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Statistics

JEE Main 2025 (Online) 2nd April Evening ShiftMCQEasy📈 Topic in 42% papers+4−1
Ques. 1JEE 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

Answer

(1) 103

Solution

Number of observations = 8

Mean: 6 + 4 + a + 8 + b + 12 + 10 + 138 = 9

53 + a + b = 72 a + b = 19

Variance: xi28 - 92 = 9.25

xi2 = 8 90.25 = 722

36 + 16 + 64 + 144 + 100 + 169 + a2 + b2 = 722

a2 + b2 = 722 - 529 = 193

2ab = (a + b)2 - (a2 + b2) = 361 - 193 = 168 ab = 84

a + b + ab = 19 + 84 = 103

Answer: (1)

Keep practising — Statistics

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

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

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