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Statistics

JEE Main 2025 (Online) 24th January Morning ShiftMCQEasy📈 Topic in 42% papers+4−1
Ques. 1JEE 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

Answer

(2) 7

Solution

Only the sums change: x=55-(4+5)+(6+8)=60 mean =6

x2=371-(16+25)+(36+64)=430

2= 43010-62=7

Answer: (2)

Keep practising — Statistics

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

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

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