,|x-y| is: [2026]

Q 25 : 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: [2026]

Q 26 : 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: [2026]

Q 27 : Let X={x∈ℕ:1≤x≤19} and for some a,b∈ℝ, Y={ax+b:x∈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 is [2026]

Q 28 : If the mean and the variance of the data Class 4–8 8–12 12–16 16–20 Frequency 3 λ 4 7 are μ and 19 respectively, then the value of λ+μ is. [2026]

Q 29 : A random variable X takes values 0, 1, 2, 3 with probabilities 2a+130, 8a-130, 4a+130, b respectively, where a,b∈R. Let μ and σ respectively be the mean and standard deviation of X such that σ2+μ2=2. Then ab is equal to : [2026]

Q 30 : 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 [2026]

Q 31 : The mean and standard deviation of the marks of 200 candidates were found to be 40 and 15 respectively. Later, it was discovered that one student's marks 40 was wrongly read as 50. The correct mean and standard deviation, respectively are

Q 33 : Let x¯, M and σ2 be respectively the mean, mode and variance of n observations x1,x2,…,xn and di=-xi-a, i=1,2,…,n where a is any number. Statement 1: Variance of d1,d2,…,dn is σ2. Statement 2: Mean and mode of d1,d2,…,dn are -x¯-a and -M-a respectively.

Q 34 : ∑i=110(xi-5)=5 and ∑i=110(xi-5)2=125, variance of x1,x2,…,x10 will be

Q 35 : The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is

Q 36 : The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and σ2 respectively. If the variance of all the 30 numbers in the two sets is 13, then 3σ2 is equal to

Q 37 : If the standard deviation of a1,a2,a3,…,a2025 is λ, then the value of standard deviation of 2a1-1,2a2-1,2a3-1,…,2a2025-1 is

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Q 11 :

Let $9 = x_{1} < x_{2} < \dots < x_{7}$ be in an A.P. with common difference $d$ . If the standard deviation of $x_{1}, x_{2}, \dots, x_{7}$ is $4$ and the mean is $\bar{x}$ , then $\bar{x} + x_{6}$ is equal to [2023]

Q 12 :

Let the six numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ be in A.P. and $a_{1} + a_{3} = 10$ . If the mean of these six numbers is $\frac{19}{2}$ and their variance is $σ^{2}$ , then $8 σ^{2}$ is equal to [2023]

Q 13 :

The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12. If the new mean of the marks is 10.2, then their new variance is equal to: [2023]

Q 14 :

Let $S$ be the set of all values of $a_{1}$ for which the mean deviation about the mean of 100 consecutive positive integers $a_{1}, a_{2}, a_{3}, \dots, a_{100}$ is $25$ . Then $S$ is [2023]

Q 15 :

Let the mean and variance of 8 numbers $x, y, 10, 12, 6, 12, 4, 8$ be 9 and 9.25 respectively. If $x > y$ , then $3 x - 2 y$ is equal to _____. [2023]

Q 16 :

Let the mean of the data

$x$ 1 3 5 7 9

Frequency $(f)$ 4 24 28 $α$ 8

be 5. If $m$ and $σ^{2}$ are respectively the mean deviation about the mean and the variance of the data, then $\frac{3 α}{m + σ^{2}}$ is equal to ________ . [2023]

Q 17 :

The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is __________. [2023]

Q 19 :

The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted and $a$ and $b$ are respectively mean and variance of remaining 6 observations, then $a + 3 b - 5$ is equal to ______. [2023]

Q 20 :

If the mean and variance of the frequency distribution

$x_{i}$ 2 4 6 8 10 12 14 16

$f_{i}$ 4 4 $α$ 15 8 $β$ 4 5

are 9 and 15.08 respectively, then the value of $α^{2} + β^{2} - α β$ is ________ . [2023]

Q 21 :

If the mean of the frequency distribution

Class : 0–10 10–20 20–30 30–40 40–50

Frequency : 2 3 $x$ 5 4

is 28, then its variance is _____ . [2023]

Q 23 :

If the variance of the frequency distribution

$x_{𝑖}$ 2 3 4 5 6 7 8

Frequency $f_{i}$ 3 6 16 $α$ 9 5 6

is 3, then $α$ is equal to _______. [2023]

Q 24 :

Let the mean and variance of 8 numbers $- 10, - 7, - 1, x, y, 9, 2, 16$ be $\frac{7}{2}$ and $\frac{293}{4}$ , respectively. Then the mean of 4 numbers $x, y, x + y + 1,$ $| x - y |$ is: [2026]

Q 26 :

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: [2026]

Q 27 :

Let $X = {x \in ℕ : 1 \leq x \leq 19}$ and for some $a, b \in ℝ, Y = {a x + b : x \in X} .$ If the mean and variance of the elements of Y are 30 and 750, respectively, then the sum of all possible values of $b$ is [2026]

Q 28 :

If the mean and the variance of the data

Class

4–8

8–12

12–16

16–20

Frequency

3

$λ$

4

7

are $μ and 19$ respectively, then the value of $λ + μ$ is. [2026]

Q 30 :

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 [2026]

Q 31 :

The mean and standard deviation of the marks of 200 candidates were found to be 40 and 15 respectively. Later, it was discovered that one student's marks 40 was wrongly read as 50. The correct mean and standard deviation, respectively are

Q 34 :

$\sum_{i = 1}^{10} (x_{i} - 5) = 5$ and $\sum_{i = 1}^{10} {(x_{i} - 5)}^{2} = 125$ , variance of $x_{1}, x_{2}, \dots, x_{10}$ will be

Q 35 :

The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is

Q 36 :

The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and $σ^{2}$ respectively. If the variance of all the 30 numbers in the two sets is 13, then 3 $σ^{2}$ is equal to

Q 37 :

If the standard deviation of $a_{1}, a_{2}, a_{3}, \dots, a_{2025}$ is $λ$ , then the value of standard deviation of $2 a_{1} - 1, 2 a_{2} - 1, 2 a_{3} - 1, \dots, 2 a_{2025} - 1$ is