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Statistics

JEE Main 2026 (Online) 4th April Morning ShiftMCQMedium📈 Topic in 14% papers+4−1
Ques. 1JEE 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

Answer

(4) 131

Solution

There are 9 numbers with mean 40, so their total is 360.

The known numbers add up to 21 + 8 + 17 + 51 + 103 + 13 + 67 = 280, hence a + b = 80.

The mean deviation about the median 21 is 26, so |xi - 21| = 9 26 = 234.

For the known numbers this sum is 0 + 13 + 4 + 30 + 82 + 8 + 46 = 183, which leaves

|a - 21| + |b - 21| = 51

If a, b 21, then a + b - 42 = 38 51; if a, b<21, then a + b<42, which is impossible. Hence b<21 a, and

(a - 21) + (21 - b) = 51 a - b = 51

Together with a + b = 80 this gives a = 1312 and b = 292.

Check of the median: 8, 13, 14.5, 17, 21, 51, 65.5, 67, 103 — the 5th value is 21, as required.

∴ 2a = 131

Answer: (4)

Keep practising — Statistics

Ques. 2JEE 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. 3JEE 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. 4JEE 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. 5JEE 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. 6JEE 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. 7JEE 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. 8JEE 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
Ques. 9JEE Main 2026, 23 January (Evening Shift)

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
Ques. 10JEE Main 2026, 22 January (Evening Shift)

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
Ques. 11JEE Main 2026, 21 January (Morning Shift)

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 :

  1. 45
  2. 35
  3. 25
  4. 13

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