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Statistics

JEE Main 2024 (Online) 9th April Morning ShiftMCQEasy📈 Topic in 14% papers+4−1
Ques. 1JEE 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

Answer

(3) 44

Solution

Total = 40 30 + x + y = 40 x + y = 10 …(i)

Cumulative frequencies: 5, 13, 18, 30, … so the 20th and 21st observations are both 18, hence median M = 18

M.D. = fi|xi - 18|40 = 5(3) + 8(2) + 5(1) + 12(0) + x(1) + y(2)40 = 1.25

36 + x + 2y = 50 x + 2y = 14 …(ii)

From (i) and (ii): y = 4, x = 6, so 4x + 5y = 24 + 20 = 44

Answer: (3)

Keep practising — Statistics

Ques. 2JEE 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. 3JEE 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. 4JEE 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. 5JEE 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
Ques. 6JEE Main 2024, 1 February (Morning Shift)

Let the median and the mean deviation about the median of 7 observations 170, 125, 230, 190, 210, a, b be 170 and 2057 respectively. Then the mean deviation about the mean of these 7 observations is :

  1. 31
  2. 28
  3. 30
  4. 32
Ques. 7JEE Main 2024, 1 February (Evening Shift)

Consider 10 observations x1, x2, , x10 such that i=110 (xi- )=2 and i=110 (xi- )2=40, where , are positive integers. Let the mean and the variance of the observations be 65 and 8425 respectively. Then is equal to :

  1. 2
  2. 1
  3. 52
  4. 32
Ques. 8JEE Main 2024, 31 January (Evening Shift)

Let the mean and the variance of 6 observations a, b, 68, 44, 48, 60 be 55 and 194, respectively. If a>b, then a+3b is

  1. 180
  2. 210
  3. 190
  4. 200
Ques. 9JEE Main 2024, 30 January (Morning Shift)

Let M denote the median of the following frequency distribution [figure] Then 20M is equal to :

  1. 104
  2. 52
  3. 208
  4. 416
Ques. 10JEE Main 2024, 30 January (Evening Shift)

The variance 2 of the data [figure] is .

Ques. 11JEE Main 2024, 29 January (Morning Shift)

If the mean and variance of the data 65,68,58,44,48,45,60, , ,60 where > , are 56 and 66.2 respectively, then 2+ 2 is equal to .

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