CBSE Class 11 Economics Measures of Central Tendency Set 03

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CHAPTER -5: MEASURES OF CENTRAL TENDENCY
QUARTILES

Very short answer questions (1 mark each)

1. What are quartiles? [1 Mark]

Answer:
1. Quartiles are the measures of central tendency that divide a given set of data into four equal parts.
2. There are three quartiles namely first quartile (\(Q_1\)), second quartile (\(Q_2\)), and third quartile (\(Q_3\)).

Teacher's Note:
a) Quartiles are partition values used in statistical analysis.
b) The second quartile (\(Q_2\)) is equal to the median.

2. Which quartile is called median? [1 Mark]

Answer:
1. The second quartile (\(Q_2\)) is called the median.
2. It divides the distribution exactly into two equal halves, similar to the median.

Teacher's Note:
a) \(Q_2\) corresponds to the 50th percentile.
b) It lies right in the middle of the arranged series.

3. Define lower quartile. [1 Mark]

Answer:
1. The lower quartile, denoted by \(Q_1\), is the first quartile.
2. It marks the value below which 25 percent of the observations lie.

Teacher's Note:
a) It is calculated using the formula \(\frac{N + 1}{4}\)th item for individual data.
b) It represents the lower boundary of the middle 50 percent of data.

4. Define upper quartile. [1 Mark]

Answer:
1. The upper quartile, denoted by \(Q_3\), is the third quartile.
2. It marks the value below which 75 percent of the observations lie.

Teacher's Note:
a) It is calculated using the formula \(\frac{3(N + 1)}{4}\)th item for individual data.
b) It forms the upper limit of the interquartile range.

Short answer questions (3/4 marks each)

1. Calculate the lower and upper quartiles from the following ungrouped data [3 Marks]
28 17 15 22 29 21 23 27 18 12 7 2 9 4 6 8 3 10 5

Answer:
1. Arrange the data in ascending order: 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, 17, 18, 21, 22, 23, 27, 28, 29.
2. Total number of items (N) = 19.
3. Lower quartile (\(Q_1\)) = Size of \(\frac{N + 1}{4}\)th item = Size of \(\frac{19 + 1}{4}\)th = 5th item = 6.
4. Upper quartile (\(Q_3\)) = Size of \(\frac{3(N + 1)}{4}\)th item = Size of \(\frac{3(20)}{4}\)th = 15th item = 22.

Teacher's Note:
a) Always arrange raw data in ascending order before locating position formulas.
b) Note that \(N\) here is odd, making position calculation straightforward.
c) Double check the count of terms to avoid indexing errors.

2. Calculate quartiles from the following discrete series: [3 Marks]

Answer:
1. Prepare cumulative frequency (c.f.) table: 4(10), 8(15), 12(20), 16(30), 20(45), 24(50), 28(60). Total N = 60.
2. \(Q_1\) position = Size of \(\frac{N + 1}{4}\)th item = Size of \(\frac{61}{4}\)th = 15.25th item.
3. Looking at c.f., 15.25 lies in cumulative frequency 15 (which corresponds to item 8) and 20 (item 12), so \(Q_1\) = 12.
4. \(Q_3\) position = Size of \(\frac{3(N + 1)}{4}\)th item = Size of \(\frac{3(61)}{4}\)th = 45.75th item, which gives \(Q_3\) = 24.

Teacher's Note:
a) Cumulative frequencies must be accurately computed by successive additions.
b) Check the exact c.f. bracket where the fractional item position falls.
c) Discrete frequency distributions use \(N + 1\) in the quartile formula.

3. Calculate quartiles from the following data [3 Marks]

Answer:
1. Cumulative frequencies: 5-15 (12), 15-25 (27), 25-36 (36), 35-45 (42), 45-55 (50). Total N = 50.
2. \(Q_1\) position = \(\frac{N}{4}\)th = \(\frac{50}{4}\) = 12.5th item, which lies in class 15-25.
3. \(Q_1 = L + \frac{\frac{N}{4} - c}{f} \times h = 15 + \frac{12.5 - 12}{15} \times 10 = 15 + \frac{5}{15}\) = 15.33.
4. \(Q_3\) position = \(\frac{3N}{4}\)th = \(\frac{150}{4}\) = 37.5th item, which lies in class 35-45; calculating gives \(Q_3 = 35 + \frac{37.5 - 36}{6} \times 10\) = 37.5.

Teacher's Note:
a) For continuous series, use \(N/4\) and \(3N/4\) instead of \(N + 1\).
b) 'L' is the lower limit of the quartile class, 'c' is cumulative frequency preceding the quartile class.
c) Ensure interval width 'h' is correctly applied in interpolation.

4. Calculate quartiles from the following data [4 Marks]
Height in inches: 58 59 60 61 62 63 64 65 66
No. of Persons: 20 18 17 21 22 25 19 20 24

Answer:
1. Set up cumulative frequency for discrete items: 58(20), 59(38), 60(55), 61(76), 62(98), 63(123), 64(142), 65(162), 66(186). Total N = 186.
2. \(Q_1\) position = Size of \(\frac{N + 1}{4}\)th = \(\frac{187}{4}\) = 46.75th item.
3. From c.f., 46.75 falls under cumulative frequency 55, corresponding to item height 60.
4. \(Q_3\) position = Size of \(\frac{3(N + 1)}{4}\)th = \(\frac{561}{4}\) = 140.25th item, which falls under c.f. 142, corresponding to item height 64.

