Math, asked by shaikhtoufik1522, 9 months ago

1: The marks of 19 students are given below:
41, 21, 38, 27,31,45,23,26,29,30,28,25,35,
42,47,50,29,31,35.
Calculate all the quartiles for the above
data.​

Answers

Answered by mysticangel8828
3

Answer:

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Step-by-step explanation:

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Answered by Devkumarr
0

Answer:

The lower , middle and upper quartile of the given data are  27 , 31, 41 respectively

Step-by-step explanation:

  • In context to the given question, we have to find the value of all the quartile values of the given data.
  • given data:

41,21,38,27,31,45,23,26,29,30,28,25,35,42,47,50,29,31,35.

Firstly by arranging the given data is ascending order;

21,23,25,26,27,28,29,29,30,31,31,35,35,38,41,42,45,47,50

There are three quartile namely, upper quartile , middle or median quartile and lower quartile

  • First or Lower Quartile (Q1) divides the distribution in such a way that one-fourth (25\%) of total items fall below it and three-fourth (75%) fall above it.

  • Middle or (Q2) or Median: divides the distribution in such a way that half (50%) of total items fall below it and half (50\%) above it.

  • Upper Quartile (Q3):Q divides the distribution in such a way that three fourth (75%) of total items fall below it and one- fourth (25%) fall above it.

Middle quartile = [N(t h)+1]/2

where n is the total no of terms

middle quartile =  [19+1]/ 2

Middle quartile = 20/2 = 10th term= 31

Middle quartile = 31

Now by finding, lower quartile

Lower quartile =  [N(t h)+1] (1/4)

where n is the total no of terms

Lower quartile =  [19+1]/ 4

Lower quartile = 20/4 = 5th term= 27

Lower quartile = 27

Lastly by finding Upper quartile

Upper quartile = [N(t h)+1] (3/4)

where n is the total no of terms

Upper quartile =[19+1] (3/4)

Upper quartile =60/4 = 15th term = 41

Upper quartile =41

Therefore, The lower , middle and upper quartile of the given data are  27 , 31, 41 respectively.

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