CBSE Class 11 Economics Measures of Dispersion Set 01

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CHAPTER -6: MEASURES OF DISPERSION

STANDARD DEVIATION

Very short answer questions (1 mark each)

1. What is meant by a measure of dispersion? [1 Mark]

Answer:
1. Measures of dispersion are statistical methods used to measure the extent of variability or scatter of the observations around the central value (average).
2. They indicate the degree of heterogeneity or uniformity in a given distribution.

Teacher's Note:
a) Dispersion measures the lack of uniformity in data values.
b) It complements measures of central tendency by showing data spread.

2. Define the following terms: [1 Mark]
a) Standard Deviation
b) Coefficient of variation
c) Lorenz-Curve
d) Variance.
e) State the formula of coefficient of variation.

Answer:
1. Standard Deviation: It is the square root of the arithmetic mean of the squared deviations of values from their arithmetic mean.
2. Coefficient of Variation: It is a relative measure of dispersion calculated as the ratio of standard deviation to the mean, expressed in percentage.
3. Lorenz Curve: It is a graphical method of studying dispersion, showing the cumulative percentage of items against the cumulative percentage of the given variable.
4. Variance: It is the square of the standard deviation, representing the mean of squared deviations from the mean.
5. Formula of Coefficient of Variation: CV = \(\frac{\sigma}{\bar{X}}\) × 100, where \(\sigma\) is standard deviation and \(\bar{X}\) is mean.

Teacher's Note:
a) Standard deviation is the most widely used absolute measure of dispersion.
b) Coefficient of variation allows comparison of variability between two different distributions.

3. What is the best measure of dispersion to check consistency of two or more distributions? [1 Mark]

Answer:
1. The coefficient of variation is the best measure to check consistency.
2. A distribution with a lower coefficient of variation is considered more consistent or uniform.

Teacher's Note:
a) Relative measures like CV are independent of units of measurement.
b) Lower variability means higher consistency.

4. What is the main difference between standard deviation and mean deviation? [1 Mark]

Answer:
1. In standard deviation, deviations are squared before finding their mean, whereas in mean deviation, absolute values of deviations are taken without squaring.
2. Standard deviation algebraically treats positive and negative signs by squaring, while mean deviation ignores signs by taking absolute values.

Teacher's Note:
a) Standard deviation gives more weight to extreme values due to squaring.
b) Mean deviation uses arithmetic mean of absolute deviations.

Short answer questions (3/4 marks each)

1. Calculate standard deviation for the following data: [3 Marks]
Marks: 12   8   17   13   15   9   18   11   6   1

Answer:
1. Number of observations (N) = 10.
2. Sum of observations (\(\sum X\)) = 12 + 8 + 17 + 13 + 15 + 9 + 18 + 11 + 6 + 1 = 110.
3. Mean (\(\bar{X}\)) = \(\frac{110}{10}\) = 11.
4. Deviations from mean (\(x = X - \bar{X}\)): 1, -3, 6, 2, 4, -2, 7, 0, -5, -10.
5. Squared deviations (\(x^2\)): 1, 9, 36, 4, 16, 4, 49, 0, 25, 100.
6. Sum of squared deviations (\(\sum x^2\)) = 1 + 9 + 36 + 4 + 16 + 4 + 49 + 0 + 25 + 100 = 244.
7. Standard deviation (\(\sigma\)) = \(\sqrt{\frac{\sum x^2}{N}}\) = \(\sqrt{\frac{244}{10}}\) = \(\sqrt{24.4}\) = 4.94.

Teacher's Note:
a) Use the direct formula based on squared deviations from the actual mean.
b) Verify arithmetic carefully during summation.
c) Final value is approximately 4.94.

