GSEB Class 11 Statistics Solutions Chapter 4 Measures of Dispersion Exercise 4

Exercise 4

Section - A

Question 1. From the following, which is the formula for coefficient of range?
(a) \(x_{\mathrm{H}} - x_{\mathrm{L}}\)
(b) \(\frac{x_{\mathrm{H}} - x_{\mathrm{L}}}{x_{\mathrm{H}} + x_{\mathrm{L}}}\)
(c) \(\frac{x_{\mathrm{H}} + x_{\mathrm{L}}}{x_{\mathrm{H}} - x_{\mathrm{L}}}\)
(d) \(x_{\mathrm{H}} - x_{\mathrm{L}}\)
Answer: (b) \(\frac{x_{\mathrm{H}}-x_{\mathrm{L}}}{x_{\mathrm{H}}+x_{\mathrm{L}}}\)
In simple words: The coefficient of range is a relative measure of dispersion, calculated by dividing the difference between the highest and lowest observations by their sum.

🎯 Exam Tip: Understand the difference between absolute and relative measures of dispersion, as coefficients are often used for comparison across different datasets.

 

Question 2. In which measure of dispersion, the absolute difference of the observation and its mean is considered?
(a) Mean Deviation
(b) Standard Deviation
(c) Range
(d) Quartile Deviation
Answer: (a) Mean Deviation
In simple words: Mean Deviation calculates dispersion by averaging the absolute differences between each data point and the mean.

🎯 Exam Tip: Be able to distinguish which measure uses absolute differences from the mean (Mean Deviation) versus squared differences (Standard Deviation).

 

Question 3. Which of the following measures is a unit-free measure?
(a) Mean Deviation
(b) Quartile Deviation
(c) Range
(d) Coefficient of Variation
Answer: (d) Coefficient of Variation
In simple words: The Coefficient of Variation is a relative measure expressed as a percentage, making it unit-free and suitable for comparing variability across datasets with different units or scales.

🎯 Exam Tip: Remember that relative measures of dispersion (like coefficients) are unit-free and ideal for comparing variability, while absolute measures retain the original data units.

 

Question 4. Which measure of dispersion is least affected by the extreme values of the observations?
(a) Range
(b) Standard Deviation
(c) Quartile Deviation
(d) Mean Deviation
Answer: (c) Quartile Deviation
In simple words: Quartile Deviation focuses on the middle 50% of data, making it less sensitive to extreme outliers compared to range or standard deviation.

🎯 Exam Tip: Quartile Deviation is robust to outliers, making it a good choice for skewed distributions or datasets with extreme values.

 

Question 5. The coefficient of variation of group A is less than the coefficient of variation of group B. Which group is more consistent with respect to variability?
(a) A
(b) B
(c) Both
(d) None of these
Answer: (a) A
In simple words: A lower coefficient of variation indicates less relative variability, meaning group A has more consistent data than group B.

🎯 Exam Tip: A smaller coefficient of variation implies greater consistency or homogeneity within a dataset, which is a key concept for comparative analysis.

 

Question 6. The weight (in kg) for 10 students are 53, 47, 60, 55, 71, 65, 61, 68, 63, 70. What is the range of the data?
(a) 17 kg
(b) 23 kg
(c) 24 kg
(d) 18 kg
Answer: (c) 24 kg
In simple words: The range is found by subtracting the minimum observed value from the maximum observed value in the dataset. For the given weights, the maximum is 71 kg and the minimum is 47 kg, so the range is 71 - 47 = 24 kg.

🎯 Exam Tip: To calculate the range, first identify the highest and lowest values in the dataset. Always include the unit in your final answer if given in the question.

 

Question 7. If the first quartile and the third quartile of a data are 30 and 50 respectively then what is the value of coefficient of quartile deviation?
(a) 0.25
(b) 50
(c) 4
(d) 20
Answer: (a) 0.25
In simple words: The coefficient of quartile deviation is calculated by dividing the quartile deviation (Q3 - Q1) by the sum of the quartiles (Q3 + Q1). With Q1 = 30 and Q3 = 50, this becomes \(\frac{50-30}{50+30} = \frac{20}{80} = 0.25\).

🎯 Exam Tip: Remember the formula for the coefficient of quartile deviation: \(\frac{Q_3 - Q_1}{Q_3 + Q_1}\). This is a relative measure useful for comparing dispersion in different datasets.

 

Question 8. What is the value of any measure of dispersion for the observations 5, 5, 5, 5, 5?
(a) 1
(b) 5
(c) 0
(d) 25
Answer: (c) 0
In simple words: When all observations in a dataset are identical, there is no variability or spread, so any measure of dispersion will be zero.

🎯 Exam Tip: Measures of dispersion quantify the spread of data; if data points are identical, there is no spread, leading to a dispersion value of zero.

 

Question 9. If mean of a variable is 10 and the coefficient of variation is 60%. What is the variance of the variable?
(a) 6
(b) 36
(c) 60
(d) 50
Answer: (b) 36
In simple words: The coefficient of variation (CV) is the standard deviation (s) divided by the mean (\(\bar{x}\)), multiplied by 100. Given CV = 60% and \(\bar{x}\) = 10, we find s = 6. The variance is then the square of the standard deviation, so variance = \(s^2 = 6^2 = 36\).

🎯 Exam Tip: Know the relationship between the coefficient of variation, standard deviation, and variance. \(CV = (\frac{s}{\bar{x}}) \times 100\), and variance = \(s^2\).

 

Question 10. If standard deviation of the series \(k_1, k_2, k_3, ..., k_n\) is 5. What will be the standard deviation of the following series?
(i) \(k_1 + 2, k_2 + 2, k_3 + 2, ..., k_n + 2\)
(ii) \(3k_1, 3k_2, 3k_3, ..., 3k_n\)
(a) (i) 7 (ii) 3
(b) (i) 5 (ii) 3
(c) (i) 5 (ii) 15
(d) (i) 7 (ii) 15
Answer: (c) (i) 5 (ii) 15
In simple words: Adding a constant to each observation does not change the standard deviation, so for series (i) it remains 5. Multiplying each observation by a constant multiplies the standard deviation by the absolute value of that constant, so for series (ii) it becomes \(3 \times 5 = 15\).

🎯 Exam Tip: Remember how transformations affect standard deviation: adding/subtracting a constant has no effect, while multiplying/dividing by a constant scales the standard deviation by that constant's absolute value.

 

Question 11. The mean and standard deviation for a variable x are 5 and 2 respectively. Now, if \(y = 3x + 4\) then what are the mean and the standard deviation of y?
(a) 19 and 6
(b) 15 and 49
(c) 19 and 10
(d) 15 and 10
Answer: (a) 19 and 6
In simple words: For a linear transformation \(y = ax + b\), the new mean is \(a \times (\text{old mean}) + b\), and the new standard deviation is \(|a| \times (\text{old standard deviation})\). Here, mean of y = \(3 \times 5 + 4 = 19\), and standard deviation of y = \(3 \times 2 = 6\).

🎯 Exam Tip: Understand the properties of mean and standard deviation under linear transformations. The mean shifts and scales, while the standard deviation only scales.

 

Question 12. The mean and standard deviation of a set of observations are 45 and 5 respectively. If a constant 5 is added to each observation, what is the coefficient of variation of the new set of the observations?
(a) 10%
(b) 50%
(c) 11.11%
(d) 900%
Answer: (a) 10%
In simple words: When a constant is added to each observation, the mean changes by that constant, but the standard deviation remains unchanged. So, the new mean is \(45 + 5 = 50\), and the standard deviation is still 5. The new coefficient of variation is then \(\frac{5}{50} \times 100\% = 10\%\).

🎯 Exam Tip: A crucial concept: adding a constant affects the mean but not the standard deviation, which then impacts the coefficient of variation.

Section - B

Answer the following question in one sentence:

Question 1. Define the range.
Answer: The range is defined as the absolute difference between the highest and lowest observations within a dataset, denoted by the symbol R, i.e., \(R = X_H - X_L\).
In simple words: Range is simply the difference between the maximum and minimum values in a dataset.

🎯 Exam Tip: The range is the simplest measure of dispersion, but it's highly sensitive to extreme values. Clearly define it with its formula.

 

Question 2. Define the quartile deviation.
Answer: Quartile deviation is a measure derived by dividing the difference between the third quartile (\(Q_3\)) and the first quartile (\(Q_1\)) of a dataset by 2, and it is represented as \(Q_d\), i.e., \(Q_d = \frac{Q_3 - Q_1}{2}\).
In simple words: Quartile deviation measures the spread of the middle half of the data, calculated as half the difference between the third and first quartiles.

🎯 Exam Tip: When defining quartile deviation, always include its formula and mention that it focuses on the central 50% of the data, making it less affected by outliers.

 

Question 3. Which types of measures of dispersion are used for comparing two or more groups in terms of the variability?
Answer: Relative measures of dispersion are utilized for comparing the variability of two or more groups.
In simple words: Relative measures like the coefficient of variation are used to compare the spread of different datasets.

🎯 Exam Tip: Emphasize that relative measures are essential for comparing variability across groups with potentially different units or scales, as they provide a standardized comparison.

 

Question 4. Which is the best measure of dispersion?
Answer: Standard deviation is generally considered the best measure of dispersion.
In simple words: Standard deviation is preferred because it considers all data points and has good mathematical properties.

🎯 Exam Tip: Standard deviation is a cornerstone of statistical analysis due to its mathematical rigor and its use in many advanced statistical methods. Mentioning these properties is key.

 

Question 5. If the data of heights of 10 students are given in centimeter, what is the unit of its variance?
Answer: If the height data for 10 students is provided in centimeters, then the unit of its variance will be (Centimeter)\(^2\).
In simple words: The unit of variance is always the square of the original data's unit.

🎯 Exam Tip: Remember that variance units are always squared units of the original data, whereas standard deviation reverts to the original units. This is a common point of confusion.

