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Statistics — JEE Previous Year Questions

Every Statistics question asked in JEE Main and JEE Advanced across the last 186 papers — 118 questions, each with its correct answer. Free to read, no account needed.

Questions

118

Papers it appeared in

118/186

Appearance rate

63%

All 118 Statistics questions

Most recent papers first.

Q1·MathematicsNumericalJEE Advanced 2026
Consider a data consisting of 10 observations x1,x2,…,x10x_{1}, x_{2}, \ldots, x_{10}x1​,x2​,…,x10​, whose mean is 5 and variance is 7. If the mean and the variance of the first 8 observations x1,x2,…,x8x_{1}, x_{2}, \ldots, x_{8}x1​,x2​,…,x8​ are 4 and 3.5, respectively, and x9<x10x_{9} < x_{10}x9​<x10​, then the value of 3x9+2x103x_{9} + 2x_{10}3x9​+2x10​ is _____.

Correct answer: 44

Step-by-step solution →
Q2·MathematicsSingle correctJEE Main 2026
A set of four observations has mean 1 and variance 13. Another set of six observations has mean 2 and variance 1. Then, the variance of all these 10 observations is equal to:
  1. (A)5.96
  2. (B)6.14
  3. (C)6.04
  4. (D)6.24

Correct answer: (C)

Step-by-step solution →
Q3·MathematicsSingle correctJEE Main 2026
Let the mean and the variance of seven observations 2, 4, α, 8, β, 12, 14, α < β, be 8 and 16 respectively. Then the quadratic equation whose roots are 3α + 2 and 2β + 1 is :
  1. (A)x2−35x+306=0x^{2} - 35x + 306 = 0x2−35x+306=0
  2. (B)x2−41x+420=0x^{2} - 41x + 420 = 0x2−41x+420=0
  3. (C)x2−45x+506=0x^{2} - 45x + 506 = 0x2−45x+506=0
  4. (D)x2−37x+342=0x^{2} - 37x + 342 = 0x2−37x+342=0

Correct answer: (B)

Step-by-step solution →
Q4·MathematicsSingle correctJEE Main 2026
A data consists of 20 observations x1_11​, x2_22​, ..., x20_{20}20​. If ∑i=120(xi+5)2=2500\sum_{i=1}^{20}(x_i + 5)^2 = 2500∑i=120​(xi​+5)2=2500 and ∑i=120(xi−5)2=100\sum_{i=1}^{20}(x_i - 5)^2 = 100∑i=120​(xi​−5)2=100, then the ratio of mean to standard deviation of this data is:
  1. (A)2 : 1
  2. (B)3 : 1
  3. (C)3 : 2
  4. (D)4 : 1

Correct answer: (B)

Step-by-step solution →
Q5·MathematicsSingle correctJEE Main 2026
The mean deviation about the mean for the data is equal to:
xix_ixi​fif_ifi​
58
76
92
102
122
156
  1. (A)4013\frac{40}{13}1340​
  2. (B)4213\frac{42}{13}1342​
  3. (C)4413\frac{44}{13}1344​
  4. (D)4613\frac{46}{13}1346​

Correct answer: (C)

Step-by-step solution →
Q6·MathematicsSingle correctJEE Main 2026
A variable X takes values 0,0,2,6,12,20,…,n(n−1)0, 0, 2, 6, 12, 20, \ldots, n(n - 1)0,0,2,6,12,20,…,n(n−1) with frequencies nC0,nC1,nC2,nC3,nC4,nC5,…,nCn^{n}C_0, ^{n}C_1, ^{n}C_2, ^{n}C_3, ^{n}C_4, ^{n}C_5, \ldots, ^{n}C_nnC0​,nC1​,nC2​,nC3​,nC4​,nC5​,…,nCn​, respectively. If the mean of this data is 60, then its median is :
  1. (A)56
  2. (B)42
  3. (C)72
  4. (D)90

Correct answer: (A)

Step-by-step solution →
Q7·MathematicsSingle correctJEE Main 2026
For 10 observations x1,x2,…,x10x_{1}, x_{2}, \ldots, x_{10}x1​,x2​,…,x10​, if ∑i=110(xi+2)2=180\sum_{i=1}^{10}(x_{i} + 2)^{2} = 180∑i=110​(xi​+2)2=180 and ∑i=110(xi−1)2=90\sum_{i=1}^{10}(x_{i} - 1)^{2} = 90∑i=110​(xi​−1)2=90, then their standard deviation is:
  1. (A)2
  2. (B)3\sqrt{3}3​
  3. (C)222\sqrt{2}22​
  4. (D)3

Correct answer: (D)

Step-by-step solution →
Q8·MathematicsSingle correctJEE Main 2026
Suppose that the mean and median of the non-negative numbers 21,8,17,a,51,103,b,13,67,(a>b)21, 8, 17, a, 51, 103, b, 13, 67, (a > b)21,8,17,a,51,103,b,13,67,(a>b), are 40 and 21, respectively. If the mean deviation about the median is 26, then 2a2a2a is equal to:
  1. (A)109109109
  2. (B)117117117
  3. (C)161161161
  4. (D)131131131

Correct answer: (D)

Step-by-step solution →
Q9·MathematicsSingle correctJEE Main 2026
The mean and variance of nnn observations are 8 and 16, respectively. If the sum of the first (n−1)(n - 1)(n−1) observations is 48 and the sum of squares of the first (n−1)(n - 1)(n−1) observations is 496, then the value of nnn is:
  1. (A)212121
  2. (B)161616
  3. (C)131313
  4. (D)777

Correct answer: (D)

Step-by-step solution →
Q10·MathematicsSingle correctJEE Main 2026
If the mean of the data is 21, then kkk is one of the roots of the equation :
Class5 − 1010 − 1515 − 2020 − 2525 − 3030 − 35
Frequency2kkk2854k+1k+1k+15
  1. (A)2x2−23x−10=02x^{2}-23x-10=02x2−23x−10=0
  2. (B)4x2−35x+24=04x^{2}-35x+24=04x2−35x+24=0
  3. (C)2x2−19x−10=02x^{2}-19x-10=02x2−19x−10=0
  4. (D)2x2−35x+98=02x^{2}-35x+98=02x2−35x+98=0

Correct answer: (C)

Step-by-step solution →
Q11·MathematicsSingle correctJEE Main 2026
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
  1. (A)5
  2. (B)4
  3. (C)6
  4. (D)7

Correct answer: (A)

Step-by-step solution →
Q12·MathematicsSingle correctJEE Main 2026
Let X={x∈N:1≤x≤19}X = \{x \in \mathbb{N} : 1 \le x \le 19\}X={x∈N:1≤x≤19} and for some a, b ∈R\in \mathbb{R}∈R, Y={ax+b:x∈X}Y = \{ax + b : x \in X\}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
  1. (A)20
  2. (B)80
  3. (C)100
  4. (D)60

Correct answer: (D)

Step-by-step solution →
Q13·MathematicsSingle correctJEE Main 2026
The mean and variance of a data of 10 observations are 10 and 2, respectively. If an observations α in this data is replaced by β, then the mean and variance become 10.1 and 1.99, respectively. Then α + β equals.
  1. (A)10
  2. (B)15
  3. (C)5
  4. (D)20

Correct answer: (D)

