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Statistics

JEE Main 2026 (Online) 24th January Evening ShiftMCQEasy📈 Topic in 42% papers+4−1
Ques. 1JEE 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

Answer

(4) 60

Solution

X = 1, 2, …, 19: mean x = 10

Variance of X = x219 - 102 = 19 20 396 19 - 100 = 130 - 100 = 30

For Y = aX + b: mean = 10a + b = 30 and variance = a2 30 = 750

⇒ a2 = 25 ⇒ a = 5

a = 5: b = 30 - 50 = -20; a = -5: b = 30 + 50 = 80

∴ Sum of all possible values of b = -20 + 80 = 60

Answer: (4)

Keep practising — Statistics

Ques. 2JEE 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. 3JEE 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. 4JEE 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. 5JEE 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. 6JEE 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. 7JEE 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. 8JEE 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. 9JEE 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. 10JEE 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
Ques. 11JEE Main 2025, 24 January (Morning Shift)

For a statistical data x1, x2, , x10 of 10 values, a student obtained the mean as 5.5 and i=110xi2=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. 5
  2. 7
  3. 9
  4. 4

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