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SUMAN MATHEWS
GRADE 11 MATH
COLLEGE MATH
PREVIEW
THIS VIDEO WOULD BE USEFUL FOR GRADE 11 STUDENTS STUDYING MATHEMATICS.
IT WOULD ALSO BE USEFUL FOR STUDENTS AT THE COLLEGE LEVEL STUDYING STATISTICS.
HERE WE SHOW HOW QUARTILES, DECILES AND PERCENTILES CAN BE CALCULATED
.
CALCULATION OF THE ABOVE IS DONE FOR DISCRETE, SIMPLE AND GROUPED
FREQUENCY DISTRIBUTIONS
IF A GIVEN SET IS ARRANGED IN ASCENDING OR DESCENDING ORDER OF MAGNITUDE,
 MEDIAN DIVIDES THE SET INTO 2 EQUAL PARTS.
 QUARTILES DIVIDE THE SET INTO 4 QUARTERS. THERE ARE 3 QUARTILES 1, 2, 3
 DECILES DIVIDE THE SET INTO 10 EQUAL PARTS. THERE ARE 9 DECILES, 1, 2,------9
PERCENTILES DIVIDE THE SET INTO 100 EQUAL PARTS.
THERE ARE 99 PERCENTILES
1, 2,  99
1 =
 + 1
4
 基瑞
FOR A DISCRETE DISTRIBUTION OR A SIMPLE FREQUENCY DISTRIBUTION
2 =
 + 1
2
 基瑞
3 = 3
 + 1
4
 基瑞
=MEDIAN
倹 =
(+1) ≠
10
value
 =
( + 1) ≠
100
p
Question 1
Calculate the 3rd quartile, 6th decile and 70th percentile
28, 17, 12, 25,26, 19, 13, 27, 21,16
We first arrange in ascending order
12, 13, 16, 17, 19, 21, 25,26, 27, 28
3 = 3
11
4
≠. Value = 8.25th value
= 8≠ p + 0.25 9≠ p  8≠ p = 26 + 0.25 27  26 = 26.25
6 =
6(11)
10
≠ p = 6.6≠ p
6 = 6≠ p + 0.6 7≠ p  6≠ p = 21 + 0.6 25  21 = 23.4
70 =
70
100
11 ≠ p = 7.77≠ p
= 25 + .77 8≠ p  7≠ p = 25 + 0.77 26  25 = 25.77
Question 2
For the following frequency distribution , find 1, 3, 6, 45
 
18 15
19 18
20 25
21 27
22 40
23 25
24 19
25 16
26 8
27 7
 
18 15
19 18
20 25
21 27
22 40
23 25
24 19
25 16
26 8
27 7
We calculate the cumulative frequencycfcf
15
33
58
85
125
150
169
185
193
200
201
4
= 50.25
From the table, we look at the cf greater than 50.25 ie 58
Value of x corresponding to 58 is 20
1 = 20
3(201)
4
= 150.75
3 = 24
6 = 22
45
201
100
= 90.45
45 = 22
6
201
10
= 120.6
QUESTION 3
For the following table, find the median, lower quartile, upper quartile, 3rd decile, 95th percentile
C I f
600-700 40
700-800 68
800-900 86
900-1000 120
1000-1100 90
1100-1200 40
1200-1300 26
For a grouped frequency distribution, we use n instead of n+1
1 =  + (

4
 

)  
Where l = lower boundary of the class in which
1 
cf = cumulative frequency of the class
preceeding this class
f = frequency of the class in which 1 
 = ゐ≠  ≠  $p
2 =  +
(

2
 

)  
3 =  + (
3
4
 )
倹 =  + (

10  )

 
 =  + (

100
 )
C I f c f
600-700 40 40
700-800 68 108
800-900 86 194
900-1000 120 314
1000-1100 90 404
1100-1200 40 444
1200-1300 26 470
470
4
= 117.5
We take 800 -900 as the class in which 1 
1 = 800 + (
117.5  108
86
)  100
= 811.04
2 =
470
2
= 235900 +
(235  194
120
)  100 = 934.17
3(470)
4
= 352.5
3 = 1000 + (
352.5314
90
)  100 =1042.78
3(470)
10
= 141
3 = 800 +
141  108
86
 100 = 838.37
95(470)
100
= 446.5
95 = 1200 +
446.5  444
26
 100 = 1209.62
FOR MORE VIDEOS ON STATISTICS, SUBSCRIBE TO MY PLAYLIST