Step-by-step solution →
Q14·MathematicsSingle correctJEE Main 2026
Let the mean and variance of 8 numbers -10, -7, -1, x, y, 9, 2, 16 be 72\frac{7}{2}27​ and 2934\frac{293}{4}4293​, respectively. Then the mean of 4 numbers x, y, x + y + 1, |x - y| is:
  1. (A)11
  2. (B)9
  3. (C)10
  4. (D)12

Correct answer: (A)

Step-by-step solution →
Q15·MathematicsSingle correctJEE Main 2026
If the mean and the variance of the data Class | 4–8 | 8–12 | 12–16 | 16–20 Frequency | 3 | λ\lambdaλ | 4 | 7 are µ and 19 respectively, then the value of λ + µ is
  1. (A)18
  2. (B)21
  3. (C)20
  4. (D)19

Correct answer: (D)

Step-by-step solution →
Q16·MathematicsSingle correctJEE Main 2026
If the mean deviation about the median of the numbers k, 2k, 3k, ...., 1000k is 500, then k2k^{2}k2 is equal to :
  1. (A)16
  2. (B)4
  3. (C)1
  4. (D)9

Correct answer: (B)

Step-by-step solution →
Q17·MathematicsSingle correctJEE Advanced 2025
Consider the following frequency distribution: Value | 4 | 5 | 8 | 9 | 6 | 12 | 11 Frequency | 5 | f1f_1f1​ | f2f_2f2​ | 2 | 1 | 1 | 3 Suppose that the sum of the frequencies is 19 and the median of this frequency distribution is 6. For the given frequency distribution, let α denote the mean deviation about the mean, β denote the mean deviation about the median, and σ2\sigma^2σ2 denote the variance. Match each entry in List-I to the correct entries in List-II. The correct option is:
List-IList-II
P.7f1+9f27f_1 + 9f_27f1​+9f2​ is equal to1.146
Q.19α is equal to2.47
R.19β is equal to3.48
S.19σ219\sigma^219σ2 is equal to4.145
5.55
  1. (A)(P) → (5), (Q) → (3), (R) → (2), (S) → (4)
  2. (B)(P) → (5), (Q) → (2), (R) → (3), (S) → (1)
  3. (C)(P) → (5), (Q) → (3), (R) → (2), (S) → (1)
  4. (D)(P) → (3), (Q) → (2), (R) → (5), (S) → (4)

Correct answer: (C)

Step-by-step solution →
Q18·MathematicsSingle correctJEE Main 2025
The mean and standard deviation of 100 observations are 40 and 5.1, respectively. By mistake one observation is taken as 50 instead of 40. If the correct mean and the correct standard deviation are μ\muμ and σ\sigmaσ respectively, then 10(μ+σ)10(\mu+\sigma)10(μ+σ) is equal to
  1. (A)445
  2. (B)451
  3. (C)447
  4. (D)449

Correct answer: (D)

Step-by-step solution →
Q19·MathematicsSingle correctJEE Main 2025
Let the mean and the standard deviation of the observation 2,3,3,4,5,7,a,b2, 3, 3, 4, 5, 7, a, b2,3,3,4,5,7,a,b be 4 and 2\sqrt22​ respectively. The mean deviation about the mode of these observations is:
  1. (A)1
  2. (B)34\dfrac{3}{4}43​
  3. (C)2
  4. (D)12\dfrac{1}{2}21​

Correct answer: (A)

Step-by-step solution →
Q20·MathematicsSingle correctJEE Main 2025
Let the Mean and Variance of five observations x1=1x_1=1x1​=1, x2=3x_2=3x2​=3, x3=ax_3=ax3​=a, x4=7x_4=7x4​=7 and x5=bx_5=bx5​=b, a>ba>ba>b, be 5 and 10 respectively. Then the Variance of the observations n+xnn+x_nn+xn​, n=1,2,…,5n=1,2,\ldots,5n=1,2,…,5 is:
  1. (A)17
  2. (B)16.4
  3. (C)17.4
  4. (D)16

Correct answer: (D)

Step-by-step solution →
Q21·MathematicsSingle correctJEE Main 2025
If the mean and variance of 6,4,a,8,b,12,10,136, 4, a, 8, b, 12, 10, 136,4,a,8,b,12,10,13 are 9 and 374\dfrac{37}{4}437​ respectively, then a+b+aba+b+aba+b+ab is equal to:
  1. (A)105
  2. (B)103
  3. (C)100
  4. (D)106

Correct answer: (B)

Step-by-step solution →
Q22·MathematicsSingle correctJEE Main 2025
Let x1,x2,…,x10x_1, x_2,\ldots,x_{10}x1​,x2​,…,x10​ be ten observations such that ∑i=110(xi−2)=30\displaystyle\sum_{i=1}^{10}(x_i-2)=30i=1∑10​(xi​−2)=30, ∑i=110(xi−β)2=98\displaystyle\sum_{i=1}^{10}(x_i-\beta)^2=98i=1∑10​(xi​−β)2=98, β>2\beta>2β>2 and their variance is 45\dfrac{4}{5}54​. If μ\muμ and σ2\sigma^2σ2 are respectively the mean and the variance of 2(x1−1)+4β, 2(x2−1)+4β, …, 2(x10−1)+4β2(x_1-1)+4\beta,\ 2(x_2-1)+4\beta,\ \ldots,\ 2(x_{10}-1)+4\beta2(x1​−1)+4β, 2(x2​−1)+4β, …, 2(x10​−1)+4β, then βμσ2\dfrac{\beta\mu}{\sigma^2}σ2βμ​ is equal to:
  1. (A)100
  2. (B)110
  3. (C)120
  4. (D)90

Correct answer: (A)

Step-by-step solution →
Q23·MathematicsSingle correctJEE Main 2025
For a statistical data x1,x2,…,x10x_1,x_2,\ldots,x_{10}x1​,x2​,…,x10​ of 10 values, a student obtained the mean as 5.5 and ∑i=110xi2=371\sum_{i=1}^{10}x_i^2=371∑i=110​xi2​=371. He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively. The variance of the corrected data is
  1. (A)777
  2. (B)444
  3. (C)999
  4. (D)555

Correct answer: (A)

Step-by-step solution →
Q24·MathematicsSingle correctJEE Main 2025
Marks obtained by all the students of class 12 are presented in a frequency distribution with classes of equal width. Let the median of this grouped data be 14 with median class interval 12-18 and median class frequency 12. If the number of students whose marks are less than 12 is 18, then the total number of students is
  1. (A)484848
  2. (B)444444
  3. (C)404040
  4. (D)525252

Correct answer: (B)

Step-by-step solution →
Q25·MathematicsIntegerJEE Main 2025
The variance of the numbers 8, 21, 34, 47, …, 3208,\ 21,\ 34,\ 47,\ \ldots,\ 3208, 21, 34, 47, …, 320, is __________.

