Table of Contents
Definition of Quartile Deviation
Quartile Deviation divides the total frequency in to four equal parts. The lower quartiles Q1 refers to the values of variates corresponding to the cumulative frequency N/4, upper quartile Q3 refers the value of variants corresponding to cumulative frequency (¾)N.\[\text{Interquartile Range}={{Q}_{3}}-{{Q}_{1}}\]\[\text{Quartile Deviation }\left( \text{QD} \right)=\frac{1}{2}\left( {{Q}_{3}}-{{Q}_{1}} \right)\]\[\text{Relative measure of dispersion coefficient of QD}=\frac{{{Q}_{3}}-{{Q}_{1}}}{{{Q}_{3}}+{{Q}_{1}}}\]\[{{Q}_{1}}=LL+\frac{\frac{N}{4}-c{{f}_{p}}}{fc}\times CI\]\[{{Q}_{2}}=LL+\frac{\frac{N}{2}-c{{f}_{p}}}{fc}\times CI\]\[{{Q}_{3}}=LL+\frac{\frac{3N}{4}-c{{f}_{p}}}{fc}\times CI\]\[LL=\text{Lower limit of the quartile class}\]\[\text{CI=Class width}\]\[fc=\text{Frequency of quartile class}\]\[\text{N=Total frequency}\]\[c{{f}_{p}}=\text{Cumulative frequency of class preceding the quartile class}\]
Note:
1. Q3 – Q1 gives the middle 50% of reading.Q3 and Q1 are also known as upper and lower limit of middle 50% of readings.
2. Quartile range is not capable of further algebraic treatment.
In the following three articles in Statistics we will discuss various types of Absolute Measure of Dispersion in details:
1. Range (R)
2. Mean Deviations (M.D.)
3. Standard Deviations (S.D.)
Merits
1. It is easy to compute and understand.
2. Rigidly defined.
3. Not affected by extreme values.
Demerits
1. Not based on all values.
2. Affected by sampling fluctuations.
3. Not capable of further algebraic treatment.
| Example 01 |
Find the quartile deviation and coefficient of quartile deviation for the given grouped data also compute middle quartile.
| Class | Frequency (f) |
| 1-10 | 3 |
| 11-20 | 16 |
| 21-30 | 26 |
| 31-40 | 31 |
| 41-50 | 16 |
| 51-60 | 8 |
| Σ f =N =100 |
Solution:
\[\frac{N}{4}=\frac{100}{4}=25\]
| Class | f | Cf |
| 1-10 | 3 | 3 |
| 11-20 | 16 | 19 |
| 21-30 | 26 | 45 ← Q1 class |
| 31-40 | 31 | 76 ← Q2 and Q3 class |
| 41-50 | 16 | 92 |
| 51-60 | 8 | 100 |
| Σ f =N =100 |
\[LL=\text{Lower limit of the quartile class}\]\[\text{CI=Class width}\]\[fc=\text{Frequency of quartile class}\]\[\text{N=Total frequency}\]\[c{{f}_{p}}=\text{Cumulative frequency of class preceding the quartile class}\]\[{{Q}_{1}}=LL+\frac{\frac{N}{4}-c{{f}_{p}}}{fc}\times CI\]\[\Rightarrow {{\text{Q}}_{1}}=20.5+\frac{\frac{100}{4}-19}{26}\times 10\]\[\Rightarrow {{\text{Q}}_{1}}=20.5+\frac{25-19}{26}\times 10=20.5+\frac{60}{26}=22.81\]\[{{Q}_{2}}=LL+\frac{\frac{N}{2}-c{{f}_{p}}}{fc}\times CI\]\[\Rightarrow {{Q}_{2}}=30.5+\frac{\frac{100}{2}-45}{31}\times 10\]\[\Rightarrow {{Q}_{2}}=30.5+\frac{50-45}{31}\times 10=30.5+\frac{50}{31}=32.11\]\[{{Q}_{3}}=LL+\frac{\frac{3N}{4}-c{{f}_{p}}}{fc}\times CI\]\[\Rightarrow {{Q}_{3}}=30.5+\frac{\frac{3\times 100}{4}-45}{31}\times 10\]\[\Rightarrow {{Q}_{3}}=30.5+\frac{75-45}{31}\times 10=30.5+\frac{300}{31}=40.18\]\[\therefore \text{Quartile Deviation }\left( \text{QD} \right)=\frac{1}{2}\left( {{Q}_{3}}-{{Q}_{1}} \right)\]\[\Rightarrow QD=\frac{1}{2}\left( 40.18-22.81 \right)=8.685\]\[\therefore C\text{oefficient of QD}=\frac{{{Q}_{3}}-{{Q}_{1}}}{{{Q}_{3}}+{{Q}_{1}}}\]\[\Rightarrow C\text{oefficient of QD=}\frac{40.18-22.81}{40.18+22.81}=\frac{17.37}{62.99}\]\[\therefore C\text{oefficient of QD=0}\text{.276}\]
| Example 02 |
Find quartile deviation from the following marks of 12 students and also coefficient of quartile deviation: 25, 30, 37, 43, 48, 54, 61, 67, 72, 80, 84, 89.
