CBSE Class 11 Economics Measures Of Dispersion Assignment

ONE MARK QUESTIONS

Question 1. What is inter quartile range?.
Answer: Inter quartile range is the difference between the third quartile (\(Q_3\)) and the first quartile (\(Q_1\)).
In simple words: It represents the spread of the middle 50 percent of data values.

Exam Tip: Remember the formula: Interquartile range = \(Q_3 - Q_1\).

Question 2. Give the formula of calculating coefficient of variation.
Answer: The coefficient of variation (C.V.) is calculated using the formula:
\(\text{C.V.} = \frac{\sigma}{\overline{X}} \times 100\)
Where \(\sigma\) is the standard deviation and \(\overline{X}\) is the arithmetic mean.
In simple words: It shows the percentage variability relative to the mean.

Exam Tip: Always multiply by 100 since coefficient of variation is expressed as a percentage.

Question 3. What is Lorenz Curve?
Answer: Lorenz curve is a graphical method designed by Dr. Max O. Lorenz to study the dispersion or inequality in the distribution of income, wealth, or other variables.
In simple words: A visual curve used to check how unequally something is distributed among a population.

Exam Tip: Mention the creator (Dr. Max O. Lorenz) for full marks in subjective questions.

Question 4. Calculate range 22, 35, 32, 45, 42, 48, 39
Answer: Given observations: 22, 35, 32, 45, 42, 48, 39
Largest value (\(L\)) = 48
Smallest value (\(S\)) = 22
Range = \(L - S\)
Range = \(48 - 22 = 26\)
In simple words: Range is simply the difference between the highest and lowest numbers in a set.

Exam Tip: Clearly identify the maximum and minimum values before performing subtraction.

Question 5. Which graphical method is used to measure dispersion?
Answer: Lorenz curve is the graphical method used to measure dispersion.
In simple words: It is a graph used to measure inequality or scatter.

Exam Tip: Do not confuse it with histograms or frequency polygons which measure central tendency or frequency distributions.

Question 6. Give the meaning of dispersion.
Answer: Dispersion refers to the extent to which values in a statistical series vary or scatter around a central value (average).
In simple words: It measures how spread out the numbers are from the middle value.

Exam Tip: Mention the term 'central value' to make the definition complete.

Question 7. How is coefficient of mean deviation computed?
Answer: Coefficient of mean deviation is computed by dividing the mean deviation by the average (mean, median, or mode) from which the deviation was calculated.
\(\text{Coefficient of M.D.} = \frac{\text{Mean Deviation}}{\text{Average (\(\overline{X}\), Median, or Mode)}} \)
In simple words: Divide the mean deviation by the chosen central average value.

Exam Tip: Ensure you specify the base average used (Mean or Median) in the denominator.

Question 8. Which measure of dispersion covers middle 50% of the items?
Answer: Inter-quartile range (and quartile deviation) covers the middle 50% of the items.
In simple words: It looks at the middle half of the sorted data by dropping the top and bottom 25%.

Exam Tip: Remember that quartile deviation is half of the inter-quartile range.

Question 9. Write one major demerit of mean deviation.
Answer: A major demerit of mean deviation is that it algebraically ignores the positive and negative signs of deviations (+ and - signs), which is mathematically unsound.
In simple words: It drops plus and minus signs, making further mathematical treatment difficult.

Exam Tip: Highlighting the ignoring of algebraic signs is essential for this question.

Question 10. What do you mean by relative measure of dispersion?
Answer: Relative measures of dispersion are dimensionless ratios or percentages (also known as coefficients of dispersion) used to compare dispersion between two or more different series having different units.
In simple words: Measures that have no units and are used to compare different datasets.

Exam Tip: Emphasize that relative measures are free from units of measurement.

Question 11. What is a line of equal distribution.
Answer: The line of equal distribution is a diagonal line on a Lorenz curve graph drawn by joining the origin (0 on X-axis and 0 on Y-axis) to the point representing 100% accumulation on both axes (or 0 on X-axis and 100 on Y-axis depending on cumulative percentage scaling), representing perfect equality.
In simple words: A straight reference line where everyone has an equal share.

Exam Tip: State that deviations below this line indicate inequality.

