GSEB Class 11 Statistics Solutions Chapter 3 Measures of Central Tendency Exercise 3

Section - A

Choose the correct option from those given below each question:

Question 1. Which average gets most affected by too large or too small values?
(a) Arithmetic mean
(b) Median
(c) Mode
(d) Geometric mean
Answer: (a) Arithmetic mean
In simple words: The arithmetic mean is highly sensitive to extreme values, meaning unusually large or small numbers can significantly distort its representation of the data's central tendency.

🎯 Exam Tip: Understanding the impact of outliers on different averages is crucial for selecting the appropriate measure of central tendency in statistical analysis.

 

Question 2. Which of the following will give us the value of median?
(a) D7
(b) Q1
(d) P50
Answer: (d) P50
In simple words: The median is the middle value in an ordered dataset, which corresponds to the 50th percentile (P50) as it divides the data into two equal halves.

🎯 Exam Tip: Remember that the median, Q2 (second quartile), and P50 all represent the same central point in a dataset.

 

Question 3. In which of the following situations, mean cannot be found?
(a) Class lengths are unequal,
(b) There are open-ended class intervals,
(c) The number of class intervals is more than 5,
(d) Inclusive type of classes are used
Answer: (b) There are open-ended class intervals,
In simple words: When class intervals are open-ended (e.g., "below 10" or "above 100"), the exact midpoint for those classes cannot be determined, making it impossible to calculate the arithmetic mean.

🎯 Exam Tip: For open-ended distributions, positional averages like the median or mode are more appropriate measures of central tendency because they don't require precise endpoints for all classes.

 

Question 4. For any set of observations, which of the following is true?
(a) \( \bar{x} \le G \)
(b) \( \bar{x} = G \)
(c) \( \bar{x} \ge G \)
(d) \( \bar{x} > G \)
Answer: (c) \( \bar{x} \ge G \)
In simple words: For any dataset of positive numbers, the arithmetic mean (\( \bar{x} \)) is always greater than or equal to the geometric mean (G). Equality holds only if all observations are identical.

🎯 Exam Tip: This relationship (\( \text{AM} \ge \text{GM} \)) is a fundamental inequality in statistics and mathematics, often tested for understanding the properties of different means.

 

Question 5. Which of the following results is true for the data that are evenly distributed around average?
(a) \( \bar{x} = M = M_0 \)
(b) \( \bar{x} > M > M_0 \)
(c) \( \bar{x} < M < M_0 \)
(d) \( \bar{x} < M > M_0 \)
Answer: (a) \( \bar{x} = M = M_0 \)
In simple words: For a perfectly symmetrical distribution, the arithmetic mean, median, and mode all coincide at the central point of the data.

🎯 Exam Tip: The equality of mean, median, and mode is a key indicator of a symmetrical distribution, such as a normal distribution.

 

Question 6. If the mean of 10 observations is 15, what is the sum of observations?
(a) 25
(b) 150
Answer: (b) 150
In simple words: To find the sum of observations, simply multiply the mean by the total number of observations: \( 15 \times 10 = 150 \).

🎯 Exam Tip: Remember the basic formula: Mean = Sum of observations / Number of observations. This can be rearranged to find any missing component.

 

Question 7. If \( \sum (x - 9) = 0 \) for data having 5 observations, then what is the value of mean?
(a) \( \bar{x} = 0 \)
(b) \( \bar{x} = 5 \)
(c) \( \bar{x} = 9 \)
(d) \( \bar{x} = 45 \)
Answer: (c) \( \bar{x} = 9 \)
In simple words: If the sum of deviations from a number is zero, that number is the mean. Here, since the sum of deviations from 9 is zero, the mean must be 9.

🎯 Exam Tip: A fundamental property of the arithmetic mean is that the sum of deviations of all observations from their mean is always zero.

 

Question 8. What is the mode of observations 7, 9, 9, 1, 7, 9, 4, 9, 1?
(a) 1
(b) 4
(c) 7
(d) 9
Answer: (d) 9
In simple words: The mode is the value that appears most frequently in a dataset. In this list of numbers, 9 occurs four times, which is more than any other number.

🎯 Exam Tip: To find the mode, it's often helpful to first arrange the data in ascending order or create a frequency count for each distinct value.

 

Question 9. In a set of 50 observations, what is the median?
(a) Value of 25th observation
(b) Value of 26th observation
(c) Value of 25.5th observation
(d) Value of 26.5th observation
Answer: (c) Value of 25.5th observation
In simple words: For an even number of observations (n=50), the median is the average of the two middle values, which are the \( (\frac{n}{2})th \) and \( (\frac{n}{2} + 1)th \) observations, or in this case, the 25th and 26th values. This is represented by the 25.5th positional value.

🎯 Exam Tip: When \( n \) is even, the median is calculated as the average of the \( (\frac{n}{2}) \) and \( (\frac{n}{2} + 1) \) terms after sorting the data. When \( n \) is odd, it's simply the \( (\frac{n+1}{2}) \) term.

 

Question 10. What is the geometric mean of 4 and 9?
(a) 4
(b) 6
(c) 6.5
Answer: (b) 6
In simple words: The geometric mean of two numbers is found by multiplying them together and then taking the square root of the product. For 4 and 9, \( \sqrt{4 \times 9} = \sqrt{36} = 6 \).

🎯 Exam Tip: The geometric mean is particularly useful for data involving growth rates or when dealing with numbers that are multiplicative in nature.

 

Question 11. If mean for a variable is 15 and its median is 20, what is the mode using empirical formula?
(a) 30
(b) 5
(c) 35
(d) 17.5
Answer: (a) 30
In simple words: Using the empirical formula (Mode = 3 Median - 2 Mean), we calculate \( 3 \times 20 - 2 \times 15 = 60 - 30 = 30 \).

🎯 Exam Tip: The empirical relationship between mean, median, and mode is a useful approximation for moderately skewed distributions, but it's not always exact for all datasets.

 

Question 12. The median of 10 observations is 14. What will be the median of observations obtained when each observation gets doubled?
(a) 10
(b) 28
(c) 7
(d) 1.4
Answer: (b) 28
In simple words: If every observation in a dataset is doubled, the median (being a positional average) will also be doubled. So, \( 14 \times 2 = 28 \).

🎯 Exam Tip: Measures of central tendency behave predictably under linear transformations (addition, subtraction, multiplication, division). If \( y = ax + b \), then \( \text{Median}(y) = a \cdot \text{Median}(x) + b \).

 

Question 13. All the observations in a data are of same value 16. What will be their mode?
(a) 8
(b) 2
(c) 16
(d) 4
Answer: (c) 16
In simple words: The mode is the value that appears most frequently. If all observations are the same value (16), then 16 is the most frequent and thus the mode.

🎯 Exam Tip: In a dataset where all values are identical, the mean, median, and mode will all be equal to that single value.

 

Question 14. Which of the following statements is false?
(a) The quartiles divide the data in 4 equal parts.
(b) The mean divides the data in 2 equal parts.
(c) The percentiles divide the data in 100 equal parts.
(d) The deciles divide the data in 10 equal parts.
Answer: (b) The mean divides the data in 2 equal parts.
In simple words: The mean does not necessarily divide the data into two equal parts based on the number of observations; that is the function of the median. The mean represents the average value, which can be affected by skewness.

🎯 Exam Tip: Positional averages (median, quartiles, deciles, percentiles) are specifically designed to divide an ordered dataset into equal parts by number of observations, whereas the mean calculates an average value.