Correct answer: 8788

Step-by-step solution →
Q26·MathematicsSingle correctJEE Main 2024
The frequency distribution of the age of students in a class of 40 students is given below. Age: 15, 16, 17, 18, 19, 20; No. of Students: 5, 8, 5, 12, xxx, yyy. If the mean deviation about the median is 1.251.251.25, then 4x+5y4x + 5y4x+5y is equal to :
  1. (A)43
  2. (B)44
  3. (C)47
  4. (D)46

Correct answer: (B)

Step-by-step solution →
Q27·MathematicsSingle correctJEE Main 2024
If the variance of the frequency distribution given by xxx: ccc, 2c2c2c, 3c3c3c, 4c4c4c, 5c5c5c, 6c6c6c with corresponding frequencies fff: 222, 111, 111, 111, 111, 111 is 160, then the value of c∈Nc\in\mathbb{N}c∈N is:
  1. (A)555
  2. (B)888
  3. (C)777
  4. (D)666

Correct answer: (C)

Step-by-step solution →
Q28·MathematicsNumericalJEE Main 2024
Let aaa, bbb, c∈Nc\in\mathbb{N}c∈N and a<b<ca<b<ca<b<c. Let the mean, the mean deviation about the mean and the variance of the 5 observations 999, 252525, aaa, bbb, ccc be 181818, 444 and 1365\dfrac{136}{5}5136​, respectively. Then 2a+b−c2a+b-c2a+b−c is equal to _______ .

Correct answer: 33

Step-by-step solution →
Q29·MathematicsSingle correctJEE Main 2024
The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. If 8 is replaced by 12, then the correct standard deviation is
  1. (A)3.86\sqrt{3.86}3.86​
  2. (B)1.81.81.8
  3. (C)3.96\sqrt{3.96}3.96​
  4. (D)1.941.941.94

Correct answer: (C)

Step-by-step solution →
Q30·MathematicsSingle correctJEE Main 2024
Let a,b∈Ra,b\in\mathbb{R}a,b∈R. Let the mean and the variance of 666 observations −3,4,7,−6,a,b-3,4,7,-6,a,b−3,4,7,−6,a,b be 222 and 232323, respectively. The mean deviation about the mean of these 666 observations is:
  1. (A)133\dfrac{13}{3}313​
  2. (B)163\dfrac{16}{3}316​
  3. (C)113\dfrac{11}{3}311​
  4. (D)143\dfrac{14}{3}314​

Correct answer: (A)

Step-by-step solution →
Q31·MathematicsSingle correctJEE Main 2024
Consider 10 observations x1,x2,…,x10x_1,x_2,\ldots,x_{10}x1​,x2​,…,x10​ such that ∑i=110(xi−α)=2\displaystyle\sum_{i=1}^{10}(x_i-\alpha)=2i=1∑10​(xi​−α)=2 and ∑i=110(xi−β)2=40\displaystyle\sum_{i=1}^{10}(x_i-\beta)^2=40i=1∑10​(xi​−β)2=40, where α,β\alpha,\betaα,β are positive integers. Let the mean and the variance of the observations be 65\dfrac{6}{5}56​ and 8425\dfrac{84}{25}2584​ respectively. The βα\dfrac{\beta}{\alpha}αβ​ is equal to:
  1. (A)2
  2. (B)32\dfrac{3}{2}23​
  3. (C)52\dfrac{5}{2}25​
  4. (D)1

Correct answer: (A)

Step-by-step solution →
Q32·MathematicsSingle correctJEE Main 2024
If the median and mean deviation about the median of 170,125,230,190,210,a,b170,125,230,190,210,a,b170,125,230,190,210,a,b be 170170170 and 2057\dfrac{205}{7}7205​ respectively. Then the mean deviation about the mean of these 777 observations is:
  1. (A)313131
  2. (B)282828
  3. (C)303030
  4. (D)323232

Correct answer: (C)

Step-by-step solution →
Q33·MathematicsSingle correctJEE Main 2024
Let the mean and the variance of 6 observations aaa, bbb, 68, 44, 48, 60 be 55 and 194, respectively. If a>ba>ba>b, then a+3ba+3ba+3b is
  1. (A)200
  2. (B)190
  3. (C)180
  4. (D)210

Correct answer: (C)

Step-by-step solution →
Q34·MathematicsSingle correctJEE Main 2024
Let MMM denote the median of the following frequency distribution. Class: 000-444, 444-888, 888-121212, 121212-161616, 161616-202020; Frequency: 333, 999, 101010, 888, 666. Then 20M20M20M is equal to:
  1. (A)416
  2. (B)104
  3. (C)52
  4. (D)208

Correct answer: (D)

Step-by-step solution →
Q35·MathematicsNumericalJEE Main 2024
The variance σ2\sigma^2σ2 of the data | xix_ixi​ | 0 | 1 | 5 | 6 | 10 | 12 | 17 | |---|---|---|---|---|---|---|---| | fif_ifi​ | 3 | 2 | 3 | 2 | 6 | 3 | 3 | is ______.

Correct answer: 29

Step-by-step solution →
Q36·MathematicsNumericalJEE Main 2024
If the mean and variance of the data 65,68,58,44,48,45,60,α,β,6065,68,58,44,48,45,60,\alpha,\beta,6065,68,58,44,48,45,60,α,β,60 where α>β\alpha>\betaα>β are 56 and 66.2 respectively, then α2+β2\alpha^2+\beta^2α2+β2 is equal to ______.

Correct answer: 6344

Step-by-step solution →
Q37·MathematicsSingle correctJEE Main 2024
If the mean and variance of five observations are 245\dfrac{24}{5}524​ and 19425\dfrac{194}{25}25194​ respectively and the mean of first four observations is 72\dfrac{7}{2}27​, then the variance of the first four observations is equal to:
  1. (A)45\dfrac{4}{5}54​
  2. (B)7712\dfrac{77}{12}1277​
  3. (C)54\dfrac{5}{4}45​
  4. (D)1054\dfrac{105}{4}4105​

Correct answer: (C)

Step-by-step solution →
Q38·MathematicsNumericalJEE Main 2024
The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12. If μ\muμ and σ2\sigma^2σ2 denote the mean and variance of the correct observations respectively, then 15(μ+μ2+σ2)15(\mu+\mu^2+\sigma^2)15(μ+μ2+σ2) is equal to __________.

Correct answer: 2521

Step-by-step solution →
Q39·MathematicsSingle correctJEE Main 2024
Let a1,a2,…,a10a_1,a_2,\ldots,a_{10}a1​,a2​,…,a10​ be 10 observations such that ∑k=110ak=50\displaystyle\sum_{k=1}^{10}a_k=50k=1∑10​ak​=50 and ∑∀k<jak aj=1100\displaystyle\sum_{\forall k<j}a_k\,a_j=1100∀k<j∑​ak​aj​=1100. Then the standard deviation of a1,a2,…,a10a_1,a_2,\ldots,a_{10}a1​,a2​,…,a10​ is equal to:
  1. (A)5
  2. (B)5\sqrt55​
  3. (C)10
  4. (D)115\sqrt{115}115​

Correct answer: (B)

Step-by-step solution →
Q40·MathematicsSingle correctJEE Advanced 2023
Consider the given data with frequency distribution xix_ixi​ 3 8 11 10 5 4 fif_ifi​ 5 2 3 2 4 4 Match each entry in List-I to the correct entries in List-II. The correct option is:
List – IList – II
P.The mean of the above data is1.2.5
Q.The median of the above data is2.5
R.The mean deviation about the mean of the above data is3.6
S.The mean deviation about the median of the above data is4.2.7
5.2.4
  1. (A)(P) → (3) (Q) → (2) (R) → (4) (S) → (5)
  2. (B)(P) → (3) (Q) → (2) (R) → (1) (S) → (5)
  3. (C)(P) → (2) (Q) → (3) (R) → (4) (S) → (1)
  4. (D)(P) → (3) (Q) → (3) (R) → (5) (S) → (5)

Correct answer: (A)

Step-by-step solution →
Q41·MathematicsSingle correctJEE Main 2023
The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is:
  1. (A)14
  2. (B)13
  3. (C)12
  4. (D)11

Correct answer: (B)

Step-by-step solution →
Q42·MathematicsNumericalJEE Main 2023
Let the mean of the data: x13579Frequency (f)42428α8\begin{array}{|c|c|c|c|c|c|}\hline x & 1 & 3 & 5 & 7 & 9 \\\hline \text{Frequency }(f) & 4 & 24 & 28 & \alpha & 8 \\\hline\end{array}xFrequency (f)​14​324​528​7α​98​​ be 555. If mmm and σ2\sigma^{2}σ2 are respectively the mean deviation about the mean and the variance of the data, then 3αm+σ2\dfrac{3\alpha}{m+\sigma^{2}}m+σ23α​ is equal to _____.