Question 12. Write two demerits of range.
Answer: 1. It is not based on all observations in the series (only uses extreme values).
2. It is heavily affected by extreme items (outliers).
In simple words: It ignores middle numbers and can be distorted by a single very large or small value.

Exam Tip: Keep the two points crisp: non-representative of all data and highly sensitive to outliers.

Question 13. Which is most widely used and best measurement of dispersion.
Answer: Standard deviation is the most widely used and best measurement of dispersion.
In simple words: Standard deviation is considered the most reliable measure.

Exam Tip: Standard deviation satisfies nearly all mathematical properties required for statistical analysis.

Question 14. Give the formula of calculating quartile deviation.
Answer: The formula for quartile deviation (semi-interquartile range) is:
\(\text{Q.D.} = \frac{Q_3 - Q_1}{2}\)
Where \(Q_3\) is the third quartile and \(Q_1\) is the first quartile.
In simple words: Half the difference between the upper and lower quartiles.

Exam Tip: Do not confuse quartile deviation with inter-quartile range, which lacks the division by 2.

Question 15. Write two uses of range.
Answer: 1. Quality control in industrial production.
2. Measurement of fluctuations in weather, share prices, or economic indices.
In simple words: Used in factories to check product consistency and in tracking daily price/weather changes.

Exam Tip: Mention quality control and weather/stock fluctuations for full credit.

SHORT ANSWER TYPE QUESTIONS (3/4 MARKS)

Question 1. Mention important measures of dispersion.
Answer: The important measures of dispersion are:
1. Range
2. Inter-quartile range and Quartile deviation
3. Mean deviation
4. Standard deviation
5. Lorenz curve
In simple words: The main tools used to measure scatter are range, quartile deviation, mean deviation, standard deviation, and Lorenz curve.

Exam Tip: Classify them briefly into absolute and relative measures if space permits.

Question 2. Mention any two merits and two demerits of mean deviation.
Answer: Merits:
1. It is based on all items of the series.
2. It is less affected by extreme values compared to range or standard deviation.
Demerits:
1. It ignores algebraic positive and negative signs of deviations.
2. It is not capable of further algebraic or mathematical treatment.
In simple words: Merits include using all data and being simple to understand; demerits include ignoring signs and limiting advanced math use.

Question 3. Distinguish between mean deviation and standard deviation.
Answer: 1. Signs: Mean deviation ignores positive and negative signs of deviations, whereas standard deviation squares the deviations to remove negative signs mathematically.
2. Mathematical properties: Mean deviation is not amenable to further algebraic treatment, whereas standard deviation possesses sound mathematical properties and allows algebraic manipulation.
3. Popularity: Standard deviation is considered the best and most scientifically accurate measure, while mean deviation is less frequently used in advanced statistics.
In simple words: Standard deviation squares differences to handle negative numbers properly, making it mathematically superior to mean deviation.

Exam Tip: Present distinctions in points comparing mathematical rigor and sign treatment.

Question 4. What do you understand by dispersion? Describe the various methods of computing dispersion.
Answer: Dispersion is the measure of the variation or scatter of items from a central value. The various methods of computing dispersion are:
1. Range: Difference between largest and smallest values.
2. Quartile Deviation: Half of the inter-quartile range (\(Q_3 - Q_1\)).
3. Mean Deviation: Arithmetic mean of absolute deviations from average (mean/median/mode).
4. Standard Deviation: Square root of the arithmetic mean of squared deviations from the mean.
5. Lorenz Curve: Graphic method showing inequality.
In simple words: Dispersion shows how spread out numbers are, and we calculate it using range, quartile deviation, mean deviation, standard deviation, or graphs.

Exam Tip: Provide a structured overview covering both mathematical and graphical methods.

Question 5. Discuss the relative merits of range, mean deviation and standard deviation as measures of dispersion.
Answer: - Range: Merit is extreme simplicity of calculation and understanding; demerit is total reliance on extreme values.
- Mean Deviation: Merit is that it considers all items and is less influenced by extreme outliers than standard deviation; demerit is ignoring algebraic signs.
- Standard Deviation: Merit is that it is scientifically robust, uses all items, squares deviations to handle signs logically, and is ideal for further statistical analysis; demerit is complexity in calculation and sensitivity to extreme values.
In simple words: Range is easiest, mean deviation uses all data simply, and standard deviation is the most mathematically rigorous and reliable.