 

Question 6. For a company making pipes, the following information of diameter (in cm) of pipes is obtained. Find the range of the diameter of the pipes.

Diameter (cm)	20-40	40-60	60-80	80-100	100-120
No. of pipes	15	40	75	20	11

Answer:
For this data:
\(X_H\) = The upper limit of the last class interval (100-120) = 120 cm.
\(X_L\) = The lower limit of the initial class interval (20-40) = 20 cm.

\( \implies \) Range \(R = X_H - X_L\)
= \(120 - 20 = 100\) cm
In simple words: To find the range for grouped data, identify the highest possible value (upper limit of the last class) and the lowest possible value (lower limit of the first class) and subtract the latter from the former.

🎯 Exam Tip: For grouped frequency distributions, the range is typically estimated as the difference between the upper boundary of the highest class and the lower boundary of the lowest class. Make sure to use class boundaries correctly.

 

Question 7. The 25th and 75th percentiles of a frequency distribution are 72.18 and 103.99 respectively. Find the quartile deviation.
Answer:
Given:
25th percentile = \(Q_1\) = 72.18.
75th percentile = \(Q_3\) = 103.99.

Quartile deviation \(Q_d = \frac{Q_3 - Q_1}{2}\)
\( = \frac{103.99 - 72.18}{2}\)
\( = \frac{31.81}{2}\)
\( = 15.905 \approx 15.91\)
In simple words: Quartile deviation is half the interquartile range. Here, it is found by taking the difference between the 75th and 25th percentiles and dividing by two.

🎯 Exam Tip: Remember that the 25th percentile is \(Q_1\) and the 75th percentile is \(Q_3\). The formula for quartile deviation is straightforward once these values are identified.

 

Question 8. Seven students of a group get 20, 20, 20, 20, 20, 20, 20 marks in a test of 25 marks. What is the standard deviation of their marks?
Answer:
The observations of marks are 20, 20, 20, 20, 20, 20, 20.
Since all observations have the same magnitude, the standard deviation, denoted by 's', will be zero.
In simple words: If all data points are identical, there is no spread, and therefore the standard deviation is zero.

🎯 Exam Tip: A fundamental property of standard deviation is that it is zero only when all data values are the same. This indicates no variability in the dataset.

 

Question 9. Find the mean deviation for the observations - 1, 0, 4.
Answer:
Observations \(x\): -1, 0, 4.

\( \implies \) Mean \(\bar{x} = \frac{-1+0+4}{3} = \frac{3}{3} = 1\)
Now, the sum of absolute deviations from the mean: \(\Sigma|x - \bar{x}|\) = \(|-1-1| + |0-1| + |4-1|\) = \(|-2| + |-1| + |3|\) = \(2 + 1 + 3 = 6\)

\( \implies \) Mean deviation \(MD = \frac{\Sigma|x - \bar{x}|}{n} = \frac{6}{3} = 2\)
In simple words: First, calculate the mean. Then, find the absolute difference between each observation and the mean, and average these absolute differences to get the mean deviation.

🎯 Exam Tip: For mean deviation, ensure you correctly calculate the mean first, then take the *absolute* differences, and finally, find their average. Watch out for negative values in the original data.

 

Question 10. Which of the following factories is more stable with respect to daily production?

	Factory A	Factory B
Average Daily Production (units)	50	48
Standard deviation (units)	10	12

Answer:
To determine the stability of factories in terms of daily production, we need to compare their coefficient of variation (CV).

**Factory A:**
Mean (\(\bar{x}\)) = Average daily production = 50
Standard deviation (\(s\)) = 10

Coefficient of variation = \(\frac{s}{\bar{x}} \times 100\)
\( = \frac{10}{50} \times 100 = 20\%\)

**Factory B:**
Mean (\(\bar{x}\)) = Average daily production = 48
Standard deviation (\(s\)) = 12

Coefficient of variation = \(\frac{s}{\bar{x}} \times 100\)
\( = \frac{12}{48} \times 100 = 25\%\)

The coefficient of variation for daily production of factory A is 20%, which is less than that of factory B (25%). Therefore, factory A exhibits more stable daily production.
In simple words: We calculate the Coefficient of Variation for both factories; the one with the lower CV is considered more stable because its relative variability is smaller.

🎯 Exam Tip: For comparing consistency or stability between two datasets with different means or units, always use the Coefficient of Variation. A lower CV indicates higher consistency.

 

Question 11. The 25th and the 75th percentiles of a data set are 20 and 36 respectively. Find the coefficient of quartile deviation of the data set.
Answer:
Given:
\(P_{25}\) = 20, which means \(Q_1\) = 20
\(P_{75}\) = 36, which means \(Q_3\) = 36

Coefficient of quartile deviation = \(\frac{Q_3 - Q_1}{Q_3 + Q_1}\)
\( = \frac{36 - 20}{36 + 20}\)
\( = \frac{16}{56}\)
\( \approx 0.29\)
In simple words: The coefficient of quartile deviation compares the interquartile range to the sum of the quartiles, providing a relative measure of dispersion based on the central 50% of the data.

🎯 Exam Tip: Ensure you differentiate between quartile deviation (\(\frac{Q_3 - Q_1}{2}\)) and the coefficient of quartile deviation (\(\frac{Q_3 - Q_1}{Q_3 + Q_1}\)). The latter is a unitless ratio for comparison.

Section - D

Answer the following questions as required:

Question 1. Explain the meaning of dispersion and state different measures of dispersion.
Answer:
A measure of average, such as the mean or median, provides a representative summary of data. However, individual observations often vary from this average. Some values are very close to the average, while others are distant, and variations among observations themselves differ. To accurately understand the dataset, it's crucial to have a measure that complements the average by indicating the spread or variability among individual observations, showing the extent and nature of these variations and how data points cluster around the average. This measure is known as dispersion.

Dispersion, therefore, quantifies how scattered the observations of a dataset are from a measure of central tendency. The measure derived from this concept is called a measure of dispersion.

Consider an example: Suppose two students, A and B, took five tests, each out of 20 marks, with the following results:

Marks	Student A	Student B
	11	10
	13	13
	13	13
	13	9
	15	20
Total marks	65	65
Average marks	\(\frac{65}{5} = 13\) marks	\(\frac{65}{5} = 13\) marks

Here, both students have the same average mark of 13. However, Student A's marks range from \(15 - 11 = 4\) marks, while Student B's marks range from \(20 - 9 = 11\) marks. Student A's marks are clustered closer to the average of 13, whereas Student B's marks are more spread out from the average. Concluding that both students are equally efficient based solely on the average is misleading. Therefore, information about the deviation of marks, obtained through dispersion, is essential.

The different measures of dispersion include:
- Range
- Quartile Deviation
- Mean Deviation
- Standard Deviation
In simple words: Dispersion measures how spread out data points are from the average. Key measures include range, quartile deviation, mean deviation, and standard deviation.

🎯 Exam Tip: When explaining dispersion, provide a clear definition, illustrate with an example showing why it's necessary (like the student marks), and then list the main absolute measures of dispersion.

 

Question 2. State the desirable characteristics of dispersion.
Answer:
The desirable characteristics of an effective measure of dispersion are as follows:
- Its definition should be clear and precise.
- It should be straightforward to comprehend and simple to calculate.
- It should be based on all observations in the dataset.
- It should be minimally affected by sampling fluctuations, making it a stable measure.
- It should be suitable for further algebraic and statistical computations.
- It should be least affected by extreme observations present in the data.
In simple words: A good measure of dispersion should be easy to understand, use all data points, be stable when sampled, and useful for further calculations.

🎯 Exam Tip: Focus on clarity, ease of computation, use of all data, stability, suitability for further analysis, and resistance to extreme values as core characteristics of a good dispersion measure.

 

Question 3. Write advantages and disadvantages of the range.
Answer:
**Advantages of Range:**
- The range is clearly defined.
- It is easy to understand and simple to compute.
- It serves as a useful measure, especially when the variability among observations in the data is minimal.

**Disadvantages of Range:**
- It does not utilize all the observations in its computation.
- It is highly affected by sampling fluctuations.
- It is not a suitable measure for algebraic operations.
- It cannot be calculated for open-ended frequency distributions.
In simple words: Range is simple to find and understand but only uses two values, making it sensitive to outliers and unsuitable for complex analysis.

🎯 Exam Tip: When discussing advantages and disadvantages, highlight the range's simplicity versus its sensitivity to extreme values and lack of reliance on all data points.

 

Question 4. Write advantages and disadvantages of the quartile deviation.
Answer:
**Advantages of Quartile Deviation:**
- Its definition is clear and it is easy to compute.
- Extreme observations do not significantly affect its value.
- Quartile deviation can also be calculated for open-ended frequency distributions.

**Disadvantages of Quartile Deviation:**
- All the observations of the data are not used in its computation.
- It is not a suitable measure for other algebraic manipulations.
- It is relatively affected by the fluctuations of sampling.
- This measure is less commonly used for advanced statistical studies.
In simple words: Quartile deviation is robust to outliers and works with open-ended data, but it doesn't use all data points and is less useful for complex math.

🎯 Exam Tip: Emphasize that quartile deviation's strength lies in its resistance to outliers and applicability to open-ended data, while its weakness comes from ignoring outer data points and limited algebraic utility.

 

Question 5. Write advantages and disadvantages of the mean deviation.
Answer:
**Advantages of Mean Deviation:**
- Its definition is clear.
- All observations are used in its computation.
- Its value is less affected by extreme observations compared to some other measures of dispersion.
- It considers the total variations of observations from their mean in its computation.
- It is a better measure than range and quartile deviation.
- It is useful as a measure of dispersion for studies in social sciences.

**Disadvantages of Mean Deviation:**
- Its computation can be difficult compared to range and quartile deviation.
- It is not a suitable measure for other algebraic manipulations due to the use of absolute values.
- It is less useful for higher study of Statistics because of its dependence on the absolute value of deviation from the mean.
- It cannot be computed for open-ended frequency distributions.
In simple words: Mean deviation uses all data and is clearer than range, but using absolute values makes it hard for further mathematical operations.