 

Question 15. The lengths (in meters) of 6 pipes manufactured by a company are as follows: 1.05, 1.15, 0.98, 1.12, 0.89, 0.95. Which of the following statements is true?
(a) Mode = 1m
(b) Mode = 1.15 m
(c) Mode 0.98 m
(d) Mode cannot be found.
Answer: (d) Mode cannot be found.
In simple words: In the given data set, each length value appears only once. Since no value repeats, there is no value that occurs most frequently, and therefore, no mode can be identified.

🎯 Exam Tip: A dataset can have one mode (unimodal), multiple modes (multimodal), or no mode at all if all observations are unique or have the same frequency.

Section - B

Answer the following questions in one sentence:

Question 1. State any one advantage of mean.
Answer: The value of the arithmetic mean considers all observations in the dataset for its calculation.
In simple words: A key benefit of the mean is that it incorporates every data point, providing a comprehensive average.

🎯 Exam Tip: Always highlight that the mean is based on all values, which makes it a robust measure for many statistical analyses, though it can be sensitive to extreme values.

 

Question 2. If observations have varying importance, which average should be used?
Answer: If observations carry varying importance, a weighted average should be utilized.
In simple words: When some data points matter more than others, a weighted average gives them higher influence on the overall average.

🎯 Exam Tip: The weighted average is essential when dealing with grouped data or when different data points contribute unequally to the total, ensuring a more accurate representation.

 

Question 3. Name any two positional averages.
Answer: Two positional averages include the Median and Quartiles.
In simple words: Positional averages locate specific points within an ordered dataset rather than calculating a sum.

🎯 Exam Tip: Positional averages are less affected by extreme values and are often used for skewed distributions or qualitative data.

 

Question 4. State the empirical relation between mean, median and mode.
Answer: The empirical relationship linking the mean, median, and mode is given by \( M_0 = 3M - 2\bar{x} \), where \( M_0 \) denotes the Mode, \( M \) represents the Median, and \( \bar{x} \) stands for the Mean.
In simple words: This formula provides an approximate way to find one of the three central tendency measures if the other two are known, especially in slightly skewed distributions.

🎯 Exam Tip: This empirical formula is valuable for estimating measures of central tendency in distributions that are not perfectly symmetrical but show a moderate degree of skewness.

 

Question 5. State the condition under which geometric mean cannot be found.
Answer: The geometric mean cannot be determined if any observation's value is either zero or negative.
In simple words: Because the geometric mean involves multiplication and roots, it requires all data points to be positive to be calculated.

🎯 Exam Tip: Always check your data for zero or negative values before attempting to calculate the geometric mean; otherwise, the calculation is undefined.

 

Question 6. Define mode.
Answer: The mode is defined as the value that appears with the highest frequency in a given dataset, and it is commonly symbolized as \( M_0 \).
In simple words: The mode is simply the most common item or value in a set of data.

🎯 Exam Tip: The mode is useful for both qualitative and quantitative data and can indicate typical values, even in distributions with multiple peaks.

 

Question 7. State the name of the statistician who gave the empirical formula between mean, median and mode.
Answer: The renowned statistician Karl Pearson is credited with formulating the empirical relationship between the mean, median, and mode.
In simple words: Karl Pearson developed the approximate formula linking mean, median, and mode.

🎯 Exam Tip: Knowing the historical context, such as key statisticians and their contributions, can enhance understanding and provide extra credit in theoretical questions.

 

Question 8. Median of 10 observations is 55. If the value of the largest observation increases from 100 to 110, find the new median.
Answer: Given \( n = 10 \) and the initial median \( M = 55 \). When the largest observation changes from 100 to 110, the median's value remains unaffected, as it is a positional average and thus impervious to changes in extreme values. Therefore, the new median also remains 55.
In simple words: The median only cares about the middle positions, so changing a value at the extreme end won't alter it, as long as the order of middle values isn't changed.

🎯 Exam Tip: This illustrates a key advantage of the median: its resistance to outliers, making it a reliable measure for skewed data where the mean might be misleading.

 

Question 9. Mean of variable x is 9. What is the mean of the variable y = x + 4?
Answer: If the mean of variable \( x \) is 9 and variable \( y \) is defined as \( y = x + 4 \), then the mean of \( y \) will be the mean of \( x \) plus 4. Therefore, \( \bar{y} = \bar{x} + 4 = 9 + 4 = 13 \).
In simple words: When you add a constant to every value in a dataset, the mean also increases by that same constant.

🎯 Exam Tip: This property of the mean under linear transformation is very useful: \( \text{If } y = ax + b, \text{ then } \bar{y} = a\bar{x} + b \). Ensure you apply the operation correctly.

 

Question 10. Find the modal value of the variable having the following frequency distribution:

x	5	10	15	20	25
f	12	48	23	10	2

Answer: The modal value (\( M_0 \)) is the observation associated with the highest frequency. In this distribution, the maximum frequency is 48, which corresponds to the observation value of 10. Thus, \( M_0 = 10 \).
In simple words: Look for the largest number in the 'frequency' row; the 'x' value above it is the mode.

🎯 Exam Tip: For discrete frequency distributions, simply identify the highest frequency and the corresponding variable value; this is your mode.

 

Question 11. Arithmetic mean of two numbers is 5. If one number is 6, find the other number.
Answer: Let the two numbers be \( x_1 \) and \( x_2 \). We are given that their arithmetic mean is 5, and one number, \( x_1 \), is 6. The formula for the mean of two numbers is \( \bar{x} = \frac{x_1 + x_2}{2} \). Substituting the given values: \( 5 = \frac{6 + x_2}{2} \)
\( \implies 10 = 6 + x_2 \)
\( \implies x_2 = 10 - 6 \)
\( \implies x_2 = 4 \) Hence, the other number is 4.
In simple words: If the average of two numbers is 5 and one is 6, the other must be 4, because \( (6+4)/2 = 5 \).

🎯 Exam Tip: This problem tests basic algebraic manipulation using the definition of the arithmetic mean. Always set up the equation carefully before solving.

 

Question 12. Find first quartile for variable with observations 15, 4, 7, 20, 2, 7, 13.
Answer: First, arrange the observations in ascending order: 2, 4, 7, 7, 13, 15, 20. Here, the total number of observations is \( n = 7 \). The position of the first quartile (\( Q_1 \)) is given by the value of the \( \left(\frac{n+1}{4}\right)th \) observation. \( Q_1 = \text{Value of } \left(\frac{7+1}{4}\right)th \text{ observation} \)
\( = \text{Value of } \left(\frac{8}{4}\right)th \text{ observation} \)
\( = \text{Value of } 2nd \text{ observation} \) Looking at the ordered data, the 2nd observation is 4. Therefore, \( Q_1 = 4 \).
In simple words: After sorting the numbers, the first quartile is the value that is one-quarter of the way through the list.

🎯 Exam Tip: Remember to always arrange the data in ascending order before calculating quartiles, deciles, or percentiles. The formula \( (\frac{n+1}{4}) \) gives the position, not the value directly.

 

Question 13. Which average can be obtained if the continuous frequency distribution has open ended classes?
Answer: If a continuous frequency distribution contains open-ended classes, the median can still be accurately determined as a measure of average.
In simple words: For data with "unlimited" categories at the ends, the median is a good choice because it doesn't need exact limits for those categories.

🎯 Exam Tip: The median is preferred for open-ended distributions because its calculation relies on the cumulative frequency and the class where the middle value falls, without needing to know the exact midpoint of the open-ended classes.

 

Question 14. If \( Q_3 = 25.75 \) for a variable, then find \( P_{75} \).
Answer: Given \( Q_3 = 25.75 \). The third quartile (\( Q_3 \)) inherently represents the 75th percentile (\( P_{75} \)) of a dataset, as both measures mark the value below which 75% of the data falls. Therefore, \( P_{75} = Q_3 = 25.75 \).
In simple words: The third quartile is the same as the 75th percentile; they both identify the point where 75% of the data is below it.