Correct answer: 8

Step-by-step solution →
Q43·MathematicsNumericalJEE Main 2023
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

Correct answer: 269

Step-by-step solution →
Q44·MathematicsSingle correctJEE Main 2023
Two dice A and B are rolled. Let the numbers obtained on A and B be α\alphaα and β\betaβ respectively. If the variance of α−β\alpha-\betaα−β is pq\dfrac{p}{q}qp​, where p and q are co-prime, then the sum of the positive divisors of p is equal to
  1. (A)36
  2. (B)48
  3. (C)31
  4. (D)72

Correct answer: (B)

Step-by-step solution →
Q45·MathematicsSingle correctJEE Main 2023
If the mean of the observations 1,2,4,5,x1,2,4,5,x1,2,4,5,x and yyy be 555 and their variance be 101010, then their mean deviation about the mean is equal to
  1. (A)103\dfrac{10}{3}310​
  2. (B)73\dfrac{7}{3}37​
  3. (C)333
  4. (D)83\dfrac{8}{3}38​

Correct answer: (D)

Step-by-step solution →
Q46·MathematicsSingle correctJEE Main 2023
Let sets A and B have 5 elements each. Let the mean of the elements in sets A and B be 5 and 8 respectively and the variance of the elements in sets A and B be 12 and 20 respectively. A new set C of 10 elements is formed by subtracting 3 from each element of A and adding 2 to each element of B. Then the sum of the mean and variance of the elements of C is _______.
  1. (A)32
  2. (B)38
  3. (C)40
  4. (D)36

Correct answer: (B)

Step-by-step solution →
Q47·MathematicsSingle correctJEE Main 2023
Let μ\muμ be the mean and σ\sigmaσ be the standard deviation of the distribution with values xi=0,1,2,3,4,5x_i=0,1,2,3,4,5xi​=0,1,2,3,4,5 and frequencies fi=k+2, 2k, k2−1, k2−1, k2+1, k−3f_i=k+2,\,2k,\,k^2-1,\,k^2-1,\,k^2+1,\,k-3fi​=k+2,2k,k2−1,k2−1,k2+1,k−3 respectively, where ∑fi=62\sum f_i=62∑fi​=62. If [x][x][x] denotes the greatest integer ≤x\le x≤x, then [μ2+σ2][\mu^2+\sigma^2][μ2+σ2] is equal to:
  1. (A)888
  2. (B)777
  3. (C)666
  4. (D)999

Correct answer: (A)

Step-by-step solution →
Q48·MathematicsNumericalJEE Main 2023
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 _________.

Correct answer: 151

Step-by-step solution →
Q49·MathematicsNumericalJEE Main 2023
Let the mean and variance of 8 numbers xxx, yyy, 10, 12, 6, 12, 4, 8, be 9 and 9.25 respectively. If x>yx>yx>y, then 3x−2y3x-2y3x−2y is equal to

Correct answer: 25

Step-by-step solution →
Q50·MathematicsSingle correctJEE Main 2023
Let the mean and variance of 12 observations be 92\dfrac{9}{2}29​ and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is mn\dfrac{m}{n}nm​, where m and n are co-prime, then m+n is equal to
  1. (A)316
  2. (B)314
  3. (C)317
  4. (D)315

Correct answer: (C)

Step-by-step solution →
Q51·MathematicsSingle correctJEE Main 2023
The mean and variance of a set of 151515 numbers are 121212 and 141414 respectively. The mean and variance of another set of 151515 numbers are 141414 and σ2\sigma^2σ2 respectively. If the variance of all the 303030 numbers in the two sets is 131313, then σ2\sigma^2σ2 is equal to:
  1. (A)999
  2. (B)121212
  3. (C)111111
  4. (D)101010

Correct answer: (D)

Step-by-step solution →
Q52·MathematicsNumericalJEE Main 2023
If the mean and variance of the frequency distribution xi=2,4,6,8,10,12,14,16x_i = 2,4,6,8,10,12,14,16xi​=2,4,6,8,10,12,14,16 with frequencies fi=4,4,α,15,8,β,4,5f_i = 4,4,\alpha,15,8,\beta,4,5fi​=4,4,α,15,8,β,4,5 are 9 and 15.08 respectively, then the value of α2+β2−αβ\alpha^2+\beta^2-\alpha\betaα2+β2−αβ is

Correct answer: 25

Step-by-step solution →
Q53·MathematicsSingle correctJEE Main 2023
Let 9=x1<x2<…<x79=x_1<x_2<\ldots<x_79=x1​<x2​<…<x7​ be in an A.P. with common difference d. If the standard deviation of x1,x2,…,x7x_1,x_2,\ldots,x_7x1​,x2​,…,x7​ is 4 and mean is xˉ\bar xxˉ, then xˉ+x6\bar x+x_6xˉ+x6​ is equal to:
  1. (A)2(9+87)2\left(9+\dfrac{8}{\sqrt7}\right)2(9+7​8​)
  2. (B)18(1+13)18\left(1+\dfrac{1}{\sqrt3}\right)18(1+3​1​)
  3. (C)252525
  4. (D)343434

Correct answer: (D)

Step-by-step solution →
Q54·MathematicsSingle correctJEE Main 2023
The mean and variance of 555 observations are 555 and 888 respectively. If 333 observations are 1,3,51, 3, 51,3,5 then the sum of cubes of the remaining two observations is
  1. (A)121612161216
  2. (B)107210721072
  3. (C)145614561456
  4. (D)179217921792

Correct answer: (B)

Step-by-step solution →
Q55·MathematicsSingle correctJEE Main 2023
Let the mean and standard deviation of marks of class A of 100 students be respectively 40 and α (>0)\alpha\,(>0)α(>0), and the mean and standard deviation of marks of class B of nnn students be respectively 55 and 30−α30-\alpha30−α. If the mean and variance of the marks of the combined class of 100+n100+n100+n students are respectively 50 and 350, then the sum of variances of classes AAA and BBB is:
  1. (A)650650650
  2. (B)450450450
  3. (C)900900900
  4. (D)500500500

Correct answer: (D)

Step-by-step solution →
Q56·MathematicsNumericalJEE Main 2023
If the variance of the frequency distribution with values xi=2,3,4,5,6,7,8x_i = 2, 3, 4, 5, 6, 7, 8xi​=2,3,4,5,6,7,8 and corresponding frequencies fi=3,6,16,α,9,5,6f_i = 3, 6, 16, \alpha, 9, 5, 6fi​=3,6,16,α,9,5,6 is 3, then α\alphaα is equal to