🎯 Exam Tip: When discussing mean deviation, highlight its inclusiveness of all data points and conceptual clarity, but also stress the mathematical challenges posed by absolute values in advanced computations.

 

Question 6. Write advantages and disadvantages of the standard deviation.
Answer:
**Advantages of Standard Deviation:**
- Its definition is clear and precise.
- All observations of the data are utilized in its computation.
- It is a more efficient measure of dispersion among other measures of dispersion.
- If means and standard deviations are known for two datasets, the combined standard deviation of a new dataset can be calculated, making it suitable for other algebraic manipulations.
- It is widely used as a basic measure in higher statistical studies.

**Disadvantages of Standard Deviation:**
- Its computation can be more difficult compared to other measures of dispersion.
- Extreme observations are given more weight in its computations due to squaring deviations.
- It may not be easy for a common person to understand.
- It cannot be obtained for open-ended frequency distributions.
In simple words: Standard deviation is precise, uses all data, and is mathematically powerful, but it can be harder to calculate and is sensitive to extreme values.

🎯 Exam Tip: Focus on standard deviation's algebraic properties and its importance in inferential statistics as key advantages. Acknowledge the computational complexity and sensitivity to outliers as drawbacks.

 

Question 7. What is standard deviation? Why is it considered as the best measure of dispersion?
Answer:
**Standard Deviation:** Standard deviation is defined as the positive square root of the mean of the squares of deviations of observations from their mean. It is commonly denoted by 's', i.e., \(s = \sqrt{\frac{\Sigma(x-\bar{x})^2}{n}}\).

It is regarded as the best measure of dispersion because it possesses the following desired characteristics of an ideal measure of dispersion:
- The definition of standard deviation is clear and precise.
- All observations in the data are incorporated in its calculation.
- It is a suitable measure for various algebraic manipulations.
- It is extensively used in higher studies of Statistics.
- It is considered the most efficient measure among all measures of dispersion.

While the value of standard deviation is significantly influenced by extreme observations and its computation is more complex compared to other dispersion measures, it still holds most of the desired characteristics of an ideal measure. Its practical utility and foundational role in dispersion analysis make it widely considered the best measure of dispersion.
In simple words: Standard deviation measures the typical distance of data points from the mean. It's considered best because it uses all data, is mathematically sound, and is fundamental in statistics.

🎯 Exam Tip: Provide the precise definition and formula for standard deviation. Then, justify its superiority by listing its key desirable properties, such as using all observations, algebraic tractability, and widespread use in advanced statistics.

 

Question 8. Write a brief note on coefficient of variation.
Answer:
Karl Pearson proposed the coefficient of variation (CV) as an appropriate relative measure of standard deviation. It is a percentage measure of standard deviation relative to the mean, expressed as:
Coefficient of variation = \(\frac{\text{Standard Deviation}}{\text{Mean}} \times 100\)
\( = \frac{s}{\bar{x}} \times 100\)

Key points about the Coefficient of Variation:
- It is an extremely useful relative measure for comparing the dispersion of two or more datasets.
- A lower coefficient of variation indicates less dispersion in the data, implying greater stability or consistency.
- Conversely, a higher coefficient of variation suggests more dispersion, indicating less stable or less consistent data.
In simple words: The coefficient of variation is a percentage that shows how much variability exists relative to the mean, making it useful for comparing consistency between different datasets.

🎯 Exam Tip: Clearly state the formula for the coefficient of variation and its primary use: comparing variability across datasets, especially when means or units differ. Emphasize the inverse relationship between CV and consistency.

Question 7. What is standard deviation? Why is it considered as the best measure of dispersion?
Answer:
Standard deviation represents the positive square root of the average of the squared deviations of data points from their mean. It is commonly denoted by 's', and its formula is: \[ s = \sqrt{\frac{\Sigma(x-\bar{x})^{2}}{n}} \] It is considered the most effective measure of dispersion because it possesses the following desirable characteristics of an ideal measure of dispersion:
• The definition of standard deviation is clear and precise.
• Its computation includes all observations within the dataset.
• It is a suitable measure for various algebraic manipulations.
• It finds extensive use in higher statistical studies.
• Among all measures of dispersion, it is the most efficient.
Although its calculation can be more complex compared to other dispersion measures and it is influenced by extreme observations, standard deviation embodies most of the desired attributes of an ideal measure. Its practical utility as a fundamental measure of dispersion makes it the best choice.
In simple words: Standard deviation shows how spread out numbers are from the average. It's considered the best measure because it uses all data points, is clear, and is useful for advanced calculations, making it highly reliable.

🎯 Exam Tip: Understanding the formula and the reasons for standard deviation's superiority is key. Be prepared to explain its characteristics in relation to other measures of dispersion for full marks.

Question 8. Write a brief note on coefficient of variation.
Answer:
Karl Pearson proposed the coefficient of variation as an appropriate relative measure of standard deviation. It expresses standard deviation as a percentage relative to the mean.
The formula for Coefficient of variation is:
Coefficient of variation = Coefficient of SD \( \times \) 100
\( \implies \) Coefficient of variation = \( \frac{s}{\bar{x}} \times 100 \)
This measure is highly valuable for comparing the dispersion of two or more datasets. A lower coefficient of variation indicates less dispersion in the data, implying greater stability and consistency. Conversely, a higher coefficient of variation suggests more dispersion, indicating less stable or consistent data.
In simple words: The coefficient of variation is a percentage that compares how much data varies relative to its average. It helps determine which dataset is more consistent when their means are different.

🎯 Exam Tip: Remember that a smaller coefficient of variation implies greater consistency in the data. This concept is crucial for comparative analysis between different datasets.

Question 9. The information of number of flowers on 100 plants of a nursery is given below. Find the quartile deviation of the number of flowers from it.

No. of flowers	11	13	15	17	19	21	23	25
No. of plants	5	8	13	20	22	18	10	4

Answer:
To find the quartile deviation, we first need to construct a cumulative frequency distribution.

No. of flowers x	No. of plants f	Cumulative frequency cf
11	5	5
13	8	13
15	13	26
17	20	46
19	22	68
21	18	86
23	10	96
25	4	100
Total	n = 100	-

First Quartile (\(Q_1\)):
\(Q_1\) = Value of \( \left(\frac{n+1}{4}\right) \)th observation
\(Q_1\) = Value of \( \left(\frac{100+1}{4}\right) \)th observation
\(Q_1\) = Value of 25.25th observation
By referring to the cumulative frequency (cf) column, the 25.25th observation falls in the class where cf is 26, which corresponds to 15 flowers.
\( \implies Q_1 = 15 \) flowers

Third Quartile (\(Q_3\)):
\(Q_3\) = Value of \( 3\left(\frac{n+1}{4}\right) \)th observation
\(Q_3\) = Value of 3 (25.25) = 75.75th observation
By referring to the cumulative frequency (cf) column, the 75.75th observation falls in the class where cf is 86, which corresponds to 21 flowers.
\( \implies Q_3 = 21 \) flowers

Coefficient of Quartile Deviation:
Coefficient of \(Q_d = \frac{Q_3-Q_1}{Q_3+Q_1}\)
\( = \frac{21-15}{21+15} = \frac{6}{36} = 0.166 \approx 0.17 \)
In simple words: First, arrange the data and find the first and third quartiles (Q1 and Q3). Then, use these values to calculate the quartile deviation, which shows the spread of the middle 50% of the data.

🎯 Exam Tip: When dealing with discrete data, ensure precise calculation of the position of quartiles. A small error in identifying the observation can lead to an incorrect Q1 or Q3 value.

Question 10. The information of number of goals in 16 matches of hockey tournament is given. Find the mean deviation of number of goals for it.

No. of goals	1	2	3	4	5
No. of matches	1	4	6	4	1

Answer:
To calculate the mean deviation, we first need to find the mean of the data.

No. of goals x	No. of matches f	f.x	\(|x-\bar{x}|\) x=3	\(f|x-\bar{x}|\)
1	1	1	2	2
2	4	8	1	4
3	6	18	0	0
4	4	16	1	4
5	1	5	2	2
Total	n = 16	\( \Sigma fx = 48 \)		\( \Sigma f|x-\bar{x}| = 12 \)

Mean (\(\bar{x}\)):
\( \bar{x} = \frac{\Sigma fx}{n} = \frac{48}{16} = 3 \) Goals

Mean deviation of the number of goals per match (MD):
\( MD = \frac{\Sigma f|x-\bar{x}|}{n} \)
\( = \frac{12}{16} = 0.75 \) Goals
In simple words: First, calculate the average number of goals. Then, for each number of goals, find how much it differs from the average, ignoring the sign. Finally, sum these differences and divide by the total number of matches to get the mean deviation.

🎯 Exam Tip: Clearly define your variables (x, f, fx, \(|x-\bar{x}|\), \(f|x-\bar{x}|\)) in the table. Ensure that the mean is calculated correctly, as it forms the basis for subsequent deviation calculations.

Question 11. In usual notations, Σd = 25, Σd² = 272, n = 100 and assumed mean = 4. Find the coefficient of variation.
Answer:
Given: n = 100, A = 4 (assumed mean), Σd = 25, Σd² = 272.

Mean (\(\bar{x}\)):
\( \bar{x} = A + \frac{\Sigma d}{n} \)
\( = 4 + \frac{25}{100} \)
\( = 4 + 0.25 \)
\( = 4.25 \)

Standard deviation (s):
\( s = \sqrt{\frac{\Sigma d^2}{n} - \left(\frac{\Sigma d}{n}\right)^2} \)
\( = \sqrt{\frac{272}{100} - \left(\frac{25}{100}\right)^2} \)
\( = \sqrt{2.72 - (0.25)^2} \)
\( = \sqrt{2.72 - 0.0625} \)
\( = \sqrt{2.6575} \)
\( \approx 1.63 \)

Coefficient of variation:
Coefficient of variation \( = \frac{s}{\bar{x}} \times 100 \)
\( = \frac{1.63}{4.25} \times 100 \)
\( = 0.3835 \times 100 \)
\( = 38.35\% \)
In simple words: Use the given sums and assumed mean to find the actual mean and standard deviation. Then, divide the standard deviation by the mean and multiply by 100 to get the coefficient of variation, which shows the relative variability.