🎯 Exam Tip: Understanding the relationships between quartiles, deciles, and percentiles (e.g., \( Q_1 = P_{25} \), \( M = Q_2 = P_{50} \), \( Q_3 = P_{75} \)) simplifies calculations and demonstrates a deeper conceptual grasp.

 

Question 15. The median of daily demand of a vendor is 15. If he sells each item for Rs. 10, find the median of his revenue?
Answer: Given the median daily demand \( M = 15 \). The selling price per item is Rs. 10. To find the median of the vendor's revenue, we multiply the median demand by the selling price per item. Median Revenue = Median Demand \( \times \) Price per Item Median Revenue = \( 15 \times \text{Rs. } 10 \) Median Revenue = Rs. 150. Hence, the median income is Rs. 150.
In simple words: If the median demand is 15 items and each sells for Rs. 10, then the median revenue is simply the median demand multiplied by the price, resulting in Rs. 150.

🎯 Exam Tip: For linear transformations like multiplication by a constant, the median (and other positional averages) transforms in the same way as the individual data points.

Section - C

Answer the following questions as required:

Question 1. Define weighted mean
Answer: When the observations within a dataset do not possess equal significance, the weighted mean is calculated by assigning a weightage to each observation that is proportional to its importance. This resulting average is termed the weighted mean and is denoted by \( \bar{x}_w \).
Suppose observations \( X_1, X_2, \ldots, X_n \) are assigned weights \( W_1, W_2, \ldots, W_n \) respectively.
The weighted mean \( \bar{x}_w \) is then calculated as:
\[ \bar{x}_w = \frac{W_1 X_1 + W_2 X_2 + \ldots + W_n X_n}{W_1 + W_2 + \ldots + W_n} \]
Which can be concisely written as:
\[ \bar{x}_w = \frac{\sum W X}{\sum W} \]
In simple words: A weighted mean is an average where some data points contribute more to the final value than others, based on their assigned "weights."

🎯 Exam Tip: The weighted mean is crucial in scenarios such as calculating grade point averages (where courses have different credit hours) or stock portfolio returns, where different investments have varying values.

 

Question 2. Explain geometric mean and state its advantages.
Answer: The geometric mean (G) of \( n \) positive and non-zero observations \( X_1, X_2, \ldots, X_n \) is defined as the \( n \)-th root of their product. Its formula is:
\[ G = \sqrt[n]{X_1 \cdot X_2 \cdot \ldots \cdot X_n} \]
Advantages of Geometric Mean:

• It is rigidly defined.
• It is based on all observations of the data.
• It is a stable measure.
• It is suitable for algebraic processes.
• It effectively balances the influence of extremely large or small observations, and its value tends to be less than the arithmetic mean.

In simple words: The geometric mean is calculated by multiplying all numbers and then taking the \( n \)-th root, and it's good for averages when numbers grow or change over time.

🎯 Exam Tip: The geometric mean is particularly valuable for averaging growth rates, ratios, or index numbers, as it minimizes the impact of extreme values more effectively than the arithmetic mean in these contexts.

 

Question 3. Explain the use of mode as a measure of central Tendency.
Answer: In everyday scenarios, we frequently use phrases like "on average" to describe typical values. For example, we might state that the average daily wage of factory workers is Rs. 110, or that a company's average daily production is 50 units, or the average lifespan in a country is 70 years. In these contexts, the "average" typically refers to the observation that occurs most frequently, which is precisely the mode. The mode provides a general value that represents the most common occurrence or characteristic within a dataset. Hence, the observation value that occurs the maximum number of times in the data is designated as the mode, commonly denoted as \( M_c \).
In simple words: The mode helps identify the most typical or frequently occurring item in a dataset, which is useful for understanding common trends in real-world situations like popular sizes or common incomes.

🎯 Exam Tip: The mode is especially useful for categorical or discrete data, and for identifying peaks in frequency distributions; it's the only measure of central tendency applicable to nominal data.

 

Question 4. Explain the positional averages briefly.
Answer: When data observations are not evenly distributed around the mean, the arithmetic mean may not accurately represent the central tendency. In such instances, to understand the data's characteristics at various points, measures of average focusing on specific positions within the ordered data are employed. These measures are known as positional averages.
The median serves as one such positional average. Additionally, quartiles, deciles, and percentiles are also utilized as positional measures of average.
In simple words: Positional averages, like the median, divide a sorted dataset into sections to find a central point, which is helpful when data isn't evenly spread.

🎯 Exam Tip: Positional averages are particularly valuable for skewed distributions or data containing outliers, as they are not influenced by extreme values, unlike the arithmetic mean.

Median and Positional Averages

The median serves as a positional average. In addition to the median, other measures of positional average include quartiles, deciles, and percentiles.

Median: The median of a dataset is the value of the observation that divides the data into two equal halves when arranged in either ascending or descending order of magnitude. It is typically denoted by M and represents the central value, signifying the average of 50% of the observations.

Quartiles: Quartiles are values that divide a sequence of observations, when arranged in increasing or decreasing order, into four equal parts. There are three quartiles, symbolically represented as \(Q_1\), \(Q_2\), and \(Q_3\).

Deciles: Deciles are values that divide the ordered observations of a given dataset into ten equal parts. There are nine deciles, denoted by \(D_1, D_2, D_3, \ldots, D_9\).

Percentiles: Percentiles are values that divide the ordered observations of a given dataset into one hundred equal parts. There are 99 percentiles, denoted by \(P_1, P_2, P_3, \ldots, P_{99}\).

Question 5. Compare mean and median as the measures of central tendency.
Answer:

Mean	Median
1. The mean is a mathematical average.	1. The median is a positional average.
2. It is computed by dividing the total sum of observations by the count of observations.	2. It represents the central value of observations when arranged in order.
3. It serves as an ideal measure of average for numerical datasets.	3. It serves as a convenient measure of average for qualitative datasets.
4. Its value is sensitive to extremely large or small observations in the dataset.	4. Its value remains unaffected by extremely large or small observations in the dataset.
5. It cannot be determined for frequency distributions with open-ended classes.	5. It can be determined for frequency distributions with open-ended classes.
6. It is a suitable measure for subsequent algebraic manipulations.	6. It is not a convenient measure of average for algebraic operations.
7. It is obtained when data observations are symmetrically distributed around a central value.	7. It is a convenient measure of average when data observations are not symmetrically distributed around a central value.

In simple words: The mean is a mathematical average, sensitive to extreme values, and good for numerical data, while the median is a positional average, unaffected by extremes, suitable for both numerical and qualitative data, and can be found with open-ended classes.

🎯 Exam Tip: When comparing measures of central tendency, clearly state whether they are mathematical or positional, their sensitivity to outliers, and their applicability to different data types or distributions (e.g., open-ended classes).

Question 6. Which average is called as optimum average? Why?
Answer:Among all the measures of average, the mean is considered the optimum average because it satisfies many characteristics of an ideal average. Consequently, it is extensively used for the analysis of statistical data.
The mean possesses the unique characteristic of being the most convenient measure for specific algebraic operations. For studying various population characteristics or comparing multiple populations, the mean is predominantly utilized as an average in advanced statistical methodologies.In simple words: The mean is considered the best average because it meets many criteria for an ideal average, is easy to use in calculations, and is widely applied in statistics for analyzing and comparing data.

🎯 Exam Tip: Focus on the mean's properties such as its mathematical definition, use in algebraic operations, and its foundational role in advanced statistical analysis to explain why it's considered "optimum."