Correct answer: 5

Step-by-step solution →
Q57·MathematicsSingle correctJEE Main 2023
Let SSS be the set of all values of a1a_1a1​ for which the mean deviation about the mean of 100 consecutive positive integers a1,a2,a3,…,a100a_1, a_2, a_3, \ldots, a_{100}a1​,a2​,a3​,…,a100​ is 252525. Then SSS is:
  1. (A)N\mathbb{N}N
  2. (B)ϕ\phiϕ
  3. (C){99}\{99\}{99}
  4. (D){9}\{9\}{9}

Correct answer: (A)

Step-by-step solution →
Q58·MathematicsNumericalJEE Main 2023
The mean and variance of 777 observations are 888 and 161616 respectively. If one observation 141414 is omitted and aaa and bbb are respectively mean and variance of remaining 666 observation, then a+3b−5a+3b-5a+3b−5 is equal to

Correct answer: 37

Step-by-step solution →
Q59·MathematicsNumericalJEE Main 2023
Let X={11,12,13,…,40,41}X=\{11,12,13,\dots,40,41\}X={11,12,13,…,40,41} and Y={61,62,63,…,90,91}Y=\{61,62,63,\dots,90,91\}Y={61,62,63,…,90,91} be the two sets of observations. If xˉ\bar xxˉ and yˉ\bar yyˉ​ are their respective means and σ2\sigma^2σ2 is the variance of all the observations in X∪YX\cup YX∪Y, then ∣xˉ+yˉ−σ2∣|\bar x+\bar y-\sigma^2|∣xˉ+yˉ​−σ2∣ is equal to _____.

Correct answer: 603

Step-by-step solution →
Q60·MathematicsSingle correctJEE Main 2023
The mean and variance of the marks obtained by the students in a test are 101010 and 444 respectively. Later, the marks of one of the students is increased from 888 to 121212. If the new mean of the marks is 10.210.210.2, then their new variance is equal to:
  1. (A)3.963.963.96
  2. (B)4.084.084.08
  3. (C)4.044.044.04
  4. (D)3.923.923.92

Correct answer: (A)

Step-by-step solution →
Q61·MathematicsSingle correctJEE Main 2023
Let the six numbers a1,a2,a3,a4,a5,a6a_1,a_2,a_3,a_4,a_5,a_6a1​,a2​,a3​,a4​,a5​,a6​ be in A.P. and a1+a3=10a_1+a_3=10a1​+a3​=10. If the mean of these six numbers is 192\dfrac{19}{2}219​ and their variance is σ2\sigma^2σ2, then 8σ28\sigma^28σ2 is equal to:
  1. (A)210210210
  2. (B)220220220
  3. (C)200200200
  4. (D)105105105

Correct answer: (A)

Step-by-step solution →
Q62·MathematicsNumericalJEE Main 2022
Let the mean and the variance of 20 observations x1, x2,…x20x_{1},\,x_{2},\ldots x_{20}x1​,x2​,…x20​ be 15 and 9, respectively. For α∈R\alpha \in Rα∈R, if the mean of (x1+α)2, (x2+α)2,…, (x20+α)2\left(x_{1}+\alpha\right)^{2},\,\left(x_{2}+\alpha\right)^{2},\ldots,\,\left(x_{20}+\alpha\right)^{2}(x1​+α)2,(x2​+α)2,…,(x20​+α)2 is 178, then the square of the maximum value of α\alphaα is equal to __________.

Correct answer: 4

Step-by-step solution →
Q63·MathematicsNumericalJEE Main 2022
Let x1x_{1}x1​, x2x_{2}x2​, x3x_{3}x3​, ….., x20x_{20}x20​ be in geometric progression with x1=3x_{1} = 3x1​=3 and the common ration 12\frac{1}{2}21​. A new data is constructed replacing each xix_{i}xi​ by (xi−i)2\left(x_{i} - i\right)^{2}(xi​−i)2. If xˉ\bar{x}xˉ is the mean of new data, then the greatest integer less than or equal to xˉ\bar{x}xˉ is __________.

Correct answer: 142

Step-by-step solution →
Q64·MathematicsNumericalJEE Main 2022
The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who took by mistake 25 instead of 15 for one observation. Then, the correct standard deviation is __________.

Correct answer: 2

Step-by-step solution →
Q65·MathematicsNumericalJEE Main 2022
The mean and standard deviation of 40 observations are 30 and 5 respectively. It was noticed that two of these observations 12 and 10 were wrongly recorded. If σ\sigmaσ is the standard deviation of the data after omitting the two wrong observations from the data, then 38σ238\sigma^{2}38σ2 is equal to____________.

Correct answer: 238

Step-by-step solution →
Q66·MathematicsSingle correctJEE Main 2022
Let the mean and the variance of 5 observations x1,x2,x3,x4,x5x_1, x_2, x_3, x_4, x_5x1​,x2​,x3​,x4​,x5​ be 245\frac{24}{5}524​ and 19425\frac{194}{25}25194​ respectively. If the mean and variance of the first 4 observation are 72\frac{7}{2}27​ and aaa respectively, then (4a+x5)(4a + x_5)(4a+x5​) is equal to:
  1. (A)13
  2. (B)15
  3. (C)17
  4. (D)18

Correct answer: (B)

Step-by-step solution →
Q67·MathematicsSingle correctJEE Main 2022
The number of values of a∈Na \in \mathbb{N}a∈N such that the variance of 3, 7, 12 aaa, 43−a43 - a43−a is a natural number is:
  1. (A)0
  2. (B)2
  3. (C)5
  4. (D)infinite

Correct answer: (A)

Step-by-step solution →
Q68·MathematicsNumericalJEE Main 2022
Suppose a class has 7 students. The average marks of these students in the mathematics examination is 62, and their variance is 20. A student fails in the examination if he/she gets less than 50 marks, then in worst case, the number of students can fail is

Correct answer: 0

Step-by-step solution →
Q69·MathematicsNumericalJEE Main 2022
The mean and standard deviation of 15 observations are found to be 8 and 3 respectively. On rechecking it was found that, in the observations, 20 was misread as 5. Then, the correct variance is equal to ______.