🎯 Exam Tip: Ensure that you correctly apply the formulas for mean and standard deviation with assumed mean. Pay close attention to the order of operations, especially squaring terms, to avoid calculation errors.

Question 12. Find the combined standard deviation using the following Information:

	Data set A		Data set B
No. of observations	50		60
Mean	113		120
Standard deviation	6		7

Answer:
Given for Data set A: \( n_1 = 50, \bar{x}_1 = 113, s_1 = 6 \)
Given for Data set B: \( n_2 = 60, \bar{x}_2 = 120, s_2 = 7 \)

Combined Mean (\(\bar{x}_c\)):
\( \bar{x}_c = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2} \)
\( = \frac{(50 \times 113) + (60 \times 120)}{50 + 60} \)
\( = \frac{5650 + 7200}{110} \)
\( = \frac{12850}{110} \)
\( = 116.82 \)

To calculate the combined standard deviation, we need the squared deviations from the combined mean:
\( d_1^2 = (\bar{x}_1 - \bar{x}_c)^2 \)
\( = (113 - 116.82)^2 \)
\( = (-3.82)^2 = 14.5924 \)

\( d_2^2 = (\bar{x}_2 - \bar{x}_c)^2 \)
\( = (120 - 116.82)^2 \)
\( = (3.18)^2 = 10.1124 \)

We also need \( s_1^2 \) and \( s_2^2 \):
\( s_1^2 = (6)^2 = 36 \)
\( s_2^2 = (7)^2 = 49 \)

Combined Standard Deviation (\(s_c\)):
\( s_c = \sqrt{\frac{n_1(s_1^2 + d_1^2) + n_2(s_2^2 + d_2^2)}{n_1 + n_2}} \)
\( = \sqrt{\frac{50(36 + 14.5924) + 60(49 + 10.1124)}{50 + 60}} \)
\( = \sqrt{\frac{50(50.5924) + 60(59.1124)}{110}} \)
\( = \sqrt{\frac{2529.62 + 3546.744}{110}} \)
\( = \sqrt{\frac{6076.364}{110}} \)
\( = \sqrt{55.23967} \)
\( \approx 7.43 \)
In simple words: First, combine the means of both datasets to find the overall mean. Then, calculate how much each individual mean deviates from this overall mean. Finally, use a formula that incorporates the individual standard deviations and these deviations to find the combined standard deviation.

🎯 Exam Tip: Be meticulous with calculations involving combined mean and standard deviation. Rounding errors in intermediate steps can significantly affect the final result, so maintain precision, especially with squared terms.

Question 13. The sum of 10 observations is 80 and the sum of their squares is 800. Find the coefficient of variation of the observations.
Answer:
Given: \( n = 10, \Sigma x = 80, \Sigma x^2 = 800 \)

Mean (\(\bar{x}\)):
\( \bar{x} = \frac{\Sigma x}{n} = \frac{80}{10} = 8 \)

Standard deviation (s):
\( s = \sqrt{\frac{\Sigma x^2}{n} - \left(\frac{\Sigma x}{n}\right)^2} \)
\( = \sqrt{\frac{800}{10} - \left(\frac{80}{10}\right)^2} \)
\( = \sqrt{80 - (8)^2} \)
\( = \sqrt{80 - 64} \)
\( = \sqrt{16} \)
\( = 4 \)

Coefficient of variation:
Coefficient of variation \( = \frac{s}{\bar{x}} \times 100 \)
\( = \frac{4}{8} \times 100 \)
\( = 0.5 \times 100 \)
\( = 50\% \)
In simple words: Calculate the average of the observations using the sum. Then, use the sums of observations and their squares to find the standard deviation. Finally, divide the standard deviation by the mean and multiply by 100 to get the coefficient of variation.

🎯 Exam Tip: Remember the direct formulas for mean and standard deviation from sums of observations and sums of squares. These are frequently used and knowing them saves time in calculations.

Section - E

Question 1. In a language spelling test of 50 marks, the frequency distribution of marks secured by 30 students is given below. Find the mean deviation of the frequency distribution.

Marks	12-16	17-21	22-26	27-31	32-36
No. of students	2	3	14	8	3

Answer:
To find the mean deviation, we will first calculate the mean (\(\bar{x}\)) of the distribution.

Marks	No. of students f	Mid value x	fx	\(|x-\bar{x}|\) (\(\bar{x}\) = 25.17)	\(f|x-\bar{x}|\)
12-16	2	14	28	11.17	22.34
17-21	3	19	57	6.17	18.51
22-26	14	24	336	1.17	16.38
27-31	8	29	232	3.83	30.64
32-36	3	34	102	8.83	26.49
Total	n = 30		\( \Sigma fx = 755 \)		\( \Sigma f|x-\bar{x}| = 114.36 \)

Mean (\(\bar{x}\)):
\( \bar{x} = \frac{\Sigma fx}{n} = \frac{755}{30} \)
\( = 25.166... \approx 25.17 \) marks

Mean deviation (MD):
\( MD = \frac{\Sigma f|x-\bar{x}|}{n} \)
\( = \frac{114.36}{30} \)
\( = 3.812 \)
\( \implies MD = 3.81 \) marks
In simple words: First, find the midpoint for each mark range and calculate the mean of the marks. Then, for each midpoint, find the absolute difference from the mean and multiply by its frequency. Sum these products and divide by the total number of students to get the mean deviation.

🎯 Exam Tip: When working with grouped frequency distributions, accurately calculating midpoints and the mean is crucial. Pay attention to rounding for the mean, as it impacts all subsequent deviation calculations.

Question 2. Find the quartile deviation of advertisement expenditure using following frequency distribution of advertising expenditure of 50 companies:

Advertisement cost (thousand Rs.)	0-5	5-15	15-30	30-40	40-60	60-100	Total
No. of companies	3	8	15	10	8	6	50

Answer:
To find the quartile deviation, we first create a cumulative frequency distribution.

Expenditure of advertisement (thousand Rs.)	No. of companies f	Cumulative frequency cf
0-5	3	3
5-15	8	11
15-30	15	26
30-40	10	36
40-60	8	44
60-100	6	50
Total	n = 50	-

First Quartile (\(Q_1\)):
\( Q_1 \text{ class} = \text{Class that includes } \left(\frac{n}{4}\right) \text{th observation} \)
\( Q_1 \text{ class} = \text{Class that includes } \left(\frac{50}{4}\right) = 12.5 \text{th observation} \)
Referring to the cf column, the 12.5th observation falls in the 15-30 class.
Here, \( L = 15, \left(\frac{n}{4}\right) = 12.5, cf = 11, f = 15, c = 15 \) (class width for 15-30)
\( Q_1 = L + \frac{\left(\frac{n}{4}\right)-cf}{f} \times c \)
\( = 15 + \frac{12.5-11}{15} \times 15 \)
\( = 15 + 1.5 = 16.5 \) thousand Rs.

Third Quartile (\(Q_3\)):
\( Q_3 \text{ class} = \text{Class that includes } 3\left(\frac{n}{4}\right) \text{th observation} \)
\( Q_3 \text{ class} = \text{Class that includes } 3(12.5) = 37.5 \text{th observation} \)
Referring to the cf column, the 37.5th observation falls in the 40-60 class.
Here, \( L = 40, 3\left(\frac{n}{4}\right) = 37.5, cf = 36, f = 8, c = 20 \) (class width for 40-60)
\( Q_3 = L + \frac{3\left(\frac{n}{4}\right)-cf}{f} \times c \)
\( = 40 + \frac{37.5-36}{8} \times 20 \)
\( = 40 + \frac{1.5 \times 20}{8} \)
\( = 40 + \frac{30}{8} \)
\( = 40 + 3.75 = 43.75 \) thousand Rs.

Quartile Deviation (\(Q_d\)):
\( Q_d = \frac{Q_3-Q_1}{2} \)
\( = \frac{43.75-16.5}{2} \)
\( = \frac{27.25}{2} = 13.625 \approx 13.63 \) thousand Rs.
In simple words: Create a cumulative frequency table, then use formulas to find the first and third quartiles (Q1 and Q3) from the grouped data. Finally, calculate the quartile deviation by subtracting Q1 from Q3 and dividing by two.

🎯 Exam Tip: Be careful with class boundaries for grouped data; ensure you use the correct lower limit (L), cumulative frequency (cf), and class frequency (f) for both Q1 and Q3. Remember to convert all currency symbols to "Rs.".

Question 3. The information of runs scored by a batsman in his 100 matches is given below. Find the standard deviation of runs scored by him from it.

Runs	0-10	10-20	20-30	30-40	40-50	50-60	60-70
No. of matches	10	15	25	25	10	10	5

Answer:
To calculate the standard deviation using the step-deviation method, we first need to find the midpoints, deviations, and apply the formula.