Question 7. The mean and mode of a variable are 5.5 and 6.4 respectively. Find its median.
Answer:Given: Mean \( \bar{x} = 5.5 \) Mode \( M_0 = 6.4 \) We need to find the Median \( M \). Using the empirical formula relating mean, median, and mode: \( M_0 = 3M - 2\bar{x} \) Substitute the given values: \( 6.4 = 3M - 2(5.5) \) \( 6.4 = 3M - 11.0 \) Now, rearrange to solve for M: \( 3M = 6.4 + 11.0 \) \( 3M = 17.4 \) \( M = \frac{17.4}{3} \) \( M = 5.8 \) Therefore, the median of the variable is 5.8.In simple words: Using the empirical relationship between mode, median, and mean, we can find the median by plugging in the given mean (5.5) and mode (6.4) into the formula \(Mode = 3 \times Median - 2 \times Mean\), which gives a median of 5.8.

🎯 Exam Tip: Remember the empirical formula \(M_0 = 3M - 2\bar{x}\) for skewed distributions. Practice rearranging it to solve for any of the three measures when the other two are given. Ensure accurate substitution and calculation.

Question 8. Geometric mean of two numbers is 8. If one number is 4, find the other number.
Answer:Given: Geometric Mean \( G = 8 \) One number \( x_1 = 4 \) Let the other number be \( x_2 \). The formula for the geometric mean of two numbers is: \( G = \sqrt{x_1 \cdot x_2} \) Substitute the given values into the formula: \( 8 = \sqrt{4 \times x_2} \) To eliminate the square root, square both sides of the equation: \( 8^2 = (\sqrt{4 \times x_2})^2 \) \( 64 = 4x_2 \) Now, solve for \( x_2 \): \( x_2 = \frac{64}{4} \) \( x_2 = 16 \) Thus, the other number is 16.In simple words: The geometric mean of two numbers is the square root of their product. Since the geometric mean is 8 and one number is 4, we square 8 (to get 64) and then divide by 4 to find the other number, which is 16.

🎯 Exam Tip: For problems involving geometric mean, recall that for 'n' numbers, it's the nth root of their product. When given the GM and some numbers, setting up the equation and solving for the unknown(s) is key, often involving raising both sides to a power.

Question 9. Mean weekly production (x) of factory is 81 units. Find the mean production cost if cost is given by y = 3x + 50.
Answer:Given the mean weekly production \( \bar{x} = 81 \) units. The production cost \( y \) is related to production \( x \) by the equation: \( y = 3x + 50 \) To find the mean production cost, we can substitute the mean production \( \bar{x} \) into the linear relationship: \( \bar{y} = 3\bar{x} + 50 \) Substitute the value of \( \bar{x} = 81 \): \( \bar{y} = 3(81) + 50 \) \( \bar{y} = 243 + 50 \) \( \bar{y} = 293 \) Therefore, the mean production cost is 293.In simple words: If the mean of a variable \(x\) is 81 and a related variable \(y\) is given by \(y = 3x + 50\), then the mean of \(y\) can be found by substituting the mean of \(x\) into the equation, resulting in a mean production cost of 293.

🎯 Exam Tip: For linear transformations of data \( (y = ax + b) \), the mean of the transformed data \( (\bar{y}) \) is simply the transformed mean of the original data \( (\bar{y} = a\bar{x} + b) \). This property simplifies calculations significantly.

Question 10. The median of observations a – 5, a + 1, a + 2, a − 3, a is 10. Find a.
Answer:The given observations are \( a – 5, a + 1, a + 2, a − 3, a \). First, arrange these observations in ascending order: \( a – 5, a – 3, a, a + 1, a + 2 \) The number of observations \( n = 5 \). For an odd number of observations, the median is the value of the \( \left(\frac{n+1}{2}\right)^{\text{th}} \) observation. So, the median is the \( \left(\frac{5+1}{2}\right)^{\text{th}} = \left(\frac{6}{2}\right)^{\text{th}} = 3^{\text{rd}} \) observation. From the ordered list, the 3rd observation is \( a \). We are given that the median is 10. Therefore, \( a = 10 \).In simple words: To find 'a', first arrange the given observations in increasing order. Since there are 5 observations, the median is the 3rd value in the ordered list, which is 'a'. As the median is given as 10, 'a' must be 10.

🎯 Exam Tip: Always arrange observations in ascending (or descending) order before finding the median. For an odd number of observations, the median is the middle value. For an even number, it's the average of the two middle values. Correct ordering is crucial for positional averages.

Question 11. The mean of marks in Mathematics of 40 students in class is 76, whereas the same for the other class of 50 students is 85. Find the mean of marks in Mathematics of students in both the classes together.
Answer:Given for the first class: Number of students \( n_1 = 40 \) Mean marks \( \bar{x}_1 = 76 \) Given for the second class: Number of students \( n_2 = 50 \) Mean marks \( \bar{x}_2 = 85 \) We need to find the combined mean of marks for both classes. The formula for combined mean \( \bar{x}_c \) is: \( \bar{x}_c = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} \) Substitute the given values: \( \bar{x}_c = \frac{(40 \times 76) + (50 \times 85)}{40 + 50} \) \( \bar{x}_c = \frac{3040 + 4250}{90} \) \( \bar{x}_c = \frac{7290}{90} \) \( \bar{x}_c = 81 \) marks Therefore, the combined mean of marks in Mathematics for students in both classes is 81.In simple words: To find the combined mean, multiply each class's student count by its mean score, sum these products, and then divide by the total number of students from both classes.

🎯 Exam Tip: When combining means from different groups, use the formula for combined mean. It's essential to multiply each group's mean by its size before summing them up and dividing by the total size of all groups. This accounts for the differing weights of each group.

Question 12. 3 students from a group of 18 students failed in the examination for the subject of Economics. The marks obtained by the 15 students who passed are as follows: 42, 65, 53, 75, 43, 50, 68, 57, 79, 48, 51, 61, 55, 70, 64. Find the median marks of all 18 students.
Answer:There are a total of 18 students. 3 students failed, and 15 students passed. Let the marks of the 3 failed students be \(x_1, x_2, x_3\). These marks would typically be lower than the passing marks. The marks of the 15 passed students are: 42, 65, 53, 75, 43, 50, 68, 57, 79, 48, 51, 61, 55, 70, 64. First, arrange the passed students' marks in ascending order: 42, 43, 48, 50, 51, 53, 55, 57, 61, 64, 65, 68, 70, 75, 79 Now, consider the marks of the 3 failed students (\(x_1, x_2, x_3\)) which are likely lower than the lowest passing score. When all 18 observations are arranged in ascending order, the sequence would be: \(x_1, x_2, x_3\), 42, 43, 48, 50, 51, 53, 55, 57, 61, 64, 65, 68, 70, 75, 79 The total number of observations \( n = 18 \). For an even number of observations, the median is the average of the \( \left(\frac{n}{2}\right)^{\text{th}} \) and \( \left(\frac{n}{2}+1\right)^{\text{th}} \) observations. Median \( M = \text{Value of } \left(\frac{18}{2}\right)^{\text{th}} \text{ observation} + \text{Value of } \left(\frac{18}{2}+1\right)^{\text{th}} \text{ observation} \) \( M = \text{Value of } 9^{\text{th}} \text{ observation} + \text{Value of } 10^{\text{th}} \text{ observation} \) From the combined ordered list: The 9th observation is 53. The 10th observation is 55. Therefore, \( M = \frac{53 + 55}{2} \) \( M = \frac{108}{2} \) \( M = 54 \) marks Hence, the median marks for all 18 students is 54.In simple words: To find the median for all 18 students, first assume the 3 failed students' marks are the lowest. Then, arrange all 18 marks in ascending order. Since there are 18 observations, the median is the average of the 9th and 10th values in the sorted list, which are 53 and 55, making the median 54.