Correct answer: 17

Step-by-step solution →
Q70·MathematicsSingle correctJEE Main 2022
The mean and variance of the data4, 5, 6, 6, 7, 8, x, y where x<yx<yx<y are 6, and 94\dfrac{9}{4}49​ respectively. Then x4+y2x^{4}+y^{2}x4+y2 is equal to
  1. (A)162
  2. (B)320
  3. (C)674
  4. (D)420

Correct answer: (B)

Step-by-step solution →
Q71·MathematicsSingle correctJEE Main 2022
The mean of the numbers a, b, 8, 5, 10 is 6 and their variance is 6.8. If M is the mean deviation of the numbers about the mean, then 25 M is equal to:
  1. (A)606060
  2. (B)555555
  3. (C)505050
  4. (D)454545

Correct answer: (A)

Step-by-step solution →
Q72·MathematicsSingle correctJEE Main 2022
The mean and standard deviation of 50 observations are 15 and 2 respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is 70. If the correct mean is 16, then the correct variance is equal to :
  1. (A)10
  2. (B)36
  3. (C)43
  4. (D)60

Correct answer: (C)

Step-by-step solution →
Q73·MathematicsNumericalJEE Main 2022
If the mean deviation about the mean of the numbers 1, 2, 3, ....., n, where n is odd, is 5(n+1)n\frac{5(n + 1)}{n}n5(n+1)​, then n is equal to ______

Correct answer: 21

Step-by-step solution →
Q74·MathematicsSingle correctJEE Main 2021
The mean and variance of 7 observations are 8 and 16 respectively. If two observations are 6 and 8, then the variance of the remaining 5 observations is :
  1. (A)925\frac{92}{5}592​
  2. (B)1345\frac{134}{5}5134​
  3. (C)53625\frac{536}{25}25536​
  4. (D)1125\frac{112}{5}5112​

Correct answer: (C)

Step-by-step solution →
Q75·MathematicsNumericalJEE Main 2021
The mean of 10 numbers 7×8, 10×10, 13×12, 16×14, ….7 \times 8,\ 10 \times 10,\ 13 \times 12,\ 16 \times 14,\ \ldots.7×8, 10×10, 13×12, 16×14, …. is ______.

Correct answer: 398

Step-by-step solution →
Q76·MathematicsNumericalJEE Main 2021
An online exam is attempted by 50 candidates out of which 20 are boys. The average marks obtained by boys is 12 with a variance 2. The variance of marks obtained by 30 girls is also 2. The average marks of all 50 candidates is 15. If μ is the average marks of girls and σ2\sigma^{2}σ2 is the variance of marks of 50 candidates, then μ+σ2\mu + \sigma^{2}μ+σ2 is equal to _________ .

Correct answer: 25

Step-by-step solution →
Q77·MathematicsNumericalJEE Main 2021
Let nnn be an odd natural number such that the variance of 1,2,3,4,...,n1, 2, 3, 4, ..., n1,2,3,4,...,n is 14. Then nnn is equal to ________.

Correct answer: 13

Step-by-step solution →
Q78·MathematicsSingle correctJEE Main 2021
The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. It was found that by mistake one data value was taken as 25 instead of 35. If α\alphaα and β\sqrt{\beta}β​ are the mean and standard deviation respectively for correct data, then (α,β)(\alpha, \beta)(α,β) is :
  1. (A)(11,26)(11, 26)(11,26)
  2. (B)(10.5,25)(10.5, 25)(10.5,25)
  3. (C)(11,25)(11, 25)(11,25)
  4. (D)(10.5,26)(10.5, 26)(10.5,26)

Correct answer: (D)

Step-by-step solution →
Q79·MathematicsNumericalJEE Main 2021
Let the mean and variance of four numbers 3, 7, x and y(x>y)y (x > y)y(x>y) be 5 and 10 respectively. Then the mean of four numbers 3+2x3 + 2x3+2x, 7+2y7 + 2y7+2y, x+yx + yx+y and x−yx - yx−y is ________.

Correct answer: 12

Step-by-step solution →
Q80·MathematicsSingle correctJEE Main 2021
Let the mean and variance of the frequency distribution x: x1_11​ = 2, x2_22​ = 6, x3_33​ = 8, x4_44​ = 9 f: 4, 4, α, β be 6 and 6.8 respectively. If x3_33​ is changed from 8 to 7, then the mean for the new data will be :
  1. (A)4
  2. (B)5
  3. (C)173\frac{17}{3}317​
  4. (D)163\frac{16}{3}316​

Correct answer: (C)

Step-by-step solution →
Q81·MathematicsSingle correctJEE Main 2021
If the mean and variance of the following data : 6,10,7,13,a,12,b,126, 10, 7, 13, a, 12, b, 126,10,7,13,a,12,b,12 are 999 and 374\frac{37}{4}437​ respectively, then (a−b)2\left(a-b\right)^2(a−b)2 is equal to :
  1. (A)32
  2. (B)24
  3. (C)12
  4. (D)16

Correct answer: (D)

Step-by-step solution →
Q82·MathematicsNumericalJEE Main 2021
Consider the following frequency distribution : Class : 10-20, 20-30, 30-40, 40-50, 50-60; Frequency : α\alphaα, 110, 54, 30, β\betaβ. If the sum of all frequencies is 584 and median is 45, then ∣α−β∣\left| \alpha - \beta \right|∣α−β∣ is equal to.......

Correct answer: 164

Step-by-step solution →
Q83·MathematicsSingle correctJEE Main 2021
The first of the two samples in a group has 100 items with mean 15 and standard deviation 3. If the whole group has 250 items with mean 15.6 and standard deviation 13.44\sqrt{13.44}13.44​, then the standard deviation of the second sample is :
  1. (A)4
  2. (B)6
  3. (C)5
  4. (D)8

Correct answer: (A)

Step-by-step solution →
Q84·MathematicsNumericalJEE Main 2021
Consider the following frequency distribution : Class : 0 – 6 6 – 12 12 – 18 18 – 24 24 – 30 Frequency : a b 12 9 5 If mean = 30922\frac{309}{22}22309​ and median = 14, then the value (a−b)2(a - b)^2(a−b)2 is equal to..................

Correct answer: 4

Step-by-step solution →
Q85·MathematicsSingle correctJEE Main 2021
The mean of 6 distinct observations is 6.5 and their variance is 10.25. If 4 out of 6 observations are 2, 4, 5 and 7, then the remaining two observations are :
  1. (A)10, 11
  2. (B)3, 18
  3. (C)1, 20
  4. (D)8, 13

Correct answer: (A)

Step-by-step solution →
Q86·MathematicsSingle correctJEE Main 2021
If the mean and variance of six observations 7, 10, 11, 15, a, b7,\ 10,\ 11,\ 15,\ a,\ b7, 10, 11, 15, a, b are 101010 and 203\frac{20}{3}320​, respectively, then the value of ∣a−b∣|a-b|∣a−b∣ is equal to
  1. (A)999
  2. (B)777
  3. (C)111111
  4. (D)111

Correct answer: (D)

Step-by-step solution →
Q87·MathematicsSingle correctJEE Main 2021
Let in a series of 2n observations, half of them are equal to a and remaining half are equal to −a. Also by adding a constant b in each of these observations, the mean and standard deviation of new set become 5 and 20, respectively. Then the value of a2+b2a^{2} + b^{2}a2+b2 is equal to :
  1. (A)425
  2. (B)650
  3. (C)250
  4. (D)925

Correct answer: (A)

Step-by-step solution →
Q88·MathematicsNumericalJEE Main 2021
The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If the mean age of the teachers in this school now is 39 years, then the age (in years) of the newly appointed teacher is_.

Correct answer: 35

Step-by-step solution →
Q89·MathematicsNumericalJEE Main 2021
Consider a set of 3n3n3n numbers having variance 4. In this set, the mean of first 2n2n2n numbers is 6 and the mean of the remaining nnn numbers is 3. A new set is constructed by adding 1 into each of first 2n2n2n numbers, and subtracting 1 from each of the remaining nnn numbers. If the variance of the new set is kkk, then 9k9k9k is equal to ______.

Correct answer: 68

Step-by-step solution →
Q90·MathematicsNumericalJEE Main 2021
Consider the statistics of two sets of observations as follows : Size Mean Variance Observation I 10 2 2 Observation II n 3 1 If the variance of the combined set of these two observations is 179\frac{17}{9}917​, then the value of n is equal to _________.