2. Calculate standard deviation for the following data: [3 Marks]

Answer:
1. Total frequency (\(\sum f\) = N) = 26 + 29 + 40 + 35 + 26 + 18 + 14 + 12 = 200.
2. Product \(fX\) values: 130, 290, 600, 700, 650, 540, 490, 480.
3. Sum of \(fX\) (\(\sum fX\)) = 3,880.
4. Mean (\(\bar{X}\)) = \(\frac{3880}{200}\) = 19.4.
5. Using formula \(\sigma = \sqrt{\frac{\sum fX^2}{N} - \left(\frac{\sum fX}{N}\right)^2}\):
6. \(fX^2\) values: 650, 2900, 9000, 14000, 16250, 16200, 17150, 19200.
7. Sum of \(fX^2\) (\(\sum fX^2\)) = 95,350.
8. \(\sigma = \sqrt{\frac{95350}{200} - (19.4)^2} = \sqrt{476.75 - 376.36} = \sqrt{100.39}\) = 10.02.

Teacher's Note:
a) Apply the shortcut formula with frequencies to simplify calculations.
b) Double check multiplication steps for \(fX\) and \(fX^2\).
c) Standard deviation comes out to approximately 10.02.

3. Calculate standard deviation for the following data: (Use assumed mean method) [4 Marks]

Answer:
1. Midpoints (m): 15, 25, 35, 45, 55, 65, 75.
2. Assumed mean (A) = 45, class interval (i) = 10.
3. Step deviations \(d = \frac{m - A}{i}\): -3, -2, -1, 0, 1, 2, 3.
4. Frequencies (f): 4, 8, 8, 16, 12, 6, 4. Total N = 58.
5. Products \(fd\): -12, -16, -8, 0, 12, 12, 12. Sum \(\sum fd\) = 10.
6. Products \(fd^2\): 36, 32, 8, 0, 12, 24, 36. Sum \(\sum fd^2\) = 148.
7. Standard deviation formula: \(\sigma = i \times \sqrt{\frac{\sum fd^2}{N} - \left(\frac{\sum fd}{N}\right)^2}\).
8. \(\sigma = 10 \times \sqrt{\frac{148}{58} - \left(\frac{10}{58}\right)^2} = 10 \times \sqrt{2.5517 - 0.0297} = 10 \times \sqrt{2.522} = 10 \times 1.588\) = 15.88.

Teacher's Note:
a) Step deviation method simplifies arithmetic when class intervals are equal.
b) Multiply by class interval 'i' at the final stage.
c) Calculation yields standard deviation equal to 15.88.
d) Always verify summation of \(fd\) and \(fd^2\) columns.

4. Calculate standard deviation for the following data: (Use assumed mean method) [4 Marks]

Answer:
1. Midpoints (m): 2, 6, 10, 14.
2. Assumed mean (A) = 6, class interval (i) = 4.
3. Step deviations \(d = \frac{m - A}{i}\): -1, 0, 1, 2.
4. Frequencies (f): 4, 8, 2, 1. Total N = 15.
5. Products \(fd\): -4, 0, 2, 2. Sum \(\sum fd\) = 0.
6. Products \(fd^2\): 4, 0, 2, 4. Sum \(\sum fd^2\) = 10.
7. Standard deviation formula: \(\sigma = i \times \sqrt{\frac{\sum fd^2}{N} - \left(\frac{\sum fd}{N}\right)^2}\).
8. \(\sigma = 4 \times \sqrt{\frac{10}{15} - 0} = 4 \times \sqrt{0.6667} = 4 \times 0.8165\) = 3.27.

Teacher's Note:
a) Step deviation method reduces large numbers to simple integers.
b) Since sum of \(fd\) is zero, calculation becomes even simpler.
c) Standard deviation equals approximately 3.27.
d) Check class boundaries and midpoints carefully.

5. Calculate standard deviation for the following data: [4 Marks]

Answer:
1. Convert cumulative frequency into simple frequency classes:
    0-10: 5
    10-20: 14 - 5 = 9
    20-30: 29 - 14 = 15
    30-40: 60 - 29 = 31
    40-50: 82 - 60 = 22
    50-60: 92 - 82 = 10
    60-70: 98 - 92 = 6
    70-80: 100 - 98 = 2
2. Midpoints (m): 5, 15, 25, 35, 45, 55, 65, 75.
3. Assumed mean (A) = 35, class interval (i) = 10.
4. Step deviations \(d\): -3, -2, -1, 0, 1, 2, 3, 4.
5. Frequencies (f): 5, 9, 15, 31, 22, 10, 6, 2. Total N = 100.
6. Products \(fd\): -15, -18, -15, 0, 22, 20, 18, 8. Sum \(\sum fd\) = 20.
7. Products \(fd^2\): 45, 36, 15, 0, 22, 40, 54, 32. Sum \(\sum fd^2\) = 244.
8. \(\sigma = 10 \times \sqrt{\frac{244}{100} - \left(\frac{20}{100}\right)^2} = 10 \times \sqrt{2.44 - 0.04} = 10 \times \sqrt{2.40}\) = 15.49.

