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Statistics · Question 17

JEE Main 2025 (Online) 7th April Morning ShiftMCQMedium📈 Topic in 45% papers+4−1
1

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

Answer

(3) 449

Solution

Incorrect: x = 100 40 = 4000

x2 = 100 (5.12 + 402 ) = 100(26.01 + 1600) = 162601

Correct: x = 4000 - 50 + 40 = 3990 = 39.9

x2 = 162601 - 2500 + 1600 = 161701

2 = 161701100 - (39.9)2 = 1617.01 - 1592.01 = 25 = 5

10( + ) = 10(39.9 + 5) = 449

Answer: (3)

JEE Main 2025 (Online) 7th April Morning Shift

Keep practising — Statistics

2

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
3

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
4

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
5

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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