Correct answer: 5

Step-by-step solution →
Q91·MathematicsSingle correctJEE Main 2021
Consider three observations a, b and c such that b = a + c. If the standard deviation of a + 2, b + 2, c + 2 is d, then which of the following is true ?
  1. (A)b2=3(a2+c2)+9d2b^{2} = 3(a^{2} + c^{2}) + 9d^{2}b2=3(a2+c2)+9d2
  2. (B)b2=a2+c2+3d2b^{2} = a^{2} + c^{2} + 3d^{2}b2=a2+c2+3d2
  3. (C)b2=3(a2+c2+d2)b^{2} = 3(a^{2} + c^{2} + d^{2})b2=3(a2+c2+d2)
  4. (D)b2=3(a2+c2)−9d2b^{2} = 3(a^{2} + c^{2}) - 9d^{2}b2=3(a2+c2)−9d2

Correct answer: (D)

Step-by-step solution →
Q92·MathematicsNumericalJEE Main 2021
Let X1X_{1}X1​, X2X_{2}X2​,……….. X18X_{18}X18​ be eighteen observation such that ∑i=118(Xi−α)=36\sum_{i=1}^{18}\left(X_{i}-\alpha\right) = 36∑i=118​(Xi​−α)=36 and ∑i=118(Xi−β)2=90\sum_{i=1}^{18}\left(X_{i}-\beta\right)^{2} = 90∑i=118​(Xi​−β)2=90, where α\alphaα and β\betaβ are distinct real numbers. If the standard deviation of these observations is 1, then the value of ∣α−β∣|\alpha - \beta|∣α−β∣ is ________________.

Correct answer: 4

Step-by-step solution →
Q93·MathematicsNumericalJEE Main 2021
If the variance of 10 natural numbers 1,1,1,…,1,k1,1,1,\ldots,1,k1,1,1,…,1,k is less than 10, then the maximum possible value of kkk is__________.

Correct answer: 11

Step-by-step solution →
Q94·MathematicsSingle correctJEE Main 2020
If ∑i=1n(xi−a)=n\displaystyle\sum_{i=1}^{n}(x_i-a) = ni=1∑n​(xi​−a)=n and ∑i=1n(xi−a)2=na, (n,a>1)\displaystyle\sum_{i=1}^{n}(x_i-a)^2 = na,\ (n,a>1)i=1∑n​(xi​−a)2=na, (n,a>1) then the standard deviation of n observations x1,x2,…,xnx_1, x_2, \ldots, x_nx1​,x2​,…,xn​ is:
  1. (A)a−1a-1a−1
  2. (B)na−1n\sqrt{a-1}na−1​
  3. (C)n(a−1)\sqrt{n(a-1)}n(a−1)​
  4. (D)a−1\sqrt{a-1}a−1​

Correct answer: (D)

Step-by-step solution →
Q95·MathematicsNumericalJEE Main 2020
Consider the data on x taking the values 0,2,4,8,…,2n0, 2, 4, 8, \ldots, 2^n0,2,4,8,…,2n with frequencies nC0^{n}C_0nC0​, nC1^{n}C_1nC1​, nC2^{n}C_2nC2​, …\ldots…, nCn^{n}C_nnCn​ respectively. If the mean of this data is 7282n\frac{728}{2^n}2n728​, then n is equal to __________.

Correct answer: 6

Step-by-step solution →
Q96·MathematicsSingle correctJEE Main 2020
The mean and variance of 7 observations are 8 and 16, respectively. If five observations are 2, 4, 10, 12, 14, then the absolute difference of the remaining two observations is:
  1. (A)2
  2. (B)4
  3. (C)1
  4. (D)3

Correct answer: (A)

Step-by-step solution →
Q97·MathematicsSingle correctJEE Main 2020
If the mean and standard deviation of the data 3, 5, 7, a, b are 5 and 2 respectively, then a and b are the roots of the equation:
  1. (A)x2−10x+19=0x^2 - 10x + 19 = 0x2−10x+19=0
  2. (B)x2−20x+18=0x^2 - 20x + 18 = 0x2−20x+18=0
  3. (C)x2−10x+18=0x^2 - 10x + 18 = 0x2−10x+18=0
  4. (D)2x2−20x+19=02x^2 - 20x + 19 = 02x2−20x+19=0

Correct answer: (A)

Step-by-step solution →
Q98·MathematicsNumericalJEE Main 2020
If the variance of the following frequency distribution : is 50, then x is equal to:
Class10 – 2020 – 3030 – 40
Frequency2x2

Correct answer: 4

Step-by-step solution →
Q99·MathematicsSingle correctJEE Main 2020
The mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations are 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is:
  1. (A)333
  2. (B)555
  3. (C)777
  4. (D)999

Correct answer: (C)

Step-by-step solution →
Q100·MathematicsSingle correctJEE Main 2020
Let xi(1≤i≤10)x_i \left( 1 \le i \le 10 \right)xi​(1≤i≤10) be ten observations of a random variable X. If ∑i=110(xi−p)=3\sum_{i=1}^{10} \left( x_i - p \right) = 3∑i=110​(xi​−p)=3 and ∑i=110(xi−p)2=9\sum_{i=1}^{10} \left( x_i - p \right)^2 = 9∑i=110​(xi​−p)2=9 where 0≠p∈R0 \ne p \in R0=p∈R, then the standard deviation of these observations is:
  1. (A)35\sqrt{\frac{3}{5}}53​​
  2. (B)45\frac{4}{5}54​
  3. (C)710\frac{7}{10}107​
  4. (D)910\frac{9}{10}109​

Correct answer: (D)

Step-by-step solution →
Q101·MathematicsSingle correctJEE Main 2020
For the frequency distribution: Variate (x): x1x_{1}x1​ x2x_{2}x2​ x3x_{3}x3​ ....x15x_{15}x15​ Frequency (f): f1f_{1}f1​ f2f_{2}f2​ f3f_{3}f3​ ....f15f_{15}f15​ Where 0<x1<x2<x3<…<x15=100<x_{1}<x_{2}<x_{3}<\ldots<x_{15}=100<x1​<x2​<x3​<…<x15​=10 and ∑i=115fi>0\sum\limits_{i=1}^{15}f_{i}>0i=1∑15​fi​>0, the standard deviation cannot be:
  1. (A)4
  2. (B)2
  3. (C)6
  4. (D)1

Correct answer: (C)

Step-by-step solution →
Q102·MathematicsSingle correctJEE Main 2020
Let the observations xix_ixi​ (1≤i≤10)(1\le i\le10)(1≤i≤10) satisfy the equation ∑i=110(xi−5)=10\displaystyle\sum_{i=1}^{10}(x_i-5)=10i=1∑10​(xi​−5)=10 and ∑i=110(xi−5)2=40\displaystyle\sum_{i=1}^{10}(x_i-5)^2=40i=1∑10​(xi​−5)2=40. If μ\muμ and λ\lambdaλ are the mean and the variance of the observations, x1−3,x2−3,……………,x10−3x_1-3,x_2-3,……………,x_{10}-3x1​−3,x2​−3,……………,x10​−3, then the ordered pair (μ,λ)(\mu,\lambda)(μ,λ) is equal to:
  1. (A)(6, 6)
  2. (B)(3, 3)
  3. (C)(3, 6)
  4. (D)(6, 3)

Correct answer: (B)

Step-by-step solution →
Q103·MathematicsSingle correctJEE Main 2020
The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11. Then the correct variance is:
  1. (A)4.02
  2. (B)3.98
  3. (C)4.01
  4. (D)3.99

Correct answer: (D)

Step-by-step solution →
Q104·MathematicsNumericalJEE Main 2020
If the mean and variance of eight numbers 3, 7, 9, 12, 13, 20, x and y be 10 and 25 respectively, then xy is equal to __________.