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  • 1. SUMAN MATHEWS GRADE 11 MATH COLLEGE MATH
  • 2. PREVIEW THIS VIDEO WOULD BE USEFUL FOR GRADE 11 STUDENTS STUDYING MATHEMATICS. IT WOULD ALSO BE USEFUL FOR STUDENTS AT THE COLLEGE LEVEL STUDYING STATISTICS. HERE WE SHOW HOW QUARTILES, DECILES AND PERCENTILES CAN BE CALCULATED . CALCULATION OF THE ABOVE IS DONE FOR DISCRETE, SIMPLE AND GROUPED FREQUENCY DISTRIBUTIONS
  • 3. IF A GIVEN SET IS ARRANGED IN ASCENDING OR DESCENDING ORDER OF MAGNITUDE, MEDIAN DIVIDES THE SET INTO 2 EQUAL PARTS. QUARTILES DIVIDE THE SET INTO 4 QUARTERS. THERE ARE 3 QUARTILES 1, 2, 3 DECILES DIVIDE THE SET INTO 10 EQUAL PARTS. THERE ARE 9 DECILES, 1, 2,------9 PERCENTILES DIVIDE THE SET INTO 100 EQUAL PARTS. THERE ARE 99 PERCENTILES 1, 2, 99
  • 4. 1 = + 1 4 基瑞 FOR A DISCRETE DISTRIBUTION OR A SIMPLE FREQUENCY DISTRIBUTION 2 = + 1 2 基瑞 3 = 3 + 1 4 基瑞 =MEDIAN 倹 = (+1) ≠ 10 value = ( + 1) ≠ 100 p
  • 5. Question 1 Calculate the 3rd quartile, 6th decile and 70th percentile 28, 17, 12, 25,26, 19, 13, 27, 21,16 We first arrange in ascending order 12, 13, 16, 17, 19, 21, 25,26, 27, 28 3 = 3 11 4 ≠. Value = 8.25th value = 8≠ p + 0.25 9≠ p 8≠ p = 26 + 0.25 27 26 = 26.25
  • 6. 6 = 6(11) 10 ≠ p = 6.6≠ p 6 = 6≠ p + 0.6 7≠ p 6≠ p = 21 + 0.6 25 21 = 23.4 70 = 70 100 11 ≠ p = 7.77≠ p = 25 + .77 8≠ p 7≠ p = 25 + 0.77 26 25 = 25.77
  • 7. Question 2 For the following frequency distribution , find 1, 3, 6, 45 18 15 19 18 20 25 21 27 22 40 23 25 24 19 25 16 26 8 27 7
  • 8. 18 15 19 18 20 25 21 27 22 40 23 25 24 19 25 16 26 8 27 7 We calculate the cumulative frequencycfcf 15 33 58 85 125 150 169 185 193 200
  • 9. 201 4 = 50.25 From the table, we look at the cf greater than 50.25 ie 58 Value of x corresponding to 58 is 20 1 = 20 3(201) 4 = 150.75 3 = 24 6 = 22 45 201 100 = 90.45 45 = 22 6 201 10 = 120.6
  • 10. QUESTION 3 For the following table, find the median, lower quartile, upper quartile, 3rd decile, 95th percentile C I f 600-700 40 700-800 68 800-900 86 900-1000 120 1000-1100 90 1100-1200 40 1200-1300 26
  • 11. For a grouped frequency distribution, we use n instead of n+1 1 = + ( 4 ) Where l = lower boundary of the class in which 1 cf = cumulative frequency of the class preceeding this class f = frequency of the class in which 1 = ゐ≠ ≠ $p 2 = + ( 2 ) 3 = + ( 3 4 )
  • 12. 倹 = + ( 10 ) = + ( 100 )
  • 13. C I f c f 600-700 40 40 700-800 68 108 800-900 86 194 900-1000 120 314 1000-1100 90 404 1100-1200 40 444 1200-1300 26 470 470 4 = 117.5 We take 800 -900 as the class in which 1 1 = 800 + ( 117.5 108 86 ) 100 = 811.04
  • 14. 2 = 470 2 = 235900 + (235 194 120 ) 100 = 934.17 3(470) 4 = 352.5 3 = 1000 + ( 352.5314 90 ) 100 =1042.78 3(470) 10 = 141
  • 15. 3 = 800 + 141 108 86 100 = 838.37 95(470) 100 = 446.5 95 = 1200 + 446.5 444 26 100 = 1209.62
  • 16. FOR MORE VIDEOS ON STATISTICS, SUBSCRIBE TO MY PLAYLIST