Teacher's Note:
a) Cumulative frequency series must be converted to exclusive frequency distribution first.
b) Verify each class frequency subtraction carefully.
c) Step deviation method makes calculation manageable for 100 items.
d) Final standard deviation is approximately 15.49.

6. Calculate co-efficient of variation of the following data. [4 Marks]

Answer:
1. For Boys:
    Mean (\(\bar{X}_1\)) = 68
    Variance (\(\sigma_1^2\)) = 9, so Standard Deviation (\(\sigma_1\)) = \(\sqrt{9}\) = 3
    Coefficient of Variation (CV₁) = \(\frac{\sigma_1}{\bar{X}_1}\) × 100 = \(\frac{3}{68}\) × 100 = 4.41%
2. For Girls:
    Mean (\(\bar{X}_2\)) = 61
    Variance (\(\sigma_2^2\)) = 4, so Standard Deviation (\(\sigma_2\)) = \(\sqrt{4}\) = 2
    Coefficient of Variation (CV₂) = \(\frac{\sigma_2}{\bar{X}_2}\) × 100 = \(\frac{2}{61}\) × 100 = 3.28%

Teacher's Note:
a) Remember that variance is the square of standard deviation, so square root is needed.
b) CV formula requires standard deviation, not variance.
c) Boys CV is 4.41% and Girls CV is 3.28%.
d) Girls' height distribution is more consistent due to lower CV.

7. The coefficient of variation of group A is 14.6%, and of group B is 36.9% and their mean are 101.2 and 101.25 respectively. Find out their standard deviation. [3 Marks]

Answer:
1. For Group A:
    CV_A = 14.6%, Mean (\(\bar{X}_A\)) = 101.2
    Formula: CV = \(\frac{\sigma}{\bar{X}}\) × 100 ⇒ \(14.6 = \frac{\sigma_A}{101.2}\) × 100
    Standard deviation (\(\sigma_A\)) = \(\frac{14.6 \times 101.2}{100}\) = 14.78
2. For Group B:
    CV_B = 36.9%, Mean (\(\bar{X}_B\)) = 101.25
    \(36.9 = \frac{\sigma_B}{101.25}\) × 100
    Standard deviation (\(\sigma_B\)) = \(\frac{36.9 \times 101.25}{100}\) = 37.36

Teacher's Note:
a) Rearrange CV formula to solve for standard deviation: \(\sigma = \frac{CV \times \bar{X}}{100}\).
b) Group A standard deviation is 14.78.
c) Group B standard deviation is 37.36.

8. The mean and standard deviation of income of workers of two factories are given below. Calculate co-efficient of variation of the following data. [3 Marks]

Answer:
1. For Factory A:
    Mean (\(\bar{X}_A\)) = 800, Standard Deviation (\(\sigma_A\)) = 120
    CV_A = \(\frac{\sigma_A}{\bar{X}_A}\) × 100 = \(\frac{120}{800}\) × 100 = 15%
2. For Factory B:
    Mean (\(\bar{X}_B\)) = 900, Standard Deviation (\(\sigma_B\)) = 135
    CV_B = \(\frac{\sigma_B}{\bar{X}_B}\) × 100 = \(\frac{135}{900}\) × 100 = 15%
3. Conclusion: Both factories have the same coefficient of variation (15%).

Teacher's Note:
a) Apply the standard coefficient of variation formula directly.
b) Factory A CV = \(\frac{120}{800}\) × 100 = 15%.
c) Factory B CV = \(\frac{135}{900}\) × 100 = 15%.