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Statistics

JEE Main 2026 (Online) 24th January Morning ShiftMCQEasy📈 Topic in 45% papers+4−1
Ques. 1JEE 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

Answer

(4) 20

Solution

For the original data, xi = 10 10 = 100, and the variance gives

xi210 - 102 = 2 xi2 = 1020

For the new data, xi' = 10 10.1 = 101, so

100 - + = 101 - = 1

The new variance gives xi'210 - (10.1)2 = 1.99, so

xi'210 = 102.01 + 1.99 = 104 xi'2 = 1040

1020 - 2 + 2 = 1040 2 - 2 = 20

( - )( + ) = 20

+ = 201 = 20

Answer: (4)

Keep practising — Statistics

Ques. 2JEE 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. 3JEE 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. 4JEE 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. 5JEE 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. 6JEE 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
Ques. 7JEE Main 2025, 7 April (Morning Shift)

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 and respectively, then 10( + ) is equal to

  1. 447
  2. 445
  3. 449
  4. 451
Ques. 8JEE Main 2025, 4 April (Evening Shift)

Let the mean and the standard deviation of the observations 2, 3, 3, 4, 5, 7, a, b be 4 and 2 respectively. Then the mean deviation about the mode of these observations is :

  1. 12
  2. 34
  3. 1
  4. 2
Ques. 9JEE Main 2025, 3 April (Evening Shift)

Let the Mean and Variance of five observations x1 = 1, x2 = 3, x3 = a, x4 = 7 and x5 = b, a > b, be 5 and 10 respectively. Then the Variance of the observations n + xn, n = 1, 2, , 5 is

  1. 17
  2. 16
  3. 16.4
  4. 17.4
Ques. 10JEE Main 2025, 2 April (Evening Shift)

If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then a + b + ab is equal to :

  1. 103
  2. 106
  3. 100
  4. 105
Ques. 11JEE Main 2025, 29 January (Morning Shift)

Let x1, x2, , x10 be ten observations such that i=110(xi - 2) = 30, i=110(xi - )2 = 98, > 2, and their variance is 45. If and 2 are respectively the mean and the variance of 2(x1 - 1) + 4 , 2(x2 - 1) + 4 , , 2(x10 - 1) + 4 , then 2 is equal to :

  1. 100
  2. 90
  3. 120
  4. 110

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