Solution:
| Sl. No. | Marks |
| 1 | 25 |
| 2 | 30 |
| 3 | 37 |
| 4 | 43 |
| 5 | 48 |
| 6 | 54 |
| 7 | 61 |
| 8 | 67 |
| 9 | 72 |
| 10 | 80 |
| 11 | 84 |
| 12 | 89 |
Here, N = 12\[{{\text{Q}}_{1}}={{\left( \frac{N+1}{4} \right)}^{th}}={{3.25}^{th}}item\]\[\Rightarrow {{\text{Q}}_{1}}={{3}^{rd}}\text{item}+0.25\text{ of item}=37+0.25\left( 43-37 \right)\]\[\therefore {{\text{Q}}_{1}}=38.5\]\[{{Q}_{3}}={{\left\{ \frac{3\left( N+1 \right)}{4} \right\}}^{th}}={{\left\{ \frac{39}{4} \right\}}^{th}}={{9.75}^{th}}\text{ item}\]\[\Rightarrow {{Q}_{3}}={{9}^{th}}\text{ item}+0.75\text{ of item}=72+0.75\left( 80-72 \right)\]\[\therefore {{Q}_{3}}=78\]\[\therefore \text{Quartile Deviation }\left( \text{QD} \right)=\frac{1}{2}\left( {{Q}_{3}}-{{Q}_{1}} \right)\]\[\Rightarrow QD=\frac{1}{2}\left( 78-38.5 \right)=19.75\]\[\therefore C\text{oefficient of QD}=\frac{{{Q}_{3}}-{{Q}_{1}}}{{{Q}_{3}}+{{Q}_{1}}}\]\[\therefore C\text{oefficient of QD}=\frac{78-38.5}{78+38.5}=0.339\]
| Example 03 |
Compute quartile deviation and its coefficient for the data given below:
| X | 58 | 59 | 60 | 61 | 62 | 63 | 64 | 65 | 66 |
| f | 15 | 20 | 32 | 35 | 33 | 22 | 20 | 10 | 8 |
Solution:
| X | f | Cf |
| 58 | 15 | 15 |
| 59 | 20 | 35 |
| 60 | 32 | 67 ←Q1 class |
| 61 | 35 | 102 |
| 62 | 33 | 135 |
| 63 | 22 | 157 ←Q3 class |
| 64 | 20 | 177 |
| 65 | 10 | 187 |
| 66 | 8 | 195 |
| N = 195 |
\[{{\text{Q}}_{1}}={{\left( \frac{N+1}{4} \right)}^{th}}\text{ size}={{\left( \frac{195+1}{4} \right)}^{th}}\text{ size}\]\[\Rightarrow {{\text{Q}}_{1}}=\text{48}.\text{7}{{\text{8}}^{th}}\text{ size and corresponding to cf 67}\]\[\therefore {{\text{Q}}_{1}}=60\]\[{{Q}_{3}}={{\left\{ \frac{3\left( N+1 \right)}{4} \right\}}^{th}}={{\left\{ \frac{3}{4}\times 196 \right\}}^{th}}={{146.33}^{th}}\text{ size}\]\[\text{It lies in 157},\text{ cf}.\text{ against cf 157},\]\[\therefore {{Q}_{3}}=63\]\[\therefore \text{Quartile Deviation }\left( \text{QD} \right)=\frac{1}{2}\left( {{Q}_{3}}-{{Q}_{1}} \right)\]\[\Rightarrow QD=\frac{1}{2}\left( 63-60 \right)=1.5\]\[\therefore C\text{oefficient of QD}=\frac{{{Q}_{3}}-{{Q}_{1}}}{{{Q}_{3}}+{{Q}_{1}}}\]\[\Rightarrow C\text{oefficient of QD}=\frac{63-60}{63+60}=\frac{3}{123}\]\[\therefore C\text{oefficient of QD}=0.024\]
Range |
Mean Deviation |
