Subject: Optional Mathematics
The measures of any set of data show the range of the data or the amount of variation is known as variability or dispersion.
It is already mentioned earlier that if the values of data in a set are widely dispersed then the different measures of central tendency cannot represent the values of the data accurately. Likewise, it has also been mentioned that if the data are close together then the measures of central tendency may represent each value of data in the set more precisely. While working on various tasks using statistics, it is necessary to find the different measures of dispersion. Here, the following four measures of dispersion are included:
The difference between the highest and the lowest values of the variable of any set of data is called the range. The range tells us how far the values of data are spread out.
In any variation,
If, the Largest item = L
the smallest item = S, then
Range (R) = L - S
Coefficient of Range
A range is an absolute value. Hence, to compare variations between two distributions we generally use the relative value which is calculated using the following formula. This relative value is called the coefficient of range.
Coefficient of Range = \(\frac{L - S}{L + S}\)
Merits of range
Demerits of range
The difference between the first quartile (Q1) and the third quartile (Q3) is known as the interquartile range. The half of the interquartile range is called semi- interquartile range or quartile deviation.
Thus, if Q1and Q3 represent the first and the third quartile respectively,
(Q.D.) = \(\frac {Q_3 - Q_1}2\)
To find the quartile deviation according to this formula, we need to find Q1and Q3. For this, let us recall the formulas to find Q1 and Q3.
For individual and discrete series:
Q1 = (\(\frac{N+1}{4}\))th term
Q3 = [\(\frac {3(N+1)}{4}\)]th term
where, N = no. of terms in the series.
Similarly,
for continuous series, the quartiles are found as follows:
Q1 = (\(\frac N4\))th term
Q3 = (\(\frac {3N}4\))th term
From this the class interval in which Q1 and Q3 lies can be found. To find the quartiles the following formulas are used:
Q1 = L + \(\frac {\frac N4- c.f.}f\) × i
where, L = lower limit of Q1 class
c.f. = cumulative frequency of the class preceding the class containing Q1
f = frequency of the class
i = class size
Similarly,
Q3 = L + \(\frac {\frac {3N}4- c.f.}f\) × i
Coefficient of Quartile Deviation
As the quartile deviation is an absolute value, we need to find the coefficient of quartile deviation to compare this with other data.
Coefficient of Quartile Deviation = \(\frac {Q_3 - Q_1}{Q_3 + Q_1}\)
Merits of Quartile Deviation
Demerits of Quartile Deviation
Find the range of the following data:
32, 28, 35, 41, 42, 38, 32, 34, 35, 32.
Soln:
Here, Largest item (L) = 42
Smallest item (S) = 28
∴ Range(R) = L - S = 42 kg - 28 kg. Ans.
Find the range of the following data:
400, 100, 200, 300, 250, 150.
Soln:
Here, Largest item (L) = Rs. 400
Smallest item (S) = Rs. 100
∴ Range(R) = L - S = Rs. 400 - Rs. 100 = Rs. 300 Ans.
Find the range of the following data:
The temperature of a certain place taken every two hours of a day are recorded as below:
8oC, 7oC, 6oC, 8oC, 10oC, 12oC,13oC, 15oC, 14oC, 11oC, 10oC,9oC and so on.
Soln:
Here, Largest item(L) = 15oC
Smallest item (S) = 6oC
∴ Range(R) = L -S = 15oC - 6oC = 9oC.
Find the range:
Marks Obtained | 40 | 45 | 50 | 55 | 60 |
No of students | 2 | 2 | 5 | 3 | 2 |
Soln:
Here, Largest mark (L) = 60
Smallest mark (S) = 40
∴ Range (R) = L - S = 60 - 40 = 20 Ans.
Find the range:
Number | 10 | 11 | 12 | 13 | 14 |
Frequency | 3 | 12 | 13 | 15 | 1 |
Soln:
Here, Largest number (L) = 14
Smallest number (S) = 10
∴ Range (R) = L - S = 14 - 10 = 4 Ans.
Find the range of the following grouped data:
Age(yrs) | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 |
No of students | 6 | 10 | 16 | 14 | 10 | 5 |
Soln:
Here, Largest C.I. (80 - 90) and
Smallest C.I (30 -40)
So, Range (R) = upper limit of largest class - lower limit of smallest class
= 90 - 30 = 60 Ans.
Find the range of following data:
Marks obtained | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
No of students | 10 | 20 | 18 | 32 | 21 |
Soln:
Here, Largest C.I (40-50)
Smallest C.I(0 - 10)
So, Range (R) = upper limit of largest class - lower limit of smallest class
= 50 - 0 = 50 Ans.