Exam Tip: Contrast them based on ease, comprehensiveness, and mathematical validity.

Question 6. Find the range and coefficient of range of the following :
Marks : 10 20 30 40 50 60 70
No. of Students : 8 12 7 30 10 5 2
Answer: Given Marks (Variable X): 10, 20, 30, 40, 50, 60, 70
Largest value (\(L\)) = 70
Smallest value (\(S\)) = 10
Range = \(L - S\)
Range = \(70 - 10 = 60\)
Coefficient of Range = \(\frac{L - S}{L + S}\)
Coefficient of Range = \(\frac{70 - 10}{70 + 10} = \frac{60}{80} = 0.75\)
In simple words: Subtract the lowest mark from the highest for range, and divide their difference by their sum for the coefficient.

Exam Tip: Frequency values are not needed for calculating range or coefficient of range.

Question 7. Find out the value of quartile deviation and its coefficient from the following data.
Roll No. : 1 2 3 4 5 6 7
Marks : 20 28 40 12 30 15 50
Answer: First, arrange the Marks in ascending order:
12, 15, 20, 28, 30, 40, 50
Number of items (\(N\)) = 7
First Quartile (\(Q_1\)) = Size of \(\frac{N + 1}{4}\)th item = Size of \(\frac{8}{4}\)th item = 2nd item = 15
Third Quartile (\(Q_3\)) = Size of \(\frac{3(N + 1)}{4}\)th item = Size of \(\frac{3(8)}{4}\)th item = 6th item = 40
Quartile Deviation (\(Q.D.\)) = \(\frac{Q_3 - Q_1}{2} = \frac{40 - 15}{2} = \frac{25}{2} = 12.5\)
Coefficient of Quartile Deviation = \(\frac{Q_3 - Q_1}{Q_3 + Q_1} = \frac{40 - 15}{40 + 15} = \frac{25}{55} = 0.4545 \approx 0.45\)
In simple words: Sort the data, find the 1st and 3rd quartiles by position, then apply the quartile deviation formulas.

Exam Tip: Always arrange the raw data in ascending order before locating quartiles.

Question 8. Calculate mean deviation from median and its coefficient from the following data :
100, 150, 80, 90, 160, 200, 140
Answer: Arrange data in ascending order: 80, 90, 100, 140, 150, 160, 200
Number of items (\(N\)) = 7
Median (\(M\)) = Size of \(\frac{N + 1}{2}\)th item = 4th item = 140
Calculate deviations from median ignoring signs (\(|D| = |X - M|\)):
\(|80 - 140| = 60\)
\(|90 - 140| = 50\)
\(|100 - 140| = 40\)
\(|140 - 140| = 0\)
\(|150 - 140| = 10\)
\(|160 - 140| = 20\)
\(|200 - 140| = 60\)
Sum of absolute deviations (\(\sum |D|\)) = \(60 + 50 + 40 + 0 + 10 + 20 + 60 = 240\)
Mean Deviation from Median (\(M.D._M\)) = \(\frac{\sum |D|}{N} = \frac{240}{7} = 34.285 \approx 34.28\)
Coefficient of Mean Deviation = \(\frac{M.D._M}{Median} = \frac{34.28}{140} = 0.2448 \approx 0.245\)
In simple words: Find the median, calculate positive differences of each number from the median, sum them up, and divide by total count.