Run	No. of matches f	Mid value x	\(d = \frac{x-A}{c}\) A=35, c=10	fd	\(fd^2 = fd \cdot d\)
0-10	10	5	-3	-30	90
10-20	15	15	-2	-30	60
20-30	25	25	-1	-25	25
30-40	25	35	0	0	0
40-50	10	45	1	10	10
50-60	10	55	2	20	40
60-70	5	65	3	15	45
Total	n = 100			\( \Sigma fd = -40 \)	\( \Sigma fd^2 = 270 \)

Standard Deviation (s) of runs:
\( s = \sqrt{\frac{\Sigma fd^2}{n} - \left(\frac{\Sigma fd}{n}\right)^2} \times c \)
\( = \sqrt{\frac{270}{100} - \left(\frac{-40}{100}\right)^2} \times 10 \)
\( = \sqrt{2.70 - (-0.40)^2} \times 10 \)
\( = \sqrt{2.70 - 0.16} \times 10 \)
\( = \sqrt{2.54} \times 10 \)
\( = 1.5937 \times 10 \)
\( \approx 15.94 \) runs
In simple words: First, create a frequency distribution table with midpoints and calculate a simplified deviation (d) from an assumed mean. Use these values to find the sums of fd and \(fd^2\). Then, apply the standard deviation formula for grouped data with the correction factor (class width 'c') to get the final standard deviation.

🎯 Exam Tip: Using the step-deviation method (assuming a mean and using 'd' and 'c') simplifies calculations for grouped data. Ensure correct application of the formula and the multiplication by 'c' at the end for the standard deviation.

Question 4. The information of marks obtained by 220 students of a college is given below. Find the quartile deviation of the marks obtained by the students.

Marks	0-9	10-19	20-29	30-39	40-49	50 or more
No. of students	30	50	64	42	29	5

Answer:
To find the quartile deviation, we first need to construct a cumulative frequency distribution. Since the classes are exclusive, we adjust them for calculations.

Marks (Exclusive)	No. of students f	Cumulative frequency cf
0-9.5	30	30
9.5-19.5	50	80
19.5-29.5	64	144
29.5-39.5	42	186
39.5-49.5	29	215
49.5 or more	5	220
Total	n = 220

First Quartile (\(Q_1\)):
\( Q_1 \text{ class} = \text{Class that includes } \left(\frac{n}{4}\right) \text{th observation} \)
\( Q_1 \text{ class} = \text{Class that includes } \left(\frac{220}{4}\right) = 55 \text{th observation} \)
Referring to the cf column, the 55th observation falls in the 9.5-19.5 class.
Here, \( L = 9.5, \left(\frac{n}{4}\right) = 55, cf = 30, f = 50, c = 10 \)
\( Q_1 = L + \frac{\left(\frac{n}{4}\right)-cf}{f} \times c \)
\( = 9.5 + \frac{55-30}{50} \times 10 \)
\( = 9.5 + \frac{25}{50} \times 10 \)
\( = 9.5 + 0.5 \times 10 \)
\( = 9.5 + 5 = 14.5 \) marks

Third Quartile (\(Q_3\)):
\( Q_3 \text{ class} = \text{Class that includes } 3\left(\frac{n}{4}\right) \text{th observation} \)
\( Q_3 \text{ class} = \text{Class that includes } 3(55) = 165 \text{th observation} \)
Referring to the cf column, the 165th observation falls in the 29.5-39.5 class.
Here, \( L = 29.5, 3\left(\frac{n}{4}\right) = 165, cf = 144, f = 42, c = 10 \)
\( Q_3 = L + \frac{3\left(\frac{n}{4}\right)-cf}{f} \times c \)
\( = 29.5 + \frac{165-144}{42} \times 10 \)
\( = 29.5 + \frac{21 \times 10}{42} \)
\( = 29.5 + \frac{210}{42} \)
\( = 29.5 + 5 = 34.5 \) marks

Quartile Deviation (\(Q_d\)):
\( Q_d = \frac{Q_3-Q_1}{2} \)
\( = \frac{34.5-14.5}{2} \)
\( = \frac{20}{2} = 10 \) marks
In simple words: Adjust class boundaries for continuity, then create a cumulative frequency table. Calculate the positions of Q1 and Q3, identify their respective classes, and use the interpolation formula to find their exact values. Finally, compute the quartile deviation using the Q1 and Q3 values.

🎯 Exam Tip: For inclusive class intervals (like 0-9), convert them to exclusive intervals (0-9.5, 9.5-19.5) before calculating quartiles to ensure accuracy. This is a common point of error.

Question 5. Goals scored by two teams in a football session are as follows. Which team is more consistent in its game?

No. of goals scored in a football match	Team A	Team B
0	15	20
1	10	10
2	7	5
3	5	4
4	3	2
5	2	1

Answer:
To determine which team is more consistent, we will compare their coefficients of variation. A lower coefficient of variation indicates greater consistency.

No. of goals scored x	Team A			Team B
	No. of matches \(f_A\)	\(f_A x\)	\(f_A x^2\)	No. of matches \(f_B\)	\(f_B x\)	\(f_B x^2\)
0	15	0	0	20	0	0
1	10	10	10	10	10	10
2	7	14	28	5	10	20
3	5	15	45	4	12	36
4	3	12	48	2	8	32
5	2	10	50	1	5	25
Total	\(n_A = 42\)	\( \Sigma f_A x = 61 \)	\( \Sigma f_A x^2 = 181 \)	\(n_B = 42\)	\( \Sigma f_B x = 45 \)	\( \Sigma f_B x^2 = 123 \)

**For Team A:**
Mean (\(\bar{x}_A\)):
\( \bar{x}_A = \frac{\Sigma f_A x}{n_A} = \frac{61}{42} \approx 1.45 \) goals

Standard Deviation (\(s_A\)):
\( s_A = \sqrt{\frac{\Sigma f_A x^2}{n_A} - \left(\frac{\Sigma f_A x}{n_A}\right)^2} \)
\( = \sqrt{\frac{181}{42} - \left(\frac{61}{42}\right)^2} \)
\( = \sqrt{4.3095 - (1.4523)^2} \)
\( = \sqrt{4.3095 - 2.1106} \)
\( = \sqrt{2.1989} \approx 1.48 \) goals

Coefficient of Variation (CV_A):
\( CV_A = \frac{s_A}{\bar{x}_A} \times 100 \)
\( = \frac{1.48}{1.45} \times 100 \)
\( \approx 1.0207 \times 100 = 102.07\% \)

**For Team B:**
Mean (\(\bar{x}_B\)):
\( \bar{x}_B = \frac{\Sigma f_B x}{n_B} = \frac{45}{42} \approx 1.07 \) goals

Standard Deviation (\(s_B\)):
\( s_B = \sqrt{\frac{\Sigma f_B x^2}{n_B} - \left(\frac{\Sigma f_B x}{n_B}\right)^2} \)
\( = \sqrt{\frac{123}{42} - \left(\frac{45}{42}\right)^2} \)
\( = \sqrt{2.9286 - (1.0714)^2} \)
\( = \sqrt{2.9286 - 1.1478} \)
\( = \sqrt{1.7808} \approx 1.33 \) goals

Coefficient of Variation (CV_B):
\( CV_B = \frac{s_B}{\bar{x}_B} \times 100 \)
\( = \frac{1.33}{1.07} \times 100 \)
\( \approx 1.243 \times 100 = 124.3\% \)

Comparing the coefficients of variation:
Team A: 102.07%
Team B: 124.3%
Since the coefficient of variation for Team A (102.07%) is less than that for Team B (124.3%), Team A's performance is more consistent.
In simple words: To see which team is more consistent, we calculate the coefficient of variation for each. This involves finding the mean and standard deviation for each team. The team with the smaller coefficient of variation is considered more consistent because its scores are relatively less spread out compared to its average.

🎯 Exam Tip: When comparing consistency between two groups, always use the coefficient of variation. Ensure precise calculation of mean and standard deviation for both groups, as minor errors can lead to incorrect conclusions about consistency.

Question 6. For a sequence of 100 observations, the mean and standard deviation are 40 and 10 respectively. In calculating these measures, two observations were taken as 30 and 70 instead of 3 and 27 by mistake. Find the corrected mean and corrected standard deviation.
Answer:
Given: \( n = 100 \), Observed mean (\(\bar{x}\)) = 40, Observed standard deviation (s) = 10.
Wrong observations: 30, 70
Correct observations: 3, 27

**Step 1: Find the original (incorrect) sum of observations (\(\Sigma x\)).**
\( \bar{x} = \frac{\Sigma x}{n} \)
\( 40 = \frac{\Sigma x}{100} \)
\( \Sigma x = 40 \times 100 = 4000 \)

**Step 2: Calculate the corrected sum of observations (Corrected \(\Sigma x\)).**
Corrected \(\Sigma x\) = Original \(\Sigma x\) - (Sum of wrong observations) + (Sum of correct observations)
Corrected \(\Sigma x\) = \( 4000 - (30 + 70) + (3 + 27) \)
Corrected \(\Sigma x\) = \( 4000 - 100 + 30 \)
Corrected \(\Sigma x\) = \( 3930 \)

**Step 3: Calculate the corrected mean (Corrected \(\bar{x}\)).**
Corrected \(\bar{x} = \frac{\text{Corrected } \Sigma x}{n} = \frac{3930}{100} = 39.3 \)

**Step 4: Find the original (incorrect) sum of squares (\(\Sigma x^2\)).**
\( s = \sqrt{\frac{\Sigma x^2}{n} - \left(\frac{\Sigma x}{n}\right)^2} \)
\( s^2 = \frac{\Sigma x^2}{n} - \left(\frac{\Sigma x}{n}\right)^2 \)
\( 10^2 = \frac{\Sigma x^2}{100} - (40)^2 \)
\( 100 = \frac{\Sigma x^2}{100} - 1600 \)
\( 100 + 1600 = \frac{\Sigma x^2}{100} \)
\( 1700 = \frac{\Sigma x^2}{100} \)
\( \Sigma x^2 = 1700 \times 100 = 170000 \)

**Step 5: Calculate the corrected sum of squares (Corrected \(\Sigma x^2\)).**
Corrected \(\Sigma x^2\) = Original \(\Sigma x^2\) - (Sum of squares of wrong observations) + (Sum of squares of correct observations)
Corrected \(\Sigma x^2\) = \( 170000 - (30^2 + 70^2) + (3^2 + 27^2) \)
Corrected \(\Sigma x^2\) = \( 170000 - (900 + 4900) + (9 + 729) \)
Corrected \(\Sigma x^2\) = \( 170000 - 5800 + 738 \)
Corrected \(\Sigma x^2\) = \( 164938 \)

**Step 6: Calculate the corrected standard deviation (Corrected s).**
Corrected \( s = \sqrt{\frac{\text{Corrected } \Sigma x^2}{n} - \left(\frac{\text{Corrected } \Sigma x}{n}\right)^2} \)
\( = \sqrt{\frac{164938}{100} - \left(\frac{3930}{100}\right)^2} \)
\( = \sqrt{1649.38 - (39.30)^2} \)
\( = \sqrt{1649.38 - 1544.49} \)
\( = \sqrt{104.89} \approx 10.24 \)
In simple words: First, use the given incorrect mean and standard deviation to find the original sums (\(\Sigma x\) and \(\Sigma x^2\)). Then, adjust these sums by subtracting the wrong values and adding the correct ones. Finally, use these corrected sums to calculate the new, accurate mean and standard deviation.

