Chapter 3 - Partition Values
EXERCISE
Q. 1. Choose the correct option :
1) Statements that do not apply to Quartiles.
a) First arrange the values in ascending or descending order.
b) Observation can be divided into 4 parts.
c) They are represented as Q\(_1\), Q\(_2\), and Q\(_3\).
d) Q\(_2\) is also known as median.
Options :
1) a
2) b and c
3) a, b and c
4) None of these
Ans: 4) None of these
2) D\(_7\)from the given data.
Data – 4, 5, 6, 7, 8, 9, 10, 11, 12
Options :
1) 7
2) 9
3) 10
4) 12
Ans: 1) 7
3) Statements related to partition values that are correct.
a) Exact divisions of percentiles into 100 parts gives 99 points
b) Deciles have total 9 parts
c) Quartiles are shown by Q\(_1\), Q\(_2\) and Q\(_3\)
d) Symbolically, Percentiles and Deciles are shown by P and D
Options :
1) a and c
2) a and b
3) a, b and c
4) a, c and d
Ans: 4) a, c and d
Q. 2. Identify the correct pairs from the given options :
| Group A | Group B |
|---|---|
1) Quartiles |
a) D\(_j\) = size of j \(\large (\frac {n\,+\,1}{10}) ^ {th\,observation}\) |
2) Deciles |
b) P\(_k\) = l + \(\large \frac {\frac{kn}{100}\,–\,cf}{f}\) × h |
3) Percentiles |
c) Q\(_i\) = l + \(\large \frac {\frac{in}{4}\,–\,cf}{f}\) × h |
Options :
1) 1 – b, 2 – c, 3 – a
2) 1 – c, 2 – a, 3 – b
3) 1 – c, 2 – b, 3 – a
4) 1 – a, 2 – b, 3 – c
Ans: 2) 1 – c, 2 – a, 3 – b
Q. 3. Give economic terms :
1) Procedure for dividing the data into equal parts.
Ans: Partitioning
2) Value that divides the series into ten equal parts.
Ans: Deciles
3) Value that divides the whole set of observations into four equal parts.
Ans: Quartiles
Q. 4. Solve the following :
1) Calculate Q\(_1\), D\(_4\) and P\(_{26}\) for the following data.
18, 24, 45, 29, 4, 7, 28, 49, 16, 26, 25, 12, 10, 9, 8
Ans:
Q\(_i\) = size of i \(\large (\frac {n\,+\,1}{4}) ^ {th\,observation}\)
∴ Q\(_1\) = size of 1 \(\large (\frac {15\,+\,1}{4}) ^ {th\,observation}\)
∴ Q\(_1\) = size of 1 \(\large (\frac {16}{4}) ^ {th\,observation}\)
∴ Q\(_1\) = size of 1 (4) \(^ {th\,observation}\)
∴ Q\(_1\) = size of 4th observation
∴ Q\(_1\) = 9
D\(_j\) = size of j \(\large (\frac {n\,+\,1}{10}) ^ {th\,observation}\)
∴ D\(_4\) = size of 4 \(\large (\frac {15\,+\,1}{10}) ^ {th\,observation}\)
∴ D\(_4\) = size of 4 \(\large (\frac {16}{10}) ^ {th\,observation}\)
∴ D\(_4\) = size of 4 (1.6) \(^ {th\,observation}\)
∴ D\(_4\) = size of 6.4th observation
∴ D\(_4\) = size of 6th observation + 0.4(7th observation – 6th observation)
∴ D\(_4\) = 12 + 0.4(16 – 12)
∴ D\(_4\) = 12 + 0.4(4)
∴ D\(_4\) = 12 + 1.6
∴ D\(_4\) = 13.6
P\(_k\) = size of k \(\large (\frac {n\,+\,1}{100}) ^ {th\,observation}\)
∴ P\(_{26}\) = size of 26 \(\large (\frac {15\,+\,1}{100}) ^ {th\,observation}\)
∴ P\(_{26}\) = size of 26 \(\large (\frac {16}{100}) ^ {th\,observation}\)
∴ P\(_{26}\) = size of 26 (0.16)\( ^ {th\,observation}\)
∴ P\(_{26}\) = size of 4.16th observation
∴ P\(_{26}\) = size of 4th observation + 0.16(5th observation – 4th observation)
∴ P\(_{26}\) = 9 + 0.16(10 – 9)
∴ P\(_{26}\) = 9 + 0.16(1)
∴ P\(_{26}\) = 9 + 0.16
∴ P\(_{26}\) = 9.16
Ans: Q\(_1\) = 9, D\(_4\) = 13.6, P\(_{26}\) = 9.16
2) Calculate Q\(_3\), D\(_5\), and P\(_{35}\) for the given data.