🎯 Exam Tip: When dealing with combined datasets where some values are implicitly lower (like failed marks), position them logically at the start of the ascending sequence. For an even 'n', the median is the average of the two central values; ensure you correctly identify these values after sorting the entire dataset.

Question 13. The mean daily sale of a company is 126.2 units. The sales on 10 days after adopting a new advertising strategy are as follows: 156, 125, 162, 153, 130, 124, 127, 142, 149, 121. Can we say that mean sale has increased by new advertising strategy?
Answer:The mean daily sales before the new advertising strategy was 126.2 units. After adopting the new advertising strategy, the sales for 10 days are: 156, 125, 162, 153, 130, 124, 127, 142, 149, 121. To determine if the mean sale has increased, calculate the mean of these new sales figures. Number of observations \( n = 10 \). Sum of new sales \( \Sigma x = 156 + 125 + 162 + 153 + 130 + 124 + 127 + 142 + 149 + 121 \) \( \Sigma x = 1389 \) Mean of new daily sales \( \bar{x}_{\text{new}} = \frac{\Sigma x}{n} \) \( \bar{x}_{\text{new}} = \frac{1389}{10} \) \( \bar{x}_{\text{new}} = 138.9 \) units Compare the new mean with the old mean: Old mean daily sales = 126.2 units New mean daily sales = 138.9 units Since \( 138.9 > 126.2 \), the mean daily sales have increased after the new advertising strategy.In simple words: By calculating the average of the new sales data, we find it to be 138.9 units. Comparing this to the original mean of 126.2 units, we can conclude that the new advertising strategy has successfully increased the mean daily sales.

🎯 Exam Tip: To assess the impact of a change (like a new strategy), always compare the 'before' and 'after' means. Clearly show the calculation for the 'after' mean and state your conclusion based on the comparison.

Section - E

Solve the following:

Question 1. The following table shows the number of units of electricity consumption of different families:

No. of units	Below 200	200-300	300-400	400-500	500 and above
No. of families	7	13	24	16	10

Find the median consumption.
Answer:First, create a cumulative frequency distribution table:

No. of units (Class Interval)	No. of families (f)	Cumulative frequency (cf)
Less than 200	7	7
200-300	13	20
300-400	24	44
400-500	16	60
500 and more	10	70
Total	\( n = 70 \)

To find the median consumption: First, determine the median class. The median class is the class interval corresponding to the \( \left(\frac{n}{2}\right)^{\text{th}} \) observation. \( \frac{n}{2} = \frac{70}{2} = 35^{\text{th}} \) observation. Looking at the cumulative frequency (cf) column, the 35th observation falls within the class 300-400 (since 20 < 35 ≤ 44). So, the median class is 300-400. Now, apply the median formula for grouped data: \( M = L + \left(\frac{\frac{n}{2} - cf}{f}\right) \times c \) Where: \( L = \) Lower boundary of the median class = 300 \( \frac{n}{2} = 35 \) \( cf = \) Cumulative frequency of the class preceding the median class = 20 \( f = \) Frequency of the median class = 24 \( c = \) Class length of the median class = \( 400 - 300 = 100 \) Substitute these values into the formula: \( M = 300 + \left(\frac{35 - 20}{24}\right) \times 100 \) \( M = 300 + \left(\frac{15}{24}\right) \times 100 \) \( M = 300 + \frac{1500}{24} \) \( M = 300 + 62.5 \) \( M = 362.5 \) units Therefore, the median electricity consumption is 362.5 units.In simple words: We first find the median position, which is the 35th observation. This falls into the 300-400 class. Using the median formula for grouped data with the lower boundary (300), cumulative frequency of the preceding class (20), frequency of the median class (24), and class length (100), we calculate the median consumption to be 362.5 units.

🎯 Exam Tip: For grouped data, always start by constructing a cumulative frequency table to identify the median class. Carefully extract \(L\), \(n/2\), \(cf\), \(f\), and \(c\) for the median formula to avoid calculation errors.

Question 2. The weekly profit and loss information of a vendor is available as follows. Find the modal profit:

Profit (thousand Rs.)	-2-0	0-2	2-4	4-6	6-8	8-10
No. of weeks	4	8	14	6	2	1

Answer:The given frequency distribution is:

Profit (thousand Rs.)	No. of weeks (f)
-2-0	4
0-2	8 (\(f_1\))
2-4	14 (\(f_m\))
4-6	6 (\(f_2\))
6-8	2
8-10	1

To find the modal profit, we first identify the modal class. The modal class is the class with the highest frequency. From the table, the maximum frequency is 14, which corresponds to the class 2-4. So, the modal class is 2-4. Now, apply the mode formula for grouped data: \( M_0 = L + \left(\frac{f_m - f_1}{2f_m - f_1 - f_2}\right) \times c \) Where: \( L = \) Lower boundary of the modal class = 2 \( f_m = \) Frequency of the modal class = 14 \( f_1 = \) Frequency of the class preceding the modal class = 8 \( f_2 = \) Frequency of the class succeeding the modal class = 6 \( c = \) Class length of the modal class = \( 4 - 2 = 2 \) Substitute these values into the formula: \( M_0 = 2 + \left(\frac{14 - 8}{2(14) - 8 - 6}\right) \times 2 \) \( M_0 = 2 + \left(\frac{6}{28 - 8 - 6}\right) \times 2 \) \( M_0 = 2 + \left(\frac{6}{28 - 14}\right) \times 2 \) \( M_0 = 2 + \left(\frac{6}{14}\right) \times 2 \) \( M_0 = 2 + \frac{12}{14} \) \( M_0 = 2 + 0.857 \) (approximately) \( M_0 \approx 2.86 \) thousand Rs. Therefore, the modal profit is approximately Rs. 2.86 thousand.In simple words: The modal profit is found by identifying the class with the highest frequency (2-4 thousand Rs.). Then, we use the mode formula for grouped data, plugging in the lower boundary, frequencies of the modal and adjacent classes, and the class length to calculate the modal profit of approximately Rs. 2.86 thousand.

🎯 Exam Tip: Clearly identify \(L\), \(f_m\), \(f_1\), \(f_2\), and \(c\) before applying the mode formula for grouped data. Double-check the denominator calculation \( (2f_m - f_1 - f_2) \) as it's a common source of error.

Question 3. The number of bags of wheat sold in a grocer's shop each day are shown in the following table:

No. of bags	25-29	30-34	35-39	40-44	45-49	50-54	55 and above
No. of days	9	17	32	24	10	5	3

Find Q1 and D4 for the number of bags sold.
Answer:First, create a cumulative frequency distribution table, ensuring to convert the inclusive classes to exclusive classes for calculation, especially since the class length is not uniform. However, for quartiles and deciles, often the boundary adjustment is implicitly handled by the formula if the class lengths are consistent. For simplicity, we'll proceed with the given classes and adjust the lower boundary \(L\) as needed.