🎯 Exam Tip: This type of problem requires careful step-by-step correction of both the sum of observations and the sum of squares. Double-check all subtractions and additions to ensure accuracy, as any error propagates through the entire calculation.

Question 7. The production function for a factory is y = 10 + 3x where x = No. of units produced and y = total cost of producing x units. The range, the quartile deviation, the mean deviation and standard deviation of daily production of the factory are 50, 5, 8 and 10 units respectively. Find the range quartile deviation mean deviation and standard deviation for total cost y from it.
Answer:
Given the relationship: \( y = 10 + 3x \)
For variable x, we have:
Range (R_x) = 50 units
Quartile Deviation (Qd_x) = 5 units
Mean Deviation (MD_x) = 8 units
Standard Deviation (s_x) = 10 units

Measures of dispersion are unaffected by a change of origin (adding or subtracting a constant) but are influenced by a change of scale (multiplying or dividing by a constant). For the given linear transformation \( y = a + bx \), where \( a=10 \) and \( b=3 \), the new measures of dispersion will be \( |b| \) times the original measures.

Range of y (R_y):
\( R_y = |3| \times R_x = 3 \times 50 = 150 \) units

Quartile Deviation of y (Qd_y):
\( Qd_y = |3| \times Qd_x = 3 \times 5 = 15 \) units

Mean Deviation of y (MD_y):
\( MD_y = |3| \times MD_x = 3 \times 8 = 24 \) units

Standard Deviation of y (s_y):
\( s_y = |3| \times s_x = 3 \times 10 = 30 \) units
In simple words: When a variable is transformed by multiplying by a constant and adding another (like y = 10 + 3x), measures of dispersion (range, quartile deviation, mean deviation, standard deviation) only change by the absolute value of the multiplication factor. The addition part (10) does not affect the spread.

🎯 Exam Tip: Remember the critical property: measures of dispersion are independent of the change of origin (addition/subtraction) but are dependent on the change of scale (multiplication/division). The coefficient 'b' in \(y = a + bx\) directly scales the dispersion measures by \(|b|\).

Section - F

Question 1. Find range, coefficient of range, quartile deviation, coefficient of quartile deviation, mean deviation and coefficient of mean deviation from the following data of number of emergency visits of 80 doctors to their patients in a town:

No. of visits	3	5	8	12	17	20	24	30	35
No. of doctors	1	3	7	15	20	13	10	7	4

Answer:
To calculate the various measures of dispersion and their coefficients, we first need to create a table including mid-values, frequencies, cumulative frequencies, and deviations from the mean.

No. of visits x	No. of doctors f	fx	\(|x-\bar{x}|\) (\(\bar{x}\) = 18.05)	\(f|x-\bar{x}|\)	Cumulative frequency cf
3	1	3	15.05	15.05	1
5	3	15	13.05	39.15	4
8	7	56	10.05	70.35	11
12	15	180	6.05	90.75	26
17	20	340	1.05	21.00	46
20	13	260	1.95	25.35	59
24	10	240	5.95	59.50	69
30	7	210	11.95	83.65	76
35	4	140	16.95	67.80	80
Total	n = 80	\( \Sigma fx = 1444 \)		\( \Sigma f|x-\bar{x}| = 472.60 \)

**1. Range (R):**
Highest observation (\(X_H\)) = 35
Lowest observation (\(X_L\)) = 3
\( R = X_H - X_L = 35 - 3 = 32 \) visits

**2. Coefficient of Range:**
Coefficient of Range \( = \frac{X_H - X_L}{X_H + X_L} \)
\( = \frac{35 - 3}{35 + 3} = \frac{32}{38} \approx 0.84 \)

**3. Quartile Deviation (\(Q_d\)):**
First Quartile (\(Q_1\)):
\( Q_1 = \text{Value of } \left(\frac{n+1}{4}\right)\text{th observation} \)
\( Q_1 = \text{Value of } \left(\frac{80+1}{4}\right) = 20.25\text{th observation} \)
From the cf column, the 20.25th observation is 12.
\( \implies Q_1 = 12 \)

Third Quartile (\(Q_3\)):
\( Q_3 = \text{Value of } 3\left(\frac{n+1}{4}\right)\text{th observation} \)
\( Q_3 = \text{Value of } 3(20.25) = 60.75\text{th observation} \)
From the cf column, the 60.75th observation is 24.
\( \implies Q_3 = 24 \)

\( Q_d = \frac{Q_3 - Q_1}{2} = \frac{24 - 12}{2} = \frac{12}{2} = 6 \) visits

**4. Coefficient of Quartile Deviation (\(Q_d\)):**
Coefficient of \(Q_d = \frac{Q_3 - Q_1}{Q_3 + Q_1} \)
\( = \frac{24 - 12}{24 + 12} = \frac{12}{36} \approx 0.33 \)

**5. Mean Deviation (MD):**
Mean (\(\bar{x}\)):
\( \bar{x} = \frac{\Sigma fx}{n} = \frac{1444}{80} = 18.05 \) visits

\( MD = \frac{\Sigma f|x-\bar{x}|}{n} = \frac{472.60}{80} \approx 5.91 \) visits

**6. Coefficient of Mean Deviation (MD):**
Coefficient of \( MD = \frac{MD}{\bar{x}} \)
\( = \frac{5.91}{18.05} \approx 0.33 \)
In simple words: This question requires calculating several measures of spread. First, identify the highest and lowest values for the range. Then, find the first and third quartiles (Q1 and Q3) for quartile deviation. Calculate the mean, and then the average absolute difference from this mean for mean deviation. Finally, compute the coefficient for each measure by dividing the measure of dispersion by the mean or sum of quartiles.

🎯 Exam Tip: When a question asks for multiple measures, calculate them systematically. Clearly label each calculation and ensure the correct formula is applied for each measure and its corresponding coefficient. Double-check your mean calculation as it influences mean deviation. Always ensure numerical accuracy, especially with rounding for final answers.

 

Question 1. Find range, coefficient of range, quartile deviation, coefficient of quartile deviation, mean deviation and coefficient of mean deviation from the following data of number of emergency visits of 80 doctors to their patients in a town:
Answer:

No. of visits x	No. of doctors f	fx	\(|x - \bar{x}|\) \( \bar{x} = 18.05\)	\(f|x - \bar{x}|\)	Cumulative frequency cf
3	1	3	15.05	15.05	1
5	3	15	13.05	39.15	4
8	7	56	10.05	70.35	11
12	15	180	6.05	90.75	26
17	20	340	1.05	21.00	46
20	13	260	1.95	25.35	59
24	10	240	5.95	59.50	69
30	7	210	11.95	83.65	76
35	4	140	16.95	67.80	80
Total	n = 80	\( \Sigma fx = 1444\)	-	\( \Sigma f|x - \bar{x}| = 472.60\)	-

Range:
The highest observation \(x_H\) is 35, and the lowest observation \(x_L\) is 3.
So, Range \(R = x_H - x_L\) \( = 35 - 3 = 32\) visits.
First Quartile:
\(Q_1\) is the value of \( \left(\frac{n+1}{4}\right)^{th}\) observation.
\( = \text{Value of } \left(\frac{80+1}{4}\right)^{th}\) observation
\( = \text{Value of } 20.25^{th}\) observation.
Referring to the cumulative frequency column, \(Q_1 = 12\).
Third Quartile:
\(Q_3\) is the value of \( 3\left(\frac{n+1}{4}\right)^{th}\) observation.
\( = \text{Value of } 3 (20.25)^{th}\) observation
\( = \text{Value of } 60.75^{th}\) observation.
Referring to the cumulative frequency column, \(Q_3 = 24\).
Quartile Deviation:
\(Q_d = \frac{Q_3 - Q_1}{2} \)
\( = \frac{24 - 12}{2} = \frac{12}{2} = 6\) Visits.
Coefficient of Quartile Deviation:
\( \text{Coefficient of } Q_d = \frac{Q_3 - Q_1}{Q_3 + Q_1} \)
\( = \frac{24 - 12}{24 + 12} = \frac{12}{36} = 0.33\)
Mean:
\( \bar{x} = \frac{\Sigma fx}{n} = \frac{1444}{80} = 18.05\) visits.
Mean Deviation:
\( \text{MD} = \frac{\Sigma f|x - \bar{x}|}{n} \)
\( = \frac{472.60}{80} = 5.91\) visits.
Coefficient of Mean Deviation:
\( \text{Coefficient of MD} = \frac{\text{MD}}{\bar{x}} \)
\( = \frac{5.91}{18.05} = 0.33\)
In simple words: This problem involves calculating various measures of dispersion for the emergency visits data. We first determine the range, quartiles, and then use these to find quartile deviation, mean, mean deviation, and their respective coefficients, providing a comprehensive view of data spread.

🎯 Exam Tip: Remember to calculate cumulative frequencies accurately for grouped data when finding quartiles. Ensure all mathematical formulas are correctly applied and show clear steps for each measure of dispersion.