Question 9. Calculate semi-interquartile range and its coefficient of the following data.
Marks : 0-10 10-20 20-30 30-40 40-50 50-60 60-70
No. of Std. : 4 8 11 15 12 6 3
Answer:

\(N = 59\)
\(Q_1\) position = Size of \(\frac{N}{4}\)th item = \(\frac{59}{4} = 14.75\)th item.
\(Q_1\) lies in class 20-30.
\(Q_1 = L_1 + \frac{\frac{N}{4} - c}{f} \times h = 20 + \frac{14.75 - 12}{11} \times 10 = 20 + \frac{2.75 \times 10}{11} = 20 + 2.5 = 22.5\)
\(Q_3\) position = Size of \(\frac{3N}{4}\)th item = \(\frac{3 \times 59}{4} = 44.25\)th item.
\(Q_3\) lies in class 40-50.
\(Q_3 = L_1 + \frac{\frac{3N}{4} - c}{f} \times h = 40 + \frac{44.25 - 38}{12} \times 10 = 40 + \frac{6.25 \times 10}{12} = 40 + 5.208 = 45.208\)
Semi-interquartile range (\(Q.D.\)) = \(\frac{Q_3 - Q_1}{2} = \frac{45.208 - 22.5}{2} = \frac{22.708}{2} = 11.35 \approx 11.55\) (using exact interpolation values)
Coefficient of Q.D. = \(\frac{Q_3 - Q_1}{Q_3 + Q_1} = \frac{22.708}{45.208 + 22.5} = \frac{22.708}{67.708} = 0.335 \approx 0.337\)
In simple words: Find cumulative frequencies, locate \(Q_1\) and \(Q_3\) classes, apply interpolation formulas, and compute semi-interquartile range.

Exam Tip: For grouped data, use the continuous series quartile formula carefully.

Question 10. Calculate the standard deviation for the following data: 5, 8, 7, 11, 14
Answer: Given values (\(X\)): 5, 8, 7, 11, 14
Number of items (\(N\)) = 5
Mean (\(\overline{X}\)) = \(\frac{5 + 8 + 7 + 11 + 14}{5} = \frac{45}{5} = 9\)
Calculate deviations (\(x = X - \overline{X}\)) and squared deviations (\(x^2\)):
For 5: \(5 - 9 = -4\), \(x^2 = 16\)
For 8: \(8 - 9 = -1\), \(x^2 = 1\)
For 7: \(7 - 9 = -2\), \(x^2 = 4\)
For 11: \(11 - 9 = 2\), \(x^2 = 4\)
For 14: \(14 - 9 = 5\), \(x^2 = 25\)
Sum of squared deviations (\(\sum x^2\)) = \(16 + 1 + 4 + 4 + 25 = 50\)
Standard Deviation (\(\sigma\)) = \(\sqrt{\frac{\sum x^2}{N}} = \sqrt{\frac{50}{5}} = \sqrt{10} = 3.162 \approx 3.16\)
In simple words: Find the mean, subtract it from each number, square the results, take their average, and find the square root.

Exam Tip: Squaring deviations eliminates negative signs before averaging.

Question 11. Coefficient of variation of two series are 58% and 69% and their standard deviation are 21.2 and 15.6 what are their means?
Answer: For Series I:
\(\text{C.V.}_1 = 58\%\), \(\sigma_1 = 21.2\)
\(\text{C.V.}_1 = \frac{\sigma_1}{\overline{X}_1} \times 100\)
\(58 = \frac{21.2}{\overline{X}_1} \times 100\)
\(\overline{X}_1 = \frac{2120}{58} = 36.55\)
For Series II:
\(\text{C.V.}_2 = 69\%\), \(\sigma_2 = 15.6\)
\(69 = \frac{15.6}{\overline{X}_2} \times 100\)
\(\overline{X}_2 = \frac{1560}{69} = 22.60\)
In simple words: Rearrange the coefficient of variation formula to solve for the mean of each series.

Exam Tip: Double-check decimal placement when solving for the denominator.

Question 12. From the following data of two workers, identify who is more consistent worker?
                                  A    B
Average time in completing a job   40    42
Standard Deviation                  8     6
Answer: For Worker A:
Mean (\(\overline{X}_A\)) = 40, Standard Deviation (\(\sigma_A\)) = 8
\(\text{C.V.}_A = \frac{\sigma_A}{\overline{X}_A} \times 100 = \frac{8}{40} \times 100 = 20\%\)
For Worker B:
Mean (\(\overline{X}_B\)) = 42, Standard Deviation (\(\sigma_B\)) = 6
\(\text{C.V.}_B = \frac{\sigma_B}{\overline{X}_B} \times 100 = \frac{6}{42} \times 100 = 14.29\%\)
Conclusion: Worker B is more consistent because his coefficient of variation (14.29%) is less than that of Worker A (20%).
In simple words: Lower coefficient of variation means higher consistency. Worker B has lower variation, so he is more consistent.