Find the range and the coefficient of the following data:
The yearly average rainfall (mm) of a town is as follow:
150, 160, 172, 185, 198, 210
Soln:
Here, Largest term (L) = 210mm
Smallest term (S) = 150mm.
∴ Range (R) = L - S = 210mm - 150mm = 60mm.
∴Cofficient of range
=\(\frac{L - S}{L + S}\) =\(\frac{210kg - 150kg}{210kg + 150kg}\) =\(\frac{60kg}{360kg}\) =\(\frac{1}{6}\) Ans.
Find the range and the coefficient of follwing data:
Weight(kg) | 30 | 32 | 35 | 38 | 32 | 38 | 41 |
No. of Students | 5 | 5 | 6 | 10 | 8 | 4 | 2 |
Soln:
Here, Largest weight (L) = 41kg
Smallest weight (S) = 30kg
∴ Range (R) = L - S = 41kg. 30kg. = 11kg.
∴ Coefficient of Range:
\(\frac{L - S}{L + S}\) =\(\frac{41kg - 30kg}{41kg + 30kg}\) = \(\frac{11kg}{71kg}\) =\(\frac{11}{71}\)
Find the range and its coefficient:
Marks obtained | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
No of students | 1 | 3 | 5 | 20 | 3 |
Soln:
Here, Largest C. I = 40 - 50
Smallest C. I = 0 - 10
So, Range (R) = Upper limit of largest class - Lower limit of smallest class.
= 50 - 0 = 50 Ans.
∴ Coeff. of Range =\(\frac{L - S}{L + S}\) =\(\frac{50 - 0}{50 + 0}\) = 1 Ans.
Find the quartile deviation and the coefficient of quartic deviation:
22, 26, 14, 30, 18, 17, 35, 41, 12, 32, 34.
Soln:
Here, arranging the given data into ascending order:
12, 14, 17, 18, 22, 26, 30, 32, 34, 35, 41.
No. of terms (n) = 11
∴ First quartile (Q1) =\(\frac{n + 1}{4}\) nth term =\(\frac{11 + 1}{4}\)nth term.
= 3th term = third term = 17
Third quartile (Q3) =\(\frac{3}{4}\)(n + 1) nth term =\(\frac{3}{4}\)(11 + 1)nth term
= 9 nthterm = 34
∴ Quartile deviation (Q. D) =\(\frac{Q3 - Q1}{2}\) =\(\frac{34 - 17}{2}\) =\(\frac{17}{2}\) = 8.5
∴Coeff. of Q. D. =\(\frac{Q3 - Q1}{Q3 + Q1}\) =\(\frac{34 - 17}{34 + 17}\) =\(\frac{17}{51}\) =\(\frac{1}{3}\). Ans.
Find the quartile deviation and the coefficient of quartile deviation:
20, 30, 45, 60, 80, 90, 110, 115, 118, 120.
Soln:
Here, no of terms (n) = 10
∴ First quartile (Q1) =\(\frac{n + 1}{4}\)nth term =\(\frac{10 + 1}{4}\)nth term
=\(\frac{11}{4}\)nth term = 2.75 nth term
= 2nd term + (3rd term - 2nd term)× 0.75
=30 +(45 - 30)×0.75 = 30 +15× 0.75
= 41.25
∴ Third quartile (Q3) =\(\frac{3}{4}\)(n + 1) nth term =\(\frac{3}{4}\) (10 + 1) nth term
=\(\frac{33}{4}\)nth term = 8.25 nth term.
= 8th term + (9th term - 8th term)× 0.25
= 115 + (118 - 115)× 0.25
= 115 + 3× 0.25
= 115 + 0.75
= 115.75
∴ Quartile deviation (Q. D) =\(\frac{Q_3 - Q_1}{2}\) =\(\frac{115.75 - 41.25}{2}\) = 37.25 Ans.
∴ Coeff of Q. D.
\(\frac{Q_3 -Q_1}{Q_3 +Q_1}\)=\(\frac{115.75-41.25}{115.75+41.25}\) =\(\frac{74.5}{157}\) =\(\frac{149}{314}\). ans
Find the quartile deviation and the coefficient of quartile deviation of following data:
The ages of students of class 7 of a certain school are recorded as below:
12, 13, 14, 13, 12, 15, 14, 15, 12, 13, 15, 12, 14, 15, 18, 13, 14, 15, 12, 15, 14, 13, 16.
