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Statistics · Question 12

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

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)

JEE Main 2026 (Online) 24th January Evening Shift

Keep practising — Statistics

2

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
3

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
4

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
5

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
JEE Main 2026 (Online) 21st January Morning Shift
6

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
JEE Main 2025 (Online) 7th April Morning Shift
7

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
JEE Main 2025 (Online) 4th April Evening Shift
8

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
JEE Main 2025 (Online) 3rd April Evening Shift
9

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
JEE Main 2025 (Online) 2nd April Evening Shift
10

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
JEE Main 2025 (Online) 29th January Morning Shift

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