Correct answer: 54

Step-by-step solution →
Q105·MathematicsNumericalJEE Main 2020
If the variance of the first n natural numbers is 10 and the variance of the first m even natural numbers is 16, then m + n is equal to _________

Correct answer: 18

Step-by-step solution →
Q106·MathematicsSingle correctJEE Main 2019
If the data x1,x2,....,x10x_{1}, x_{2}, ...., x_{10}x1​,x2​,....,x10​ is such that the mean of first four of these is 11, the mean of the remaining six is 16 and the sum of squares of all of these is 2,000; then the standard deviation of this data is :
  1. (A)222\sqrt{2}22​
  2. (B)2
  3. (C)4
  4. (D)2\sqrt{2}2​

Correct answer: (B)

Step-by-step solution →
Q107·MathematicsSingle correctJEE Main 2019
If for some x ∈\in∈ R, the frequency distribution of the marks obtained by 20 students in a test is Then the mean of the marks is:
Marks2357
Frequency(x+1)2(x+1)^{2}(x+1)22x − 5x2−3xx^{2}-3xx2−3xx
  1. (A)2.8
  2. (B)3.2
  3. (C)2.5
  4. (D)3.0

Correct answer: (A)

Step-by-step solution →
Q108·MathematicsSingle correctJEE Main 2019
If both the means and the standard deviation of 50 observations x1x_1x1​, x2x_2x2​, ………, x50x_{50}x50​ are equal to 16, then the mean of (x1−4)2(x_1 - 4)^{2}(x1​−4)2, (x2−4)2(x_2 - 4)^{2}(x2​−4)2, …., (x50−4)2(x_{50} - 4)^{2}(x50​−4)2 is
  1. (A)400
  2. (B)380
  3. (C)525
  4. (D)480

Correct answer: (A)

Step-by-step solution →
Q109·MathematicsSingle correctJEE Main 2019
If the standard deviation of the numbers −1,0,1,k-1,0,1,k−1,0,1,k is 5\sqrt{5}5​ where k>0k>0k>0, then k is equal to:
  1. (A)4534\sqrt{\dfrac{5}{3}}435​​
  2. (B)6\sqrt{6}6​
  3. (C)262\sqrt{6}26​
  4. (D)21032\sqrt{\dfrac{10}{3}}2310​​

Correct answer: (C)

Step-by-step solution →
Q110·MathematicsSingle correctJEE Main 2019
The mean and the median of the following ten numbers in increasing order 10, 22, 26, 29, 34, x, 42, 67, 70, y are 42 and 35 respectively, then yx\dfrac{y}{x}xy​ is equal to
  1. (A)73\dfrac{7}{3}37​
  2. (B)94\dfrac{9}{4}49​
  3. (C)72\dfrac{7}{2}27​
  4. (D)83\dfrac{8}{3}38​

Correct answer: (A)

Step-by-step solution →
Q111·MathematicsSingle correctJEE Main 2019
A student score the following marks in five tests : 45, 54, 41, 57, 43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is:
  1. (A)103\frac{10}{3}310​
  2. (B)1003\frac{100}{3}3100​
  3. (C)1003\frac{100}{\sqrt{3}}3​100​
  4. (D)103\frac{10}{\sqrt{3}}3​10​

Correct answer: (D)

Step-by-step solution →
Q112·MathematicsSingle correctJEE Main 2019
The mean and variance of seven observations are 8 and 16, respectively. If 5 of the observations are 2, 4, 10, 12, 14, then the product of the remaining two observations is:
  1. (A)40
  2. (B)45
  3. (C)49
  4. (D)48

Correct answer: (D)

Step-by-step solution →
Q113·MathematicsSingle correctJEE Main 2019
If the sum of the deviations of 50 observations from 30 is 50, then the mean of these observations is:
  1. (A)30
  2. (B)51
  3. (C)50
  4. (D)31

Correct answer: (D)

Step-by-step solution →
Q114·MathematicsSingle correctJEE Main 2019
The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4; then the absolute value of the difference of the other two observations, is :
  1. (A)7
  2. (B)5
  3. (C)1
  4. (D)3

Correct answer: (A)

Step-by-step solution →
Q115·MathematicsSingle correctJEE Main 2019
The outcome of each of 30 items was observed; 10 items gave an outcome 12−d\dfrac{1}{2}-d21​−d each, 10 items gave outcome 12\dfrac{1}{2}21​ each and the remaining 10 items gave outcome 12+d\dfrac{1}{2}+d21​+d each. If the variance of this outcome data is 43\dfrac{4}{3}34​ then ∣d∣|d|∣d∣ equals:
  1. (A)23\dfrac{2}{3}32​
  2. (B)2
  3. (C)52\dfrac{\sqrt{5}}{2}25​​
  4. (D)2\sqrt{2}2​

Correct answer: (D)

Step-by-step solution →
Q116·MathematicsSingle correctJEE Main 2019
The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is:
  1. (A)10 : 3
  2. (B)4 : 9
  3. (C)5 : 8
  4. (D)6 : 7

Correct answer: (B)

Step-by-step solution →
Q117·MathematicsSingle correctJEE Main 2019
A data consists of n observations: x1,x2,......,xnx_{1},x_{2},......,x_{n}x1​,x2​,......,xn​. If ∑i=1n(xi+1)2=9n\displaystyle\sum_{i=1}^{n}(x_{i}+1)^{2}=9ni=1∑n​(xi​+1)2=9n and ∑i=1n(xi−1)2=5n\displaystyle\sum_{i=1}^{n}(x_{i}-1)^{2}=5ni=1∑n​(xi​−1)2=5n, then the standard deviation of this data is:
  1. (A)5
  2. (B)5\sqrt{5}5​
  3. (C)7\sqrt{7}7​
  4. (D)2

Correct answer: (B)

Step-by-step solution →
Q118·MathematicsSingle correctJEE Main 2019
5 students of a class have an average height 150 cm and variance 18 cm2cm^{2}cm2. A new student, whose height is 156 cm, joined them. The variance (in cm2cm^{2}cm2) of the height of these six students is:
  1. (A)16
  2. (B)22
  3. (C)20
  4. (D)18

Correct answer: (C)

Step-by-step solution →

Statistics — frequently asked

How many questions from Statistics appear in JEE?

Statistics has appeared in 118 of the last 186 JEE Main and JEE Advanced papers — about 63% of them — contributing 118 questions in total across those papers.

Is Statistics an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 63% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Statistics questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Mathematics chapters

  • Three Dimensional Geometry 344
  • Matrices and Determinants 342
  • Sets, Relations and Functions 313
  • Sequence and Series 287
  • Definite Integration 267
  • Vector Algebra 245
  • Differential Equations 240
  • Probability 226

All 26 Mathematics chapters →

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