Soln:
Frequency table:
Age(years) (X) | Frequency (f) No. of student | Cumulative frequency (c. f) |
12 | 5 | 5 |
13 | 5 | 5+5=10 |
14 | 5 | 10+5=15 |
15 | 6 | 15+6=21 |
16 | 1 | 21+1=22 |
18 | 1 | 22+1=23 |
N = 23 |
No. of term N = 6
First quartile (Q1) = nth term \(\frac{N + 1}{4}\) th term =\(\frac{23 + 1}{4}\) th term =\(\frac{24}{4}\) th term = 6 th term = 13.
∴ Third quartile (Q3) =\(\frac{3}{4}\) (N + 1) th term =\(\frac{3}{4}\)× 24 th term = 18 th term = 15.
∴ Quartile deviation (Q. D) =\(\frac{Q_3 - Q_1}{2}\) =\(\frac{15 - 13}{2}\) =\(\frac{2}{2}\) = 1.
∴ Coeff. of Q .D =\(\frac{Q_3 - Q_1}{Q_3 + Q_1}\) =\(\frac{15 -13}{15 +13}\) =\(\frac{2}{28}\) =\(\frac{1}{14}\) = 0.07 Ans.
The marks obtained by 45 students of class 9 of a school in the Compulsory Mathematics Examination are recorded as below. Find the quartile deviation.
Marks obtained | 50 | 60 | 75 | 82 | 90 | 91 |
No. of Students | 16 | 12 | 8 | 5 | 3 | 1 |
Soln:
From the given data:
Marks (x) | No. of students (f) | Cumulative frequency (c. f) |
50 | 16 | 16 |
60 | 12 | 28 |
75 | 8 | 36 |
82 | 5 | 41 |
90 | 3 | 44 |
91 | 1 | 45 |
N = 45 |
No. of terms N = 45
First quartile(Q3) =\(\frac{N + 1}{4}\) th term =\(\frac{45 + 1}{4}\) th term =\(\frac{23}{4}\) th term = 11.5 th term = 50
[ Here of greater then 11.5 in 16 which mark is 50.]
Third quartile (Q3) =\(\frac{3}{4}\) (N + 1) th term =\(\frac{3}{4}\) (45 + 1) th term
=\(\frac{69}{2}\) th term = 34.5 th term = 75.
∴ Quartile deviation (Q. D) =\(\frac{Q_3 - Q _1}{2}\) =\(\frac{75 - 50}{2}\) =\(\frac{25}{2}\) = 12.5 Ans.
Find the quartile deviation from following data:
Marks Obtained | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
No. of students | 5 | 15 | 10 | 8 | 6 | 2 |
Soln:
Marks (x) | No. of students (f) | Cumulative frequency (c. f) |
20-30 | 5 | 5 |
30-40 | 15 | 20 |
40-50 | 10 | 30 |
50-60 | 8 | 38 |
60-70 | 6 | 44 |
70-80 | 2 | 46 |
N = 46 |
C. I lying Q1 is
=\(\frac{N}{4}\) th term =\(\frac{46}{4}\) th term = 11.5 th term
= 30-40
[ Here C. F. just greater than 11.5 is 20 which is in 30-40]
Using formula,
Q1 =L +\(\frac{\frac{N}{4} - c.f.}{f}\)× i
Here, L = 30, \(\frac{N}{4}\) = 11.5, c. f. = 5, f = 15 and i = 10
∴ Q1 = 30 +\(\frac{11.5 - 5}{15}\)× 10 = 30 + 4.33 = 34.33
C. I. lying Q3 is
=\(\frac{3}{4}\) N th term =\(\frac{3}{4}\)× 46 th term =\(\frac{69}{2}\) th term = 34.5 th term = 50-60
Using formula, Q3 = L +\(\frac{\frac{N}{4} - c.f.}{f}\)× i
Here, L = 50, \(\frac{3}{4}\)×N =\(\frac{3}{4}\)× 46 =\(\frac{69}{2}\)
c. f. = 30, f = 8, i = 10
∴Q3 = 50 +\(\frac{\frac{69}{2} - 30}{8}\)× 10 = 50 +\(\frac{9}{16}\)× 10 =50 +\(\frac{9}{8}\)×5 =\(\frac{445}{8}\) = 55.625
∴ Quartile deviation (Q. D) =\(\frac{Q_3 - Q_1}{2}\) =\(\frac{55.62 -34.33}{2}\) =\(\frac{21.29}{2}\) = 10.645 Ans.
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