No. of bags (Class Interval)	No. of days (f)	Cumulative frequency (cf)
25-29	9	9
30-34	17	26
35-39	32	58
40-44	24	82
45-49	10	92
50-54	5	97
55 and more	3	100
Total	\( n = 100 \)

**First Quartile (Q1):** First, determine the Q1 class. The Q1 class is the class interval corresponding to the \( \left(\frac{n}{4}\right)^{\text{th}} \) observation. \( \frac{n}{4} = \frac{100}{4} = 25^{\text{th}} \) observation. Referring to the cumulative frequency (cf) column, the 25th observation falls within the class 30-34 (since 9 < 25 ≤ 26). So, the Q1 class is 30-34. For exclusive classes, this becomes 29.5-34.5. Now, apply the quartile formula for grouped data: \( Q_1 = L + \left(\frac{\frac{n}{4} - cf}{f}\right) \times c \) Where: \( L = \) Lower boundary of the Q1 class = 29.5 (for exclusive class) \( \frac{n}{4} = 25 \) \( cf = \) Cumulative frequency of the class preceding the Q1 class = 9 \( f = \) Frequency of the Q1 class = 17 \( c = \) Class length of the Q1 class = \( 34.5 - 29.5 = 5 \) Substitute these values into the formula: \( Q_1 = 29.5 + \left(\frac{25 - 9}{17}\right) \times 5 \) \( Q_1 = 29.5 + \left(\frac{16}{17}\right) \times 5 \) \( Q_1 = 29.5 + \frac{80}{17} \) \( Q_1 = 29.5 + 4.7058 \) (approximately) \( Q_1 \approx 34.21 \) bags Hence, the first quartile \( Q_1 \) is approximately 34.21 bags. **Fourth Decile (D4):** First, determine the D4 class. The D4 class is the class interval corresponding to the \( 4\left(\frac{n}{10}\right)^{\text{th}} \) observation. \( 4\left(\frac{n}{10}\right) = 4\left(\frac{100}{10}\right) = 4(10) = 40^{\text{th}} \) observation. Referring to the cumulative frequency (cf) column, the 40th observation falls within the class 35-39 (since 26 < 40 ≤ 58). So, the D4 class is 35-39. For exclusive classes, this becomes 34.5-39.5. Now, apply the decile formula for grouped data: \( D_4 = L + \left(\frac{4\frac{n}{10} - cf}{f}\right) \times c \) Where: \( L = \) Lower boundary of the D4 class = 34.5 (for exclusive class) \( 4\frac{n}{10} = 40 \) \( cf = \) Cumulative frequency of the class preceding the D4 class = 26 \( f = \) Frequency of the D4 class = 32 \( c = \) Class length of the D4 class = \( 39.5 - 34.5 = 5 \) Substitute these values into the formula: \( D_4 = 34.5 + \left(\frac{40 - 26}{32}\right) \times 5 \) \( D_4 = 34.5 + \left(\frac{14}{32}\right) \times 5 \) \( D_4 = 34.5 + \frac{70}{32} \) \( D_4 = 34.5 + 2.1875 \) (approximately) \( D_4 \approx 36.69 \) bags Hence, the fourth decile \( D_4 \) is approximately 36.69 bags.In simple words: To find the first quartile (Q1), we locate the 25th observation, which falls in the 30-34 class. Using the quartile formula, Q1 is approximately 34.21 bags. For the fourth decile (D4), we locate the 40th observation, which falls in the 35-39 class. Using the decile formula, D4 is approximately 36.69 bags.

🎯 Exam Tip: Always construct a cumulative frequency table first. When dealing with inclusive class intervals for quartiles/deciles, remember to adjust to exclusive class boundaries (\(L\)) for accurate calculations if the formula expects continuous data. Double-check all parameters \(L, n/k, cf, f, c\).

Question 4. The heights of students of a college are given in the following table. Find the mean height of the students.

Height (in cm)	150-155	155-160	160-165	165-170	170-175	175-180	180-185
No. of students	8	10	20	17	15	4	1

Answer:To find the mean height, we can use the assumed mean method. Let's create a frequency distribution table with mid-values, deviation, and frequency-deviation products. Let assumed mean \( A = 167.5 \) (mid-value of 165-170 class) Class length \( c = 5 \)

Height (in cm)	No. of students (f)	Mid value (x)	\( d = \frac{x - A}{c} \)	\( f \cdot d \)
150-155	8	152.5	-3	-24
155-160	10	157.5	-2	-20
160-165	20	162.5	-1	-20
165-170	17	167.5	0	0
170-175	15	172.5	1	15
175-180	4	177.5	2	8
180-185	1	182.5	3	3
Total	\( n = \Sigma f = 75 \)			\( \Sigma fd = -38 \)

The formula for mean using the assumed mean method is: \( \bar{x} = A + \left(\frac{\Sigma fd}{n}\right) \times c \) Substitute the values: \( \bar{x} = 167.5 + \left(\frac{-38}{75}\right) \times 5 \) \( \bar{x} = 167.5 - \frac{38 \times 5}{75} \) \( \bar{x} = 167.5 - \frac{190}{75} \) \( \bar{x} = 167.5 - 2.5333 \) (approximately) \( \bar{x} \approx 164.97 \) cm Therefore, the mean height of the students is approximately 164.97 cm.In simple words: To find the mean height, we used the assumed mean method. We calculated the mid-value for each height range, then found the deviations from an assumed mean (167.5) and multiplied by frequency. Summing these and applying the formula \(Mean = Assumed Mean + (\Sigma fd / n) \times c\), we get a mean height of approximately 164.97 cm.

🎯 Exam Tip: When calculating mean for grouped data, the assumed mean method simplifies calculations for larger frequencies. Choose an assumed mean near the center of the data. Ensure accurate calculation of mid-values, deviations, and the \( \Sigma fd \) sum. Remember the \( \times c \) factor in the formula.

Question 5. The monthly Income (In thousand Rs.) of 130 persons living in a certain area is as follows:

Income (thousand Rs.)	Less than 4	4-8	8-12	12-20	20-28	28-36
No. of persons	6	14	31	35	28	16

Find the median of Income.
Answer:The given frequency distribution has unequal class lengths. First, we need to set up the class intervals and their cumulative frequencies.

Income (thousand Rs.)	No. of persons (f)	Cumulative frequency (cf)
0-4	6	6
4-8	14	20
8-12	31	51
12-20	35	86
20-28	28	114
28-36	16	130
Total	\( n = 130 \)

To find the median income: First, determine the median class. The median class is the class interval corresponding to the \( \left(\frac{n}{2}\right)^{\text{th}} \) observation. \( \frac{n}{2} = \frac{130}{2} = 65^{\text{th}} \) observation. Referring to the cumulative frequency (cf) column, the 65th observation falls within the class 12-20 (since 51 < 65 ≤ 86). So, the median class is 12-20. Now, apply the median formula for grouped data: \( M = L + \left(\frac{\frac{n}{2} - cf}{f}\right) \times c \) Where: \( L = \) Lower boundary of the median class = 12 \( \frac{n}{2} = 65 \) \( cf = \) Cumulative frequency of the class preceding the median class = 51 \( f = \) Frequency of the median class = 35 \( c = \) Class length of the median class = \( 20 - 12 = 8 \) Substitute these values into the formula: \( M = 12 + \left(\frac{65 - 51}{35}\right) \times 8 \) \( M = 12 + \left(\frac{14}{35}\right) \times 8 \) \( M = 12 + \frac{112}{35} \) \( M = 12 + 3.2 \) \( M = 15.2 \) thousand Rs. Therefore, the median income is Rs. 15.2 thousand.In simple words: First, calculate the cumulative frequencies to find the median position (\(n/2 = 65\)). This places the median in the 12-20 thousand Rs. class. Using the median formula for grouped data, with \(L=12\), \(cf=51\), \(f=35\), and \(c=8\), we calculate the median income to be Rs. 15.2 thousand.

🎯 Exam Tip: For grouped data with unequal class intervals, the median formula remains the same, but accurately identifying \(L\), \(f\), and \(c\) for the median class is crucial. Always verify the class boundaries for proper calculation of \(c\).