Ans:
Q\(_i\) = size of i \(\large (\frac {n\,+\,1}{4}) ^ {th\,observation}\)
∴ Q\(_3\) = size of 3 \(\large (\frac {79\,+\,1}{4}) ^ {th\,observation}\)
∴ Q\(_3\) = size of 3 \(\large (\frac {80}{4}) ^ {th\,observation}\)
∴ Q\(_3\) = size of 3 (20) \(^ {th\,observation}\)
∴ Q\(_3\) = size of 60th observation
Size of 60th observation lies in cf 67, hence quartile value = ₹5 Lakhs
∴ Q\(_3\) = ₹5 Lakhs
D\(_j\) = size of j \(\large (\frac {n\,+\,1}{10}) ^ {th\,observation}\)
∴ D\(_5\) = size of 5 \(\large (\frac {79\,+\,1}{10}) ^ {th\,observation}\)
∴ D\(_5\) = size of 5 \(\large (\frac {80}{10}) ^ {th\,observation}\)
∴ D\(_5\) = size of 5 (8) \(^ {th\,observation}\)
∴ D\(_5\) = size of 40th observation
Size of 40th observation lies in cf 52, hence decile value = ₹4 Lakhs
∴ D\(_5\) = ₹4 Lakhs
P\(_k\) = size of k \(\large (\frac {n\,+\,1}{100}) ^ {th\,observation}\)
∴ P\(_{35}\) = size of 35 \(\large (\frac {79\,+\,1}{100}) ^ {th\,observation}\)
∴ P\(_{35}\) = size of 35 \(\large (\frac {80}{100}) ^ {th\,observation}\)
∴ P\(_{35}\) = size of 35 (0.80) \(^ {th\,observation}\)
∴ P\(_{35}\) = size of 28th observation
Size of 28th observation lies in cf 52, hence percentile value = ₹4 Lakhs
∴ P\(_{35}\) = ₹4 Lakhs
Ans: Q\(_3\) = ₹5 Lakhs, D\(_5\) = ₹4 Lakhs, P\(_{35}\) = ₹4 Lakhs
3) Find out P\(_{50}\) for the following data.
Ans:
P\(_k\) = size of k \(\large (\frac {n}{100}) ^ {th\,observation}\)
∴ P\(_{50}\) = size of 50 \(\large (\frac {60}{100}) ^ {th\,observation}\)
∴ P\(_{50}\) = size of 50 (0.60) \(^ {th\,observation}\)
∴ P\(_{50}\) = size of 30th observation
Size of 30th observation lies in cf 45
Hence l = 60, cf = 20, n = 60, h = 20
P\(_i\) = l + \(\large (\frac {\frac {i\,×\,n}{100}\,–\,cf}{f})\) × h
∴ P\(_{50}\) = 60 + \(\large (\frac {\frac {50\,×\,60}{100}\,–\,20}{25})\) × 20
∴ P\(_{50}\) = 60 + \(\large (\frac {30\,–\,20}{25})\) × 20
∴ P\(_{50}\) = 60 + \(\large (\frac {10}{25})\) × 20
∴ P\(_{50}\) = 60 + \(\large (\frac {200}{25})\)
∴ P\(_{50}\) = 60 + 8
∴ P\(_{50}\) = 68
Ans: P\(_{50}\) = 68
4) Calculate Q\(_3\) for the following data.
Ans:
Q\(_3\) = size of 3 \(\large (\frac {n}{4}) ^ {th\,observation}\)
∴ Q\(_3\) = size of 3 \(\large (\frac {250}{4}) ^ {th\,observation}\)
∴ Q\(_3\) = size of 3 (62.5) \(^ {th\,observation}\)
∴ Q\(_3\) = size of 187.5th observation
Size of 187.5th observation lies in cf 200
Hence, quartile value is 50 – 60
∴ l = 50, f = 32, cf = 168, n = 250, h = 10
Q\(_i\) = l + \(\large (\frac {\frac {i\,×\,n}{4}\,–\,cf}{f})\) × h
∴ Q\(_3\) = 50 + \(\large (\frac {\frac {3\,×\,250}{4}\,–\,168}{32})\) × 10
∴ Q\(_3\) = 50 + \(\large (\frac {187.5\,–\,168}{32})\) × 10
∴ Q\(_3\) = 50 + \(\large (\frac {19.5}{32})\) × 10
∴ Q\(_3\) = 50 + \(\large (\frac {195}{32})\)
∴ Q\(_3\) = 50 + 6.09375
∴ Q\(_3\) = 56.09375
Ans: The Q\(_3\) is approximately 56.09375 lakh.