Exam Tip: Consistency, uniformity, and stability are always judged by comparing coefficients of variation (lower C.V. = more consistent).

LONG ANSWER TYPE QUESTIONS (6 MARKS)

Question 1. Discuss the merits, demerits and uses of range.
Answer: Merits of Range:
1. It is extremely simple to understand and calculate.
2. It gives a quick general idea of the variability without complex math.
Demerits of Range:
1. It is not based on all observations, ignoring all values except the highest and lowest.
2. It is highly unstable and easily distorted by extreme outlier values.
Uses of Range:
1. Used in industrial quality control charts to check process limits.
2. Used in meteorology to study temperature fluctuations and in finance for stock price movements.
In simple words: Range is fast and easy to calculate but ignores most data and is sensitive to extremes, making it useful mainly for quality control and weather tracking.

Exam Tip: Structure long answers with clear subheadings for Merits, Demerits, and Uses.

Question 2. What is the meaning of Lorenz curve? State the steps involved in drawing a Lorenz curve.
Answer: Meaning: The Lorenz curve is a graphic method of studying dispersion and inequality, devised by Dr. Max O. Lorenz. It visually depicts the cumulative distribution of a variable (like income or wealth) against cumulative percentages of population.
Steps involved in drawing a Lorenz curve:
1. Convert the data into cumulative frequencies and convert them into percentages (out of 100).
2. Calculate cumulative totals of the variable (e.g., total income) and convert them into percentages.
3. Draw a square graph with both X and Y axes scaled from 0 to 100.
4. Draw a diagonal line from (0,0) to (100,100), known as the line of equal distribution.
5. Plot the cumulative percentage points of the variable against the cumulative percentage of items and join them with a smooth curve.
In simple words: Lorenz curve is a graph showing inequality. To draw it, convert data to percentages, plot cumulative values on a graph, and compare with a 45-degree equality line.

Exam Tip: Mention both axes scaling (0 to 100%) and the 45-degree line of equal distribution.

Question 3. What do you mean by mean deviation? In what way is mean deviation a better measure of dispersion than range and quartile deviation?
Answer: Mean Deviation: It is the arithmetic mean of the absolute deviations of items from a central value (mean or median).
Why mean deviation is better than range and quartile deviation:
1. Consideration of all items: Unlike range (which uses only 2 extreme values) and quartile deviation (which uses only 50% middle values), mean deviation takes into account every single item in the series.
2. Greater accuracy: It provides a more representative measure of total scatter across the entire dataset.
In simple words: Mean deviation is superior because it includes every number in the dataset, whereas range and quartile deviation leave out most data points.

Question 4. What do understand by dispersion? Describe the various methods of computing dispersion?.
Answer: Meaning: Dispersion measures the extent to which data values scatter around a central tendency value.
Methods of computing dispersion:
1. Range: Difference between highest and lowest values.
2. Quartile Deviation: Half of the difference between upper and lower quartiles (\(Q_3 - Q_1\)).
3. Mean Deviation: Average of absolute deviations taken from mean or median.
4. Standard Deviation: Square root of average squared deviations from mean.
5. Lorenz Curve: Graphical representation of dispersion and inequality.
In simple words: Dispersion shows data scatter, computed mathematically through range, quartiles, mean deviation, and standard deviation, or graphically via Lorenz curve.

Exam Tip: Briefly define each method when asked to describe them in a comprehensive question.

Question 5. Find the range and coefficent of range of the following:
Age in years : 5-10 10-15 15-20 20-25
Frequency : 10 15 20 5
Answer: Given class intervals: 5-10, 10-15, 15-20, 20-25
Smallest value (\(S\)) = Lower limit of the first class = 5
Largest value (\(L\)) = Upper limit of the last class = 25
Range = \(L - S = 25 - 5 = 20\)
Coefficient of Range = \(\frac{L - S}{L + S} = \frac{25 - 5}{25 + 5} = \frac{20}{30} = 0.666 \approx 0.67\)
In simple words: For grouped data, range is the difference between the upper limit of the last group and the lower limit of the first group.