 

Question 2. Find the percentage of observations lying within the limits \( \bar{x} \pm s\) using the following distribution of credit days taken by the merchants:

Credit days x	No. of merchants f	\(d = \frac{(x-A)}{c}\) \(A = 15\)	fd	\(fd^2 = fd \cdot d\)
12	5	-3	-15	45
13	10	-2	-20	40
14	25	-1	-25	25
15	65	0	0	0
16	45	1	45	45
17	35	2	70	140
18	8	3	24	72
19	7	4	28	112
Total	n = 200	-	\( \Sigma fd = 107\)	\( \Sigma fd^2 = 479\)

Answer:Mean:
Given assumed mean \(A = 15\), class length \(c = 1\).
\( \bar{x} = A + \frac{\Sigma fd}{n} \times c\)
\( = 15 + \frac{107}{200} \)
\( = 15 + 0.535 \)
\( = 15.535\) days.
Standard deviation:
\( s = \sqrt{\frac{\Sigma fd^2}{n} - \left(\frac{\Sigma fd}{n}\right)^2} \times c \)
\( = \sqrt{\frac{479}{200} - \left(\frac{107}{200}\right)^2} \times 1 \)
\( = \sqrt{2.395 - (0.535)^2} \)
\( = \sqrt{2.395 - 0.286} \)
\( = \sqrt{2.109} \)
\( = 1.452\) days.
Now, let's find the limits \( \bar{x} \pm s \):
\( \bar{x} - s = 15.535 - 1.452 = 14.083 \)
\( \bar{x} + s = 15.535 + 1.452 = 16.987 \)
The observations lying outside the interval [14.083, 16.987] are those with values of 12, 13, 14, 17, 18, 19.
Frequencies for these observations:
For x < 14.083: \(f(12) + f(13) + f(14) = 5 + 10 + 25 = 40\)
For x > 16.987: \(f(17) + f(18) + f(19) = 35 + 8 + 7 = 50\)
Total observations outside the interval = \(40 + 50 = 90\).
Total number of observations \(n = 200\).
Number of observations within \( \bar{x} \pm s\) = \( \text{Total observations} - \text{Observations outside the interval} \)
\( = 200 - 90 = 110\).
Percentage of observations within \( \bar{x} \pm s = \frac{110}{200} \times 100 = 55\% \).
In simple words: This solution calculates the mean and standard deviation of credit days. Then, it identifies how many observations fall within one standard deviation from the mean, expressed as a percentage, to show data concentration.

🎯 Exam Tip: When dealing with discrete data for \( \bar{x} \pm s\) interval, carefully sum the frequencies of observations falling within the calculated lower and upper bounds. Be mindful of inclusive/exclusive boundaries if the data were continuous.

 

Question 3. Find an appropriate measure of dispersion from the following data. Also find its relative measure.

Marks	No. of students f	Cumulative frequency cf
Less than 10	2	2
10-20	4	6
20-30	10	16
30-40	3	19
More than 40	1	20
Total	n = 20	-

Answer:The provided frequency distribution is open-ended ("Less than 10", "More than 40"). For such distributions, Quartile Deviation is considered an appropriate measure of dispersion.
First Quartile (\(Q_1\)):
\(Q_1\) class = Class including the \( \left(\frac{n}{4}\right)^{th}\) observation.
\( = \text{Class including the } \left(\frac{20}{4}\right)^{th} = 5^{th}\) observation.
Referring to the cumulative frequency column, the \(Q_1\) class is 10-20.
Here, Lower limit \(L = 10\), \( \frac{n}{4} = 5\), cumulative frequency of preceding class \(cf = 2\), frequency of \(Q_1\) class \(f = 4\), class length \(c = 10\).
\(Q_1 = L + \frac{\left(\frac{n}{4}\right)-cf}{f} \times c\)
\( = 10 + \frac{5-2}{4} \times 10 \)
\( = 10 + \frac{3 \times 10}{4} \)
\( = 10 + 7.5 = 17.5\) marks.
Third Quartile (\(Q_3\)):
\(Q_3\) class = Class including the \( 3\left(\frac{n}{4}\right)^{th}\) observation.
\( = \text{Class including the } 3(5)^{th} = 15^{th}\) observation.
Referring to the cumulative frequency column, the \(Q_3\) class is 20-30.
Here, Lower limit \(L = 20\), \( 3\left(\frac{n}{4}\right) = 15\), cumulative frequency of preceding class \(cf = 6\), frequency of \(Q_3\) class \(f = 10\), class length \(c = 10\).
\(Q_3 = L + \frac{3\left(\frac{n}{4}\right)-cf}{f} \times c\)
\( = 20 + \frac{15-6}{10} \times 10 \)
\( = 20 + 9 = 29\) marks.
Quartile Deviation (\(Q_d\)):
\(Q_d = \frac{Q_3 - Q_1}{2}\)
\( = \frac{29 - 17.5}{2} = \frac{11.5}{2} = 5.75\) marks.
Relative measure of quartile deviation (Coefficient of Quartile Deviation):
\( \text{Coefficient of } Q_d = \frac{Q_3 - Q_1}{Q_3 + Q_1} \)
\( = \frac{29 - 17.5}{29 + 17.5} = \frac{11.5}{46.5} = 0.25\)
In simple words: For this open-ended data, we chose quartile deviation to measure spread. We found the first and third quartiles, then calculated the quartile deviation and its coefficient to understand the data's variability.

🎯 Exam Tip: Always identify if the frequency distribution is open-ended. For such cases, quartile deviation is the most appropriate measure of dispersion. Pay close attention to class boundaries and cumulative frequencies in calculations.

 

Question 4. The information of the salary of 200 employees of company is given below. Find the standard deviation of the salary of the employees:

Salary (thousand Rs.)	No. of persons
Less than 10	5
Less than 20	17
Less than 30	47
Less than 40	92
Less than 50	142
Less than 60	179
Less than 70	200

Answer:The provided data is a 'less than' type cumulative frequency distribution. To calculate the standard deviation, we first need to convert it into a simple frequency distribution. The class length is 10 (e.g., 20-10 = 10, 30-20 = 10).
Here is the converted frequency distribution:

Salary (thousand Rs.)	No. of workers f	Mid value x	\(d = \frac{(x-A)}{c}\) \(A = 35, c = 10\)	fd	\(fd^2 = fd \cdot d\)
0-10	5	5	-3	-15	45
10-20	\(17-5 = 12\)	15	-2	-24	48
20-30	\(47-17 = 30\)	25	-1	-30	30
30-40	\(92-47 = 45\)	35	0	0	0
40-50	\(142-92 = 50\)	45	1	50	50
50-60	\(179-142 = 37\)	55	2	74	148
60-70	\(200-179 = 21\)	65	3	63	189
Total	n = 200	-	-	\( \Sigma fd = 118\)	\( \Sigma fd^2 = 510\)

Mean:
\( \bar{x} = A + \frac{\Sigma fd}{n} \times c \)
\( = 35 + \frac{118}{200} \times 10 \)
\( = 35 + \frac{1180}{200} \)
\( = 35 + 5.9 = 40.9\) thousand Rs.
Standard deviation:
\( s = \sqrt{\frac{\Sigma fd^2}{n} - \left(\frac{\Sigma fd}{n}\right)^2} \times c \)
\( = \sqrt{\frac{510}{200} - \left(\frac{118}{200}\right)^2} \times 10 \)
\( = \sqrt{2.55 - (0.59)^2} \times 10 \)
\( = \sqrt{2.55 - 0.3481} \times 10 \)
\( = \sqrt{2.2019} \times 10 \)
\( = 1.4838 \times 10 \)
\( = 14.84\) thousand Rs.
In simple words: We converted the given cumulative frequency data to a regular frequency distribution. Then, using an assumed mean and step deviation method, we calculated the mean and standard deviation for the employees' salaries.

🎯 Exam Tip: When given a 'less than' or 'more than' type cumulative frequency distribution, always convert it to a simple frequency distribution first. The step deviation method simplifies calculations for grouped data with equal class intervals.

 

Question 5. The following is a distribution of closing prices (in Rs.) of shares of 100 different small scale industries on a certain day. Find the mean deviation of the closing prices of shares.

Price (Rs.)	No. of industries f
0-10	3
10-20	8
20-30	15
30-40	20
40-50	25
50-60	10
60-70	9
70-80	6
80-90	4

Answer:To find the mean deviation, we first need to calculate the mean of the data.

Price (Rs.)	No. of industries f	Mid value x	fx	\(|x - \bar{x}|\) \( \bar{x} = 42.6\)	\(f|x - \bar{x}|\)
0-10	3	5	15	37.6	112.8
10-20	8	15	120	27.6	220.8
20-30	15	25	375	17.6	264.0
30-40	20	35	700	7.6	152.0
40-50	25	45	1125	2.4	60.0
50-60	10	55	550	12.4	124.0
60-70	9	65	585	22.4	201.6
70-80	6	75	450	32.4	194.4
80-90	4	85	340	42.4	169.6
Total	n = 100	-	\( \Sigma fx = 4260\)	-	\( \Sigma f|x - \bar{x}| = 1499.2\)

Mean:
\( \bar{x} = \frac{\Sigma fx}{n} = \frac{4260}{100} = 42.6\) Rs.
Mean deviation:
\( \text{MD} = \frac{\Sigma f|x - \bar{x}|}{n} \)
\( = \frac{1499.2}{100} = 14.99\) Rs.
In simple words: To find the mean deviation, we first calculated the mean of the share prices. Then, we summed the absolute differences of each price from the mean, weighted by their frequencies, and divided by the total number of industries.

🎯 Exam Tip: Always calculate the mean accurately as the first step for mean deviation. Remember to use the absolute values of deviations from the mean to prevent positive and negative deviations from cancelling each other out.

