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Statistics

JEE Main 2025 (Online) 7th April Morning ShiftMCQMedium📈 Topic in 42% papers+4−1
Ques. 1JEE 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

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)

Keep practising — Statistics

Ques. 2JEE 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. 3JEE 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. 4JEE 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. 5JEE 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. 6JEE 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
Ques. 7JEE Main 2025, 23 January (Morning Shift)

Marks obtains by all the students of class 12 are presented in a freqency distribution with classes of equal width. Let the median of this grouped data be 14 with median class interval 12-18 and median class frequency 12. If the number of students whose marks are less than 12 is 18 , then the total number of students is :

  1. 52
  2. 44
  3. 40
  4. 48
Ques. 8JEE Main 2025, 23 January (Evening Shift)

The variance of the numbers 8, 21, 34, 47, , 320 is .

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