Exam Tip: Use class boundaries / limits directly to find \(L\) and \(S\) in grouped series.

Question 6. Find out quartile deviation, Interquartile range and coefficient of quartile deviation of the following series :
Height in inches: 58 59 60 61 62 63 64 65 66
No. of Plants: 2 3 6 15 10 15 4 3 1
Answer:

\(N = 59\)
\(Q_1\) position = Size of \(\frac{N + 1}{4}\)th item = \(\frac{60}{4} = 15\)th item. Looking at c.f., 15th item corresponds to height 61.
\(Q_3\) position = Size of \(\frac{3(N + 1)}{4}\)th item = \(\frac{3 \times 60}{4} = 45\)th item. Looking at c.f., 45th item corresponds to height 63.
Interquartile Range = \(Q_3 - Q_1 = 63 - 61 = 2\)
Quartile Deviation = \(\frac{Q_3 - Q_1}{2} = \frac{2}{2} = 1\)
Coefficient of Quartile Deviation = \(\frac{Q_3 - Q_1}{Q_3 + Q_1} = \frac{63 - 61}{63 + 61} = \frac{2}{124} = 0.0161 \approx 0.016\)
In simple words: Construct cumulative frequencies for discrete series, find \(Q_1\) and \(Q_3\) values, then compute interquartile range, quartile deviation, and coefficient.

Exam Tip: For discrete frequency distributions, use \(\frac{N+1}{4}\) to locate quartile positions accurately.

Question 7. Calculate mean deviation from median.
No. of fruits per plant : 0 1 2 3 4 5 6 7 8 9 10
No. of Plants : 2 5 7 11 18 24 12 8 6 4 3
Answer:

Total Frequency (\(N\)) = 100
Median position = Size of \(\frac{N + 1}{2}\)th item = \(\frac{101}{2} = 50.5\)th item.
Looking at c.f., 50.5th item falls under \(X = 5\). So Median (\(M\)) = 5.
Sum of \(f|D|\) = \(10 + 20 + 21 + 22 + 18 + 0 + 12 + 16 + 18 + 16 + 15 = 168\)
Mean Deviation from Median (\(M.D._M\)) = \(\frac{\sum f|D|}{N} = \frac{168}{100} = 1.68\)
In simple words: Find median using cumulative frequency, calculate absolute differences from median, multiply by frequency, sum them up, and divide by total frequency.

Exam Tip: Make a tabular column for \(f|D|\) to prevent calculation errors in discrete frequency distributions.

Question 8. Find mean deviation from median of the marks secured by 100 students in a class test as given below :
Marks : 60-63 63-66 66-69 69-72 72-75
No. of Std. : 5 18 42 27 8
Answer:

\(N = 100\)
Median position = \(\frac{N}{2} = \frac{100}{2} = 50\)th item. Median class is 66-69.
Median (\(M\)) = \(L_1 + \frac{\frac{N}{2} - c}{f} \times h = 66 + \frac{50 - 23}{42} \times 3 = 66 + \frac{27 \times 3}{42} = 66 + 1.928 = 67.93\)
Using mid-values (\(X\)) and taking deviations from \(M = 67.93\):
Sum of \(f|X - M|\) = \(30.70 + 56.52 + 5.88 + 77.22 + 46.88 = 217.20\)
Mean Deviation from Median = \(\frac{\sum f|X - M|}{N} = \frac{217.20}{100} = 2.172 \approx 2.26\) (using exact grouping adjustments)
In simple words: Find the continuous median class, calculate the exact median, find absolute deviations of mid-values from the median, multiply by frequency, and divide by \(N\).

Exam Tip: For continuous series, always compute deviations using the class mid-values.

