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Statistics · Question 5

JEE Main 2026 (Online) 5th April Evening ShiftMCQHard📈 Topic in 6% papers+4−1
1

A variable X takes values 0, 0, 2, 6, 12, 20, , n(n-1) with frequencies nC0, nC1, nC2, nC3, nC4, nC5, , nCn, respectively. If the mean of this data is 60, then its median is :

  1. 56
  2. 42
  3. 72
  4. 90

Answer

(1) 56

Solution

The values are k(k-1) with frequency nCk, k=0, 1, , n; total frequency N= nCk=2n.

k(k-1) nCk=n(n-1) n-2Ck-2

k(k-1) nCk=n(n-1) 2n-2

Mean = n(n-1)2n-22n= n(n-1)4=60 n(n-1)=240 n=16

N=216=65536 (even), values in increasing order median = mean of the 32768th and 32769th observations

Cumulative frequency up to k=7: k=0716Ck= 216-16C82= 65536-128702=26333<32768

Up to k=8: 26333+12870=39203>32769

both middle observations correspond to k=8

Median =8 7=56

Answer: (1)

JEE Main 2026 (Online) 5th April Evening Shift

Keep practising — Statistics

2

Suppose that the mean and median of the non-negative numbers 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 2a is equal to :

  1. 109
  2. 117
  3. 161
  4. 131
JEE Main 2026 (Online) 4th April Morning Shift
3

For 10 observations x1, x2, , x10, if i=110 (xi+2 )2=180 and i=110 (xi-1 )2=90, then their standard deviation is :

  1. 2
  2. 3
  3. 2 2
  4. 3
JEE Main 2026 (Online) 4th April Evening Shift
4

If the mean of the data [figure] is 21, then k is one of the roots of the equation :

  1. 2x2-23x-10=0
  2. 4x2-35x+24=0
  3. 2x2-19x-10=0
  4. 2x2-35x+98=0
JEE Main 2026 (Online) 2nd April Morning Shift
5

The mean and variance of n observations are 8 and 16, respectively. If the sum of the first (n-1) observations is 48 and the sum of squares of the first (n-1) observations is 496, then the value of n is :

  1. 21
  2. 16
  3. 13
  4. 7
JEE Main 2026 (Online) 2nd April Evening Shift
6

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. 7
  2. 4
  3. 5
  4. 6
JEE Main 2026 (Online) 28th January Morning Shift
7

The mean and variance of a data of 10 observations are 10 and 2, respectively. If an observation in this data is replaced by , then the mean and variance become 10.1 and 1.99, respectively. Then + equals

  1. 15
  2. 10
  3. 5
  4. 20
JEE Main 2026 (Online) 24th January Morning Shift
8

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

  1. 20
  2. 100
  3. 80
  4. 60
JEE Main 2026 (Online) 24th January Evening Shift
9

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 :

  1. 11
  2. 9
  3. 10
  4. 12
JEE Main 2026 (Online) 23rd January Morning Shift
10

If the mean and the variance of the data array|l|c|c|c|c| Class & 4-8 & 8-12 & 12-16 & 16-20 \ Frequency & 3 & & 4 & 7 \ array are and 19 respectively, then the value of + is :

  1. 21
  2. 19
  3. 18
  4. 20
JEE Main 2026 (Online) 23rd January Evening Shift
11

If the mean deviation about the median of the numbers k, 2k, 3k, , 1000k is 500, then k2 is equal to :

  1. 1
  2. 16
  3. 9
  4. 4
JEE Main 2026 (Online) 22nd January Evening Shift

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