 

Question 6. The information of daily wages (in Rs.) of 230 workers of a factory is given below. Calculate the coefficient of variation from the following data for the daily wages of workers:

Daily wages (Rs.)	No. of workers
Less than 100	12
Less than 200	30
Less than 300	65
Less than 400	107
Less than 500	157
Less than 600	202
Less than 700	222
Less than 800	230

Answer:The given distribution is a 'less than' type cumulative frequency distribution. We need to convert it into a simple frequency distribution. The class length is 100.
Here is the converted frequency distribution:

Daily wages (Rs.)	No. of workers f	Mid value x	\(d = \frac{(x-A)}{c}\) \(A = 450, c = 100\)	fd	\(fd^2 = fd \cdot d\)
0-100	12	50	-4	-48	192
100-200	\(30-12 = 18\)	150	-3	-54	162
200-300	\(65-30 = 35\)	250	-2	-70	140
300-400	\(107-65 = 42\)	350	-1	-42	42
400-500	\(157-107 = 50\)	450	0	0	0
500-600	\(202-157 = 45\)	550	1	45	45
600-700	\(222-202 = 20\)	650	2	40	80
700-800	\(230-222 = 8\)	750	3	24	72
Total	n = 230	-	-	\( \Sigma fd = -105\)	\( \Sigma fd^2 = 733\)

Mean:
\( \bar{x} = A + \frac{\Sigma fd}{n} \times c \)
\( = 450 + \frac{-105}{230} \times 100 \)
\( = 450 - \frac{10500}{230} \)
\( = 450 - 45.65 \)
\( = 404.35\) Rs.
Standard deviation:
\( s = \sqrt{\frac{\Sigma fd^2}{n} - \left(\frac{\Sigma fd}{n}\right)^2} \times c \)
\( = \sqrt{\frac{733}{230} - \left(\frac{-105}{230}\right)^2} \times 100 \)
\( = \sqrt{3.1869 - (-0.4565)^2} \times 100 \)
\( = \sqrt{3.1869 - 0.2084} \times 100 \)
\( = \sqrt{2.9785} \times 100 \)
\( = 1.7258 \times 100 \)
\( = 172.58\) Rs.
Coefficient of variation:
\( \text{Coefficient of variation} = \frac{s}{\bar{x}} \times 100 \)
\( = \frac{172.58}{404.35} \times 100 \)
\( = 0.4268 \times 100 \)
\( = 42.68\% \)
In simple words: First, the cumulative frequency distribution was converted into a standard frequency table. Then, the mean and standard deviation of daily wages were computed. Finally, the coefficient of variation was calculated by dividing the standard deviation by the mean and multiplying by 100 to express relative dispersion.

🎯 Exam Tip: Remember that coefficient of variation is a relative measure, useful for comparing variability between different datasets. Ensure to convert cumulative frequencies correctly and use the assumed mean method for efficiency in calculations.

 

Question 7. The marks obtained by two students A and B in 10 sets of examinations are given below: Which student is more consistent in his study?

Set	1	2	3	4	5	6	7	8	9	10
Marks of A	44	80	76	48	52	72	68	56	60	64
Marks of B	48	75	54	60	63	69	72	51	57	56

Answer:To determine which student is more consistent, we will compare their coefficients of variation. A lower coefficient of variation indicates greater consistency.
**Student A:**

Marks x	\( (x-\bar{x}) \) \( \bar{x} = 62\)	\( (x-\bar{x})^2\)
44	-18	324
80	18	324
76	14	196
48	-14	196
52	-10	100
72	10	100
68	6	36
56	-6	36
60	-2	4
64	2	4
\( \Sigma x = 620\)	\( \Sigma(x-\bar{x}) = 0\)	\( \Sigma(x-\bar{x})^2 = 1320\)

Mean for Student A:
\( \bar{x}_A = \frac{\Sigma x}{n} = \frac{620}{10} = 62\) marks.
Standard deviation for Student A:
\( s_A = \sqrt{\frac{\Sigma(x-\bar{x})^2}{n}} \)
\( = \sqrt{\frac{1320}{10}} = \sqrt{132} = 11.49\) marks.
Coefficient of variation for Student A:
\( \text{CV}_A = \frac{s_A}{\bar{x}_A} \times 100 \)
\( = \frac{11.49}{62} \times 100 = 0.1853 \times 100 = 18.53\% \)
**Student B:**

Marks x	\( (x-\bar{x}) \) \( \bar{x} = 60.5\)	\( (x-\bar{x})^2\)
48	-12.5	156.25
75	14.5	210.25
54	-6.5	42.25
60	-0.5	0.25
63	2.5	6.25
69	8.5	72.25
72	11.5	132.25
51	-9.5	90.25
57	-3.5	12.25
56	-4.5	20.25
\( \Sigma x = 605\)	\( \Sigma(x-\bar{x}) = 0\)	\( \Sigma(x-\bar{x})^2 = 742.50\)

Mean for Student B:
\( \bar{x}_B = \frac{\Sigma x}{n} = \frac{605}{10} = 60.5\) marks.
Standard deviation for Student B:
\( s_B = \sqrt{\frac{\Sigma(x-\bar{x})^2}{n}} \)
\( = \sqrt{\frac{742.50}{10}} = \sqrt{74.25} = 8.62\) marks.
Coefficient of variation for Student B:
\( \text{CV}_B = \frac{s_B}{\bar{x}_B} \times 100 \)
\( = \frac{8.62}{60.5} \times 100 = 0.1425 \times 100 = 14.25\% \)
Comparison:
Since \( \text{CV}_B (14.25\%) < \text{CV}_A (18.53\%)\), Student B is more consistent in their study performance.
In simple words: To compare consistency, we calculated the mean, standard deviation, and then the coefficient of variation for each student. Student B has a lower coefficient of variation, indicating more consistent performance in their examinations.

🎯 Exam Tip: Consistency is best evaluated using the Coefficient of Variation when comparing datasets with different means. A smaller CV indicates less relative variability and thus greater consistency.

 

Question 8. The following are the distributions of weights (in kg) for the students of two groups A and B. Find the coefficient of variation of each group. Which group has greater relative variation?

Weight (kg)	20-30	30-40	40-50	50-60	60-70
Group A	7	10	20	18	7
Group B	5	9	21	15	6

Answer:To find the coefficient of variation for each group and compare their relative variation, we will calculate the mean and standard deviation for both Group A and Group B.
**Group A:**

Weight kg	Mid value x	No. of students f	\(d = \frac{(x-A)}{c}\) \(A = 45, c = 10\)	fd	\(fd^2 = fd \cdot d\)
20-30	25	7	-2	-14	28
30-40	35	10	-1	-10	10
40-50	45	20	0	0	0
50-60	55	18	1	18	18
60-70	65	7	2	14	28
Total	-	n = 62	-	\( \Sigma fd = 8\)	\( \Sigma fd^2 = 84\)

Mean for Group A:
\( \bar{x}_A = A + \frac{\Sigma fd}{n} \times c \)
\( = 45 + \frac{8}{62} \times 10 \)
\( = 45 + \frac{80}{62} \)
\( = 45 + 1.29 \)
\( = 46.29\) kg.
Standard deviation for Group A:
\( s_A = \sqrt{\frac{\Sigma fd^2}{n} - \left(\frac{\Sigma fd}{n}\right)^2} \times c \)
\( = \sqrt{\frac{84}{62} - \left(\frac{8}{62}\right)^2} \times 10 \)
\( = \sqrt{1.3548 - (0.1290)^2} \times 10 \)
\( = \sqrt{1.3548 - 0.0166} \times 10 \)
\( = \sqrt{1.3382} \times 10 \)
\( = 1.1568 \times 10 \)
\( = 11.57\) kg.
Coefficient of variation for Group A:
\( \text{CV}_A = \frac{s_A}{\bar{x}_A} \times 100 \)
\( = \frac{11.57}{46.29} \times 100 \)
\( = 0.2500 \times 100 \)
\( = 25\% \)
**Group B:**

Weight kg	Mid value x	No. of students f	\(d = \frac{(x-A)}{c}\) \(A = 45, c = 10\)	fd	\(fd^2 = fd \cdot d\)
20-30	25	5	-2	-10	20
30-40	35	9	-1	-9	9
40-50	45	21	0	0	0
50-60	55	15	1	15	15
60-70	65	6	2	12	24
Total	-	n = 56	-	\( \Sigma fd = 8\)	\( \Sigma fd^2 = 68\)

Mean for Group B:
\( \bar{x}_B = A + \frac{\Sigma fd}{n} \times c \)
\( = 45 + \frac{8}{56} \times 10 \)
\( = 45 + \frac{80}{56} \)
\( = 45 + 1.43 \)
\( = 46.43\) kg.
Standard deviation for Group B:
\( s_B = \sqrt{\frac{\Sigma fd^2}{n} - \left(\frac{\Sigma fd}{n}\right)^2} \times c \)
\( = \sqrt{\frac{68}{56} - \left(\frac{8}{56}\right)^2} \times 10 \)
\( = \sqrt{1.2143 - (0.1429)^2} \times 10 \)
\( = \sqrt{1.2143 - 0.0204} \times 10 \)
\( = \sqrt{1.1939} \times 10 \)
\( = 1.0927 \times 10 \)
\( = 10.93\) kg.
Coefficient of variation for Group B:
\( \text{CV}_B = \frac{s_B}{\bar{x}_B} \times 100 \)
\( = \frac{10.93}{46.43} \times 100 \)
\( = 0.2354 \times 100 \)
\( = 23.54\% \)
Comparison of relative variation:
Since \( \text{CV}_A (25\%) > \text{CV}_B (23.54\%)\), Group A has a greater relative variation in weights compared to Group B.
In simple words: We calculated the mean, standard deviation, and coefficient of variation for both Group A and Group B. By comparing their coefficients of variation, we found that Group A has a higher coefficient, indicating more relative variability in student weights than Group B.

🎯 Exam Tip: Coefficient of Variation is crucial for comparing variability between datasets with differing units or scales, like comparing weight distributions. Always explicitly state which group has higher or lower variation based on the CV values.