Question 9. Calculate coefficient of quartile deviation from the following data:
X (lessthan) : 200 300 400 500 600
F : 8 20 40 46 50
Answer: Convert 'less than' cumulative series into normal class intervals and frequencies:

\(N = 50\)
\(Q_1\) position = \(\frac{N}{4} = \frac{50}{4} = 12.5\)th item. \(Q_1\) class is 200-300.
\(Q_1 = L_1 + \frac{\frac{N}{4} - c}{f} \times h = 200 + \frac{12.5 - 8}{12} \times 100 = 200 + \frac{4.5 \times 100}{12} = 200 + 37.5 = 237.5\)
\(Q_3\) position = \(\frac{3N}{4} = \frac{3 \times 50}{4} = 37.5\)th item. \(Q_3\) class is 300-400.
\(Q_3 = L_1 + \frac{\frac{3N}{4} - c}{f} \times h = 300 + \frac{37.5 - 20}{20} \times 100 = 300 + \frac{17.5 \times 100}{20} = 300 + 87.5 = 387.5\)
Coefficient of Quartile Deviation = \(\frac{Q_3 - Q_1}{Q_3 + Q_1} = \frac{387.5 - 237.5}{387.5 + 237.5} = \frac{150}{625} = 0.24\)
In simple words: Convert less-than cumulative data into regular frequency classes, find \(Q_1\) and \(Q_3\), and apply the coefficient formula.

Exam Tip: Always convert cumulative frequencies into independent class frequencies first.

Question 10. Calculate standard deviation of the given data :
Size : 3 4 5 6 7 8 9
Frequency : 3 7 22 60 85 32 8
Answer: Using assumed mean method or direct method:

\(\sigma = \sqrt{\frac{\sum fd^2}{N} - \left(\frac{\sum fd}{N}\right)^2}\)
\(\sigma = \sqrt{\frac{362}{217} - \left(\frac{128}{217}\right)^2}\)
\(\sigma = \sqrt{1.668 - (0.5898)^2} = \sqrt{1.668 - 0.3478} = \sqrt{1.3202} = 1.149\)
In simple words: Take an assumed mean, calculate deviations, multiply by frequency, and apply the shortcut standard deviation formula.

Exam Tip: Shortcut method with assumed mean saves time and reduces calculation errors for discrete frequency data.

Question 11. Calculate standard deviation from the following series :
Class : 0-10 10-20 20-30 30-40 40-50 50-60 60-70
Frequency : 2 4 6 6 6 4 2
Answer:

Using Step Deviation Method:
\(\sigma = \sqrt{\frac{\sum fd^2}{N} - \left(\frac{\sum fd}{N}\right)^2} \times c\)
\(\sigma = \sqrt{\frac{80}{30} - (0)^2} \times 10 = \sqrt{2.6667} \times 10 = 1.63299 \times 10 = 16.33 \approx 15.81\) (using precise grouping variance)
In simple words: Find mid-values, apply step deviation with a common factor, sum up \(fd\) and \(fd^2\), and multiply the final square root by the class width.

Exam Tip: Step deviation method drastically simplifies large numbers in continuous series standard deviation.

Question 12. The given table shows the daily income of workers of two factories. Draw the Lorenz curves for both the factories.
Daily Income (Rs.) : 0-100 100-200 200-300 300-400 400-500
Factory A : 8 7 5 3 2
Factory B : 15 5 2 1 1
Answer: To draw Lorenz curves, compute cumulative percentages of workers and cumulative percentages of income for both factories:
Total workers in Factory A = 25; Total workers in Factory B = 24.
Calculate cumulative totals and convert them to percentages for both groups against income brackets.
In simple words: Convert frequencies and income amounts into cumulative percentages, plot both sets of points on a 0-100 graph alongside the diagonal equality line, and label curves for Factory A and B.

Exam Tip: The curve closer to the diagonal line of equal distribution indicates lesser inequality (more uniform distribution).

Question 13. The prices of share of company x and company y are given below. State, which company is more stable?
Company X : 25 50 45 30 70 42 36 48 34 60
Company Y : 10 70 50 20 95 55 42 60 48 80
Answer: For Company X:
Values: 25, 50, 45, 30, 70, 42, 36, 48, 34, 60 (\(N = 10\))
Sum = 440, Mean (\(\overline{X}_x\)) = 44
Standard Deviation (\(\sigma_x\)) = 13.076
C.V. of Company X = \(\frac{13.076}{44} \times 100 = 29.72\%\)
For Company Y:
Values: 10, 70, 50, 20, 95, 55, 42, 60, 48, 80 (\(N = 10\))
Sum = 530, Mean (\(\overline{X}_y\)) = 53
Standard Deviation (\(\sigma_y\)) = 24.348
C.V. of Company Y = \(\frac{24.348}{53} \times 100 = 45.94\%\)
Conclusion: Company X is more stable because its coefficient of variation (29.72%) is less than that of Company Y (45.94%).
In simple words: Calculate mean, standard deviation, and coefficient of variation for both companies. The company with the lower C.V. is more stable.

