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Statistics

JEE Main 2026 (Online) 6th April Evening ShiftMCQMedium📈 Topic in 45% papers+4−1
Ques. 1JEE Main 2026, 6 April (Evening Shift)

Let the mean and the variance of seven observations 2, 4, , 8, , 12, 14, < , be 8 and 16 respectively. Then the quadratic equation whose roots are 3 +2 and 2 +1 is :

  1. x2-35x+306=0
  2. x2-41x+420=0
  3. x2-45x+506=0
  4. x2-37x+342=0

Answer

(2) x2-41x+420=0

Solution

Mean = 8, so 2 + 4 + + 8 + + 12 + 147 = 8

40 + + = 56 + = 16

Variance = 16, so xi27 - 82 = 16 xi2 = 560

4 + 16 + 64 + 144 + 196 + 2 + 2 = 560 2 + 2 = 136

2 = ( + )2 - ( 2 + 2 ) = 256 - 136 = 120 = 60

So α, β are the roots of t2 - 16t + 60 = 0, i.e. t = 6, 10, and < , = 6, = 10.

The required roots are 3 + 2 = 20 and 2 + 1 = 21, with sum = 41 and product = 420.

Required equation: x2 - 41x + 420 = 0

Answer: (2)

Keep practising — Statistics

Ques. 2JEE Main 2026, 5 April (Morning Shift)

The mean deviation about the mean for the data arraycccccccxi & 5 & 7 & 9 & 10 & 12 & 15\ fi & 8 & 6 & 2 & 2 & 2 & 6 array is equal to:

  1. 4013
  2. 4213
  3. 4413
  4. 4613
Ques. 3JEE Main 2026, 5 April (Evening Shift)

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
Ques. 4JEE Main 2026, 4 April (Morning Shift)

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
Ques. 5JEE Main 2026, 4 April (Evening Shift)

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
Ques. 6JEE Main 2026, 2 April (Morning Shift)

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
Ques. 7JEE Main 2026, 2 April (Evening Shift)

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
Ques. 8JEE Main 2026, 28 January (Morning Shift)

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
Ques. 9JEE Main 2026, 24 January (Morning Shift)

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
Ques. 10JEE Main 2026, 24 January (Evening Shift)

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
Ques. 11JEE Main 2026, 23 January (Morning Shift)

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

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