Teacher's Note:
a) This is a discrete series presented horizontally.
b) Always construct a vertical frequency and cumulative frequency table to avoid errors.
c) Check that the sum of frequencies matches total N = 186.
d) Verify both quartile positions carefully against the cumulative frequency column.

5. Calculate quartiles from the following data [3 Marks]
Items: Less than 5 10 15 20 25 30
Number: 4 6 3 8 12 7

Answer:
1. Convert 'less than' cumulative frequencies into simple class intervals and frequencies: 0-5 (4), 5-10 (2), 10-15 (3), 15-20 (5), 20-25 (4), 25-30 (22). Total N = 40.
2. Alternatively, use the given cumulative frequencies directly: c.f. = 4, 10, 13, 21, 33, 40.
3. \(Q_1\) position = \(\frac{N}{4}\)th = \(\frac{40}{4}\) = 10th item, which gives \(Q_1\) = 10.
4. \(Q_3\) position = \(\frac{3N}{4}\)th = \(\frac{120}{4}\) = 30th item, which falls in interval 20-25, giving \(Q_3 = 20 + \frac{30 - 21}{12} \times 5\) = 23.75.

Teacher's Note:
a) 'Less than' series numbers are already cumulative frequencies.
b) Form regular continuous classes before applying the interpolation formula.
c) Verify subtraction between consecutive less than values to get simple frequencies.

6. Calculate quartiles from the following data [3 Marks]
Items: More than 10 20 30 40 50 60
Number: 50 46 40 30 18 7

Answer:
1. Convert 'more than' data into standard classes and frequencies: 10-20 (4), 20-30 (6), 30-40 (10), 40-50 (12), 50-60 (11), 60-70 (7). Total N = 50.
2. Compute cumulative frequency: 4, 10, 20, 32, 43, 50.
3. \(Q_1\) position = \(\frac{N}{4}\)th = \(\frac{50}{4}\) = 12.5th item, lying in class 30-40.
4. Calculate \(Q_1 = 30 + \frac{12.5 - 10}{10} \times 10\) = 32.5, and \(Q_3 = 50 + \frac{37.5 - 43}{...}\) following respective class boundaries.

Teacher's Note:
a) 'More than' cumulative frequencies must be converted to normal frequencies first.
b) Subtract descending values carefully to get correct frequencies for each class.
c) Check that total frequency equals the first value of the 'more than' series.

7. Calculate quartiles from the following data [4 Marks]
Less than: 10 20 30 40 50 60 70 80
Number: 15 35 60 84 96 127 198 250

Answer:
1. Given data is a 'less than' cumulative frequency distribution with N = 250.
2. Convert into classes and simple frequencies: 0-10 (15), 10-20 (20), 20-30 (25), 30-40 (24), 40-50 (12), 50-60 (31), 60-70 (71), 70-80 (52).
3. \(Q_1\) position = \(\frac{N}{4}\)th = \(\frac{250}{4}\) = 62.5th item, which lies in class 30-40.
4. \(Q_1 = 30 + \frac{62.5 - 60}{24} \times 10\) = 31.04; similarly, \(Q_3\) position = \(\frac{3(250)}{4}\) = 187.5th item, yielding \(Q_3\) in class 60-70.

Teacher's Note:
a) Recognize that 'less than' cumulative frequencies are directly usable for positioning.
b) Subtract consecutive cumulative frequencies to find exact frequencies for interpolation.
c) Watch out for unequal class intervals if present in complex data sets.
d) Double check arithmetic during fraction interpolation steps.

8. Calculate quartiles and verify the result on a graph [4 Marks]
Classes: 0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40
Frequency: 7 10 20 13 17 10 14 9

Answer:
1. Prepare cumulative frequency table: 0-5 (7), 5-10 (17), 10-15 (37), 15-20 (50), 20-25 (67), 25-30 (77), 30-35 (91), 35-40 (100). Total N = 100.
2. \(Q_1\) position = \(\frac{N}{4}\)th = \(\frac{100}{4}\) = 25th item, lying in class 10-15.
3. \(Q_1 = 10 + \frac{25 - 17}{20} \times 5 = 10 + \frac{40}{20}\) = 12.
4. \(Q_3\) position = \(\frac{3N}{4}\)th = 75th item, lying in class 25-30; calculating gives \(Q_3 = 25 + \frac{75 - 77}{...}\) and verified via ogive curve on graph paper.

Teacher's Note:
a) Graphical verification involves drawing a cumulative frequency curve (less than ogive).
b) Locate \(N/4\) and \(3N/4\) on the vertical axis and drop perpendiculars to read quartiles.
c) Algebraic calculation must match the graphical reading approximately.
d) Ensure proper scale is chosen when plotting ogives on graph sheets.