Question 6. The data about population (in thousands) of 70 villages in a district is given in the following table :

Population (in thousands)	0-10	10-20	20-30	30-40	40-50
No. of villages	6	18	22	15	9

Find the mode for the population using graphical method.
Answer:To find the mode graphically, we will draw a histogram of the given frequency distribution.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक हिस्टोग्राम को दर्शाता है जो विभिन्न जनसंख्या श्रेणियों (हजारों में) में गाँवों की संख्या के वितरण को प्रस्तुत करता है। X-अक्ष जनसंख्या (हजारों में) दिखाता है, जबकि Y-अक्ष गाँवों की संख्या (आवृत्ति) दर्शाता है। सबसे ऊँची पट्टी (20-30 हजार) मॉडल वर्ग को इंगित करती है, और इसके शीर्ष को पड़ोसी पट्टियों के शीर्ष कोनों से जोड़कर और प्रतिच्छेदन बिंदु से X-अक्ष पर लंबवत रेखा खींचकर मोड का अनुमान लगाया जाता है। From the histogram: The modal class is 20-30 (the class with the highest frequency, 22). By drawing lines from the top corners of the highest bar to the adjacent corners of the neighboring bars, their intersection point projects onto the X-axis, giving the mode. From the provided histogram, the mode (OQ) is observed to be 23 thousand. Hence, the mode of the population is 23 thousand.In simple words: The mode is found graphically by drawing a histogram of the population data. The modal class is the one with the tallest bar (20-30 thousand). By drawing lines from the top corners of this bar to the adjacent corners of the neighboring bars, their intersection point reveals the mode on the X-axis, which is 23 thousand.

🎯 Exam Tip: For finding mode graphically, construct a histogram, identify the tallest bar (modal class), and then draw lines from the top corners of this bar to the top opposite corners of the adjacent bars. The x-coordinate of the intersection point of these lines on the x-axis is the mode.

Question 7. The marks obtained by 60 students in an examination are as follows. Find the mean marks of the students.

Marks	0-10	10-20	20-25	25-30	30-35	35-40
No. of students	3	8	20	16	9	4

Answer:The class intervals are not uniform. For the first two classes, the class length \( c = 10 \). For the subsequent classes, \( c = 5 \). To calculate the mean, we can use the step-deviation method (a variation of the assumed mean method) with an appropriate assumed mean. Let assumed mean \( A = 22.5 \) (mid-value of the 20-25 class). Let \( c \) be the common class interval for the deviations, but since the class lengths are not uniform, we will use a modified approach where \( d = x - A \) and then sum \( f \cdot d \) or use the actual class lengths in the \( \frac{x-A}{c} \) formula. Here, the solution provided uses \( d = \frac{x-A}{5} \) implying \( c=5 \) is used as a common divisor, which is generally acceptable for simplification as long as the mid-points are correctly calculated.

Marks (Class Interval)	No. of students (f)	Mid value (x)	\( d = \frac{x - A}{5} \) (with \( A = 22.5 \))	\( f \cdot d \)
0-10	3	5	\( \frac{5 - 22.5}{5} = -3.5 \)	-10.5
10-20	8	15	\( \frac{15 - 22.5}{5} = -1.5 \)	-12.0
20-25	20	22.5	\( \frac{22.5 - 22.5}{5} = 0 \)	0
25-30	16	27.5	\( \frac{27.5 - 22.5}{5} = 1 \)	16.0
30-35	9	32.5	\( \frac{32.5 - 22.5}{5} = 2 \)	18.0
35-40	4	37.5	\( \frac{37.5 - 22.5}{5} = 3 \)	12.0
Total	\( n = \Sigma f = 60 \)			\( \Sigma fd = 23.5 \)

The formula for mean using the step-deviation method is: \( \bar{x} = A + \left(\frac{\Sigma fd}{n}\right) \times c \) Where \( c \) is the common factor used in deviation calculation (here, 5). Substitute the values: \( \bar{x} = 22.5 + \left(\frac{23.5}{60}\right) \times 5 \) \( \bar{x} = 22.5 + \frac{117.5}{60} \) \( \bar{x} = 22.5 + 1.9583 \) (approximately) \( \bar{x} \approx 24.46 \) marks Therefore, the mean marks of the students are approximately 24.46.In simple words: To calculate the mean, we use the assumed mean method. We find the mid-point for each mark range, calculate the deviation from an assumed mean (22.5) divided by 5, then multiply by frequency. Summing these and applying the formula \(Mean = Assumed Mean + (\Sigma fd / n) \times c\) (where \(c=5\)), we get a mean mark of approximately 24.46.

🎯 Exam Tip: When class intervals are uneven, ensure correct calculation of mid-values and deviations. If using the step-deviation method, the 'c' in the formula is the common divisor used for simplifying deviations, not necessarily the class length of the modal class. Choose an assumed mean wisely to simplify computations.

Question 8. A survey was conducted for 50 employees in an office regarding their computer usage time. The details are shown In the following table:

Time (hours)	5-5.5	5.5-6	6-6.5	6.5-7	7-7.5	7.5-8	8-8.5	8.5-9
No. of employees	1	3	5	11	15	9	4	2

Find the quartiles Q1 and Q3 for the time for the usage of computer.
Answer:First, create a cumulative frequency distribution table:

Time of use of computers (hours)	No. of employees (f)	Cumulative frequency (cf)
5-5.5	1	1
5.5-6	3	4
6-6.5	5	9
6.5-7	11	20
7-7.5	15	35
7.5-8	9	44
8-8.5	4	48
8.5-9	2	50
Total	\( n = 50 \)

**First Quartile (Q1):** First, determine the Q1 class. The Q1 class is the class interval corresponding to the \( \left(\frac{n}{4}\right)^{\text{th}} \) observation. \( \frac{n}{4} = \frac{50}{4} = 12.5^{\text{th}} \) observation. Referring to the cumulative frequency (cf) column, the 12.5th observation falls within the class 6.5-7 (since 9 < 12.5 ≤ 20). So, the Q1 class is 6.5-7. Now, apply the quartile formula for grouped data: \( Q_1 = L + \left(\frac{\frac{n}{4} - cf}{f}\right) \times c \) Where: \( L = \) Lower boundary of the Q1 class = 6.5 \( \frac{n}{4} = 12.5 \) \( cf = \) Cumulative frequency of the class preceding the Q1 class = 9 \( f = \) Frequency of the Q1 class = 11 \( c = \) Class length of the Q1 class = \( 7 - 6.5 = 0.5 \) Substitute these values into the formula: \( Q_1 = 6.5 + \left(\frac{12.5 - 9}{11}\right) \times 0.5 \) \( Q_1 = 6.5 + \left(\frac{3.5}{11}\right) \times 0.5 \) \( Q_1 = 6.5 + \frac{1.75}{11} \) \( Q_1 = 6.5 + 0.15909 \) (approximately) \( Q_1 \approx 6.66 \) hours Hence, the first quartile \( Q_1 \) is approximately 6.66 hours. **Third Quartile (Q3):** First, determine the Q3 class. The Q3 class is the class interval corresponding to the \( 3\left(\frac{n}{4}\right)^{\text{th}} \) observation. \( 3\left(\frac{n}{4}\right) = 3\left(\frac{50}{4}\right) = 3(12.5) = 37.5^{\text{th}} \) observation. Referring to the cumulative frequency (cf) column, the 37.5th observation falls within the class 7.5-8 (since 35 < 37.5 ≤ 44). So, the Q3 class is 7.5-8. Now, apply the quartile formula for grouped data: \( Q_3 = L + \left(\frac{3\frac{n}{4} - cf}{f}\right) \times c \) Where: \( L = \) Lower boundary of the Q3 class = 7.5 \( 3\frac{n}{4} = 37.5 \) \( cf = \) Cumulative frequency of the class preceding the Q3 class = 35 \( f = \) Frequency of the Q3 class = 9 \( c = \) Class length of the Q3 class = \( 8 - 7.5 = 0.5 \) Substitute these values into the formula: \( Q_3 = 7.5 + \left(\frac{37.5 - 35}{9}\right) \times 0.5 \) \( Q_3 = 7.5 + \left(\frac{2.5}{9}\right) \times 0.5 \) \( Q_3 = 7.5 + \frac{1.25}{9} \) \( Q_3 = 7.5 + 0.13888 \) (approximately) \( Q_3 \approx 7.64 \) hours Hence, the third quartile \( Q_3 \) is approximately 7.64 hours.In simple words: For Q1, we find the 12.5th observation, which falls in the 6.5-7 hour class. Using the quartile formula, Q1 is approximately 6.66 hours. For Q3, we find the 37.5th observation, which is in the 7.5-8 hour class. Using the formula again, Q3 is approximately 7.64 hours.

