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Statistics

JEE Main 2025 (Online) 3rd April Evening ShiftMCQMedium📈 Topic in 45% papers+4−1
Ques. 1JEE 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

Answer

(2) 16

Solution

Mean = 5: 1 + 3 + a + 7 + b = 25 a + b = 14

Variance = 10: xi25 - 52 = 10 xi2 = 175

1 + 9 + a2 + 49 + b2 = 175 a2 + b2 = 116

2ab = 142 - 116 = 80 ab = 40

so a = 10, b = 4 (as a > b).

The new observations n + xn are 2, 5, 13, 11, 9, with mean = 405 = 8 and

y2 = 4 + 25 + 169 + 121 + 81 = 400

Variance = 4005 - 82 = 80 - 64 = 16

Answer: (2)

Keep practising — Statistics

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

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

Ques. 7JEE Main 2024, 9 April (Morning Shift)

The frequency distribution of the age of students in a class of 40 students is given below. [figure] If the mean deviation about the median is 1.25, then 4x + 5y is equal to :

  1. 43
  2. 46
  3. 44
  4. 47
Ques. 8JEE Main 2024, 9 April (Evening Shift)

If the variance of the frequency distribution array|c|c|c|c|c|c|c| x & c & 2c & 3c & 4c & 5c & 6c \ f & 2 & 1 & 1 & 1 & 1 & 1 \ array is 160, then the value of c N is

  1. 5
  2. 8
  3. 6
  4. 7
Ques. 9JEE Main 2024, 8 April (Evening Shift)

Let a, b, c N and a < b < c. Let the mean, the mean deviation about the mean and the variance of the 5 observations 9, 25, a, b, c be 18, 4 and 1365, respectively. Then 2a + b - c is equal to

Ques. 10JEE Main 2024, 6 April (Morning Shift)

The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is

  1. 1.94
  2. 3.96
  3. 3.86
  4. 1.8
Ques. 11JEE Main 2024, 4 April (Morning Shift)

Let , R. Let the mean and the variance of 6 observations -3, 4, 7,-6, , be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is :

  1. 163
  2. 113
  3. 143
  4. 133

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