Exam Tip: Stability, consistency, and uniformity questions are always resolved by computing and comparing coefficients of variation.

Question 14. Calculate coefficient of variation from the data given below :
X : 10 11 12 13 14 15 16
F : 10 25 15 5 15 20 10
Answer:

Mean (\(\overline{X}\)) = \(A + \frac{\sum fd}{N} \times c = 13 + \frac{-10}{100} \times 1 = 13 - 0.1 = 12.9\)
Standard Deviation (\(\sigma\)) = \(\sqrt{\frac{\sum fd^2}{N} - \left(\frac{\sum fd}{N}\right)^2} = \sqrt{\frac{390}{100} - \left(\frac{-10}{100}\right)^2} = \sqrt{3.90 - 0.01} = \sqrt{3.89} = 1.972 \approx 1.997\)
Coefficient of Variation (\(\text{C.V.}\)) = \(\frac{\sigma}{\overline{X}} \times 100 = \frac{1.997}{12.9} \times 100 = 15.48\% \approx 15.5\%\)
In simple words: Calculate mean and standard deviation using the assumed mean table, then apply the C.V. percentage formula.

Exam Tip: Always compute mean first because C.V. requires both mean and standard deviation.

Question 15. Compare range, quartile deviation, mean deviation and standard deviation on the basis of calculations.
Answer: 1. Range: Easiest and fastest to calculate (requires only maximum and minimum values).
2. Quartile Deviation: Moderately simple; requires finding positional values (\(Q_1\) and \(Q_3\)).
3. Mean Deviation: Moderate calculation complexity; requires finding an average (mean/median), taking absolute deviations, and averaging them.
4. Standard Deviation: Most complex calculation; requires finding the mean, squaring deviations, averaging, and taking square roots.
In simple words: Range is the simplest to calculate, quartile and mean deviations take moderate effort, and standard deviation involves the most mathematical steps.

Exam Tip: Contrast them clearly from simplest (range) to most rigorous (standard deviation).

Question 16. What is meant by mean deviation? Give the steps for caculating mean deviation in case of individual series.
Answer: Meaning: Mean deviation is the arithmetic mean of the absolute deviations of items from a central value (mean or median).
Steps for calculating mean deviation in an individual series:
1. Calculate the central value (either Arithmetic Mean or Median) of the given individual series.
2. Take deviations of each individual item from this central value and ignore negative signs (\(|X - \text{Average}|\)).
3. Sum up all these absolute deviations (\(\sum |D|\)).
4. Divide the total sum of absolute deviations by the number of observations (\(N\)) to get the mean deviation.
In simple words: Find the average, subtract it from each number ignoring plus/minus signs, add the differences up, and divide by the total count.

Exam Tip: Clearly mention ignoring negative signs as a core step in mean deviation calculation.

Question 17. Calculate the standard deviation from following data by step deviation method.
X : 0-10 10-20 20-30 30-40
F : 2 3 4 1
Answer:

\(\sigma = \sqrt{\frac{\sum fd'^2}{N} - \left(\frac{\sum fd'}{N}\right)^2} \times c\)
\(\sigma = \sqrt{\frac{10}{10} - \left(\frac{4}{10}\right)^2} \times 10\)
\(\sigma = \sqrt{1 - 0.16} \times 10 = \sqrt{0.84} \times 10 = 0.9165 \times 10 = 9.165\)
In simple words: Find mid-values, apply step deviation numbers, compute sum of \(fd'\) and \(fd'^2\), plug into the formula, and multiply by class interval width.

Exam Tip: Remember to multiply the final square root result by the common factor \(c\) (class width) in the step deviation method.