5) Calculate D\(_7\) for the following data.
Ans:
D\(_j\) = size of j \(\large (\frac {n}{10}) ^ {th\,observation}\)
∴ D\(_7\) = size of 7 \(\large (\frac {250}{10}) ^ {th\,observation}\)
∴ D\(_7\) = size of 7 (25) \(^ {th\,observation}\)
∴ D\(_7\) = size of 175th observation
Size of 175th observation lies in cf 200
Hence, decile class is 50 – 60
∴ l = 50, f = 32, cf = 168, n = 250, h = 10
D\(_i\) = l + \(\large (\frac {\frac {j\,×\,n}{10}\,–\,cf}{f})\) × h
∴ D\(_7\) = 50 + \(\large (\frac {\frac {7\,×\,250}{10}\,–\,168}{32})\) × 10
∴ D\(_7\) = 50 + \(\large (\frac {175\,–\,168}{32})\) × 10
∴ D\(_7\) = 50 + \(\large (\frac {7}{32})\) × 10
∴ D\(_7\) = 50 + \(\large (\frac {70}{32})\)
∴ D\(_7\) = 50 + 2.1875
∴ D\(_7\) = 52.1875
Ans: The D\(_7\) is approximately 52.1875 crores.
6) Calculate P\(_{15}\) for the following data.
Ans:
P\(_k\) = size of k \(\large (\frac {n}{100}) ^ {th\,observation}\)
∴ P\(_{15}\) = size of 15 \(\large (\frac {100}{100}) ^ {th\,observation}\)
∴ P\(_{15}\) = size of 15 (1) \(^ {th\,observation}\)
∴ P\(_{15}\) = size of 15th observation
Size of 15th observation lies in cf 15
Hence l = 10, f = 10, cf = 5, n = 100, h = 10
P\(_i\) = l + \(\large (\frac {\frac {i\,×\,n}{100}\,–\,cf}{f})\) × h
∴ P\(_{15}\) = 10 + \(\large (\frac {\frac {15\,×\,100}{100}\,–\,5}{10})\) × 10
∴ P\(_{15}\) = 10 + \(\large (\frac {15\,–\,5}{10})\) × 10
∴ P\(_{15}\) = 10 + \(\large (\frac {10}{10})\) × 10
∴ P\(_{15}\) = 10 + 10
∴ P\(_{15}\) = 20
Ans: P\(_{15}\) = 20 lakhs
Q. 5. State with reasons whether you agree or disagree with the following statements :
1) Partition values have application only in theory but not in practice.
Ans: No, I do not agree with the statement.
Partition values like quartiles help economists understand key details about income, sales, and stock levels. Also, deciles and percentiles are useful to study things like poverty levels, income differences, and household wealth in more detail. So, partition values are very helpful in the study of Economics
2) Average can misinterpret the representative value.
Ans: Yes, I agree with the statement.
An average is not always based on all the values because it is a positional average and can change with different samples. It sometimes gives more weight to bigger values and less to smaller ones. So, the average may not always show the true or correct picture of the data.
3) Median is also known as the second quartile.
Ans: Yes, I agree with the statement.
The median is the middle value in an arranged set of data.
It divides the data into two equal parts — half of the values are below it and half are above it. The median is not affected by very high or very low values. Since both median and second quartile (Q\(_2\)) show the middle value, we can say:
Median = Q\(_2\) = n\(\large (\frac {n}{2})\).
Q. 6. Answer the following questions on the basis of the given table :
1) Write the formula of Q\(_1\) and Q\(_3\).
Ans:
Q\(_1\) = size of 1 \(\large (\frac {n\,+\,1}{4}) ^ {th\,observation}\)
Q\(_3\) = size of 3 \(\large (\frac {n\,+\,1}{4}) ^ {th\,observation}\)
2) Find out the cumulative frequency of the last value in the above data.
Ans: The cumulative frequency of the last value in the above data is 38.
3) Find out the value of ‘n’ in the above data.
Ans: The value of ‘n’ in the above data is 38.