🎯 Exam Tip: When calculating quartiles for grouped data, ensure accurate identification of the quartile class using \( n/4 \) or \( 3n/4 \). Pay attention to the lower class boundary (L), cumulative frequency of the preceding class (cf), frequency of the quartile class (f), and the class length (c) for precise results.

Question 8. Find the quartiles Q1 and Q3 for the time for the usage of computer.
Answer:Third Quartile: The third quartile class (Q3 class) is identified by the \( \left(\frac{3n}{4}\right) \)th observation.
In this case, \( \frac{3 \times 50}{4} = 37.5 \)th observation.
Referring to the cumulative frequency (cf) column, the 37.5th observation falls within the 7.5 - 8 class.
Therefore, the Q3 class is 7.5 - 8.
Using the formula for Q3 for grouped data:
\( Q_3 = L + \frac{\left(\frac{3n}{4}\right)-cf}{f} \times c \)
Here, L (lower boundary of Q3 class) = 7.5
\( \left(\frac{3n}{4}\right) \) = 37.5
cf (cumulative frequency of the class preceding Q3 class) = 35
f (frequency of Q3 class) = 9
c (class width) = 0.5
Substituting these values into the formula:
\( Q_3 = 7.5 + \frac{37.5 - 35}{9} \times 0.5 \)
\( = 7.5 + \frac{2.5 \times 0.5}{9} \)
\( = 7.5 + \frac{1.25}{9} \)
\( = 7.5 + 0.1388... \)
\( \approx 7.5 + 0.14 \)
\( Q_3 \approx 7.64 \) hours.
Thus, the third quartile Q3 is approximately 7.64 hours.
In simple words: The third quartile (Q3) indicates that 75% of employees use computers for 7.64 hours or less. This value helps understand the upper usage patterns.

🎯 Exam Tip: Remember to correctly identify the quartile class and apply the formula accurately for grouped data. Pay attention to class boundaries and cumulative frequencies.

Section - F

Solve The Following:

Question 1. The data about marks scored by 55 students from a school are given below:

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70
No. of students	4	7	11	14	9	7	3

(i) If 30 % students failed in the examination, what are the passing marks?
(ii) If top 5% students are to be selected for scholarship, find the lowest marks for scholarship?
Answer:First, let's create a cumulative frequency distribution table:

Marks	No. of students f	Cumulative frequency cf
0-10	4	4
10-20	7	11
20-30	11	22
30-40	14	36
40-50	9	45
50-60	7	52
60-70	3	55
Total	n = 55

(i) If 30% of students failed, their marks are at or below the 3rd decile (D3).
To find D3, we calculate the position: \( \text{Position of D3} = \frac{3n}{10} \)th observation.
\( \text{Position of D3} = \frac{3 \times 55}{10} = 16.5 \)th observation.
Referring to the cumulative frequency column, the 16.5th observation falls in the 20-30 class.
Therefore, the D3 class is 20-30.
Using the formula for D3 for grouped data:
\( D_3 = L + \frac{\left(\frac{3n}{10}\right)-cf}{f} \times c \)
Here, L (lower boundary of D3 class) = 20
\( \left(\frac{3n}{10}\right) \) = 16.5
cf (cumulative frequency of the class preceding D3 class) = 11
f (frequency of D3 class) = 11
c (class width) = 10
Substituting these values into the formula:
\( D_3 = 20 + \frac{16.5 - 11}{11} \times 10 \)
\( = 20 + \frac{5.5}{11} \times 10 \)
\( = 20 + 0.5 \times 10 \)
\( = 20 + 5 \)
\( D_3 = 25 \) marks.
Hence, the maximum marks of a student who failed are 25. Therefore, the passing marks required for the examination are 26.
(ii) Scholarship is given to the top 5% of students (those with maximum marks). This corresponds to the 95th percentile (P95).
To find P95, we calculate the position: \( \text{Position of P95} = \frac{95n}{100} \)th observation.
\( \text{Position of P95} = \frac{95 \times 55}{100} = 52.25 \)th observation.
Referring to the cumulative frequency column, the 52.25th observation falls in the 60-70 class.
Therefore, the P95 class is 60-70.
Using the formula for P95 for grouped data:
\( P_{95} = L + \frac{\left(\frac{95n}{100}\right)-cf}{f} \times c \)
Here, L (lower boundary of P95 class) = 60
\( \left(\frac{95n}{100}\right) \) = 52.25
cf (cumulative frequency of the class preceding P95 class) = 52
f (frequency of P95 class) = 3
c (class width) = 10
Substituting these values into the formula:
\( P_{95} = 60 + \frac{52.25 - 52}{3} \times 10 \)
\( = 60 + \frac{0.25}{3} \times 10 \)
\( = 60 + \frac{2.5}{3} \)
\( = 60 + 0.8333... \)
\( \approx 60 + 0.83 \)
\( P_{95} \approx 60.83 \) marks.
Thus, the minimum marks required for scholarship are 60.83 marks.
In simple words: 25 marks is the threshold for failing, meaning students need 26 or more to pass. For a scholarship, a student needs at least 60.83 marks to be in the top 5%.

🎯 Exam Tip: For percentiles and deciles in grouped data, correctly identify the class interval based on the \( \frac{kn}{10} \) or \( \frac{kn}{100} \) position, then apply the interpolation formula. Ensure careful calculation of L, cf, f, and c.

Question 2. Two brands of tyres are to be compared for their mean life. The following data are available:

Life (thousand km)	10-15	15-20	20-25	25-30	30-35	35-40
No. of tyres of brand A	4	7	10	5	3	1
No. of tyres of brand B	5	8	15	9	6	2

On the basis of mean, which brand of tyres is better?
Answer:To compare the brands based on their mean life, we calculate the mean for each brand using the step-deviation method.
We use an assumed mean (A) = 22.5 and class width (c) = 5.

Life (thousand km)	Mid value x	d = \(\frac{x-A}{c}\) A = 22.5, c = 5	Brand A No. of tyres f	f.d	Brand B No. of tyres f	f.d
10-15	12.5	-2	4	-8	5	-10
15-20	17.5	-1	7	-7	8	-8
20-25	22.5	0	10	0	15	0
25-30	27.5	1	5	5	9	9
30-35	32.5	2	3	6	6	12
35-40	37.5	3	1	3	2	6
Total			n = 30	\( \Sigma fd = -1 \)	n = 45	\( \Sigma fd = 9 \)

**For Brand A tyres:**
Mean \( \bar{x}_A = A + \frac{\Sigma fd}{n} \times c \)
Given A = 22.5, \( \Sigma fd = -1 \), n = 30, and c = 5.
\( \bar{x}_A = 22.5 + \frac{-1}{30} \times 5 \)
\( = 22.5 - \frac{5}{30} \)
\( = 22.5 - 0.166... \)
\( \approx 22.5 - 0.17 \)
\( \bar{x}_A \approx 22.33 \) thousand km.
**For Brand B tyres:**
Mean \( \bar{x}_B = A + \frac{\Sigma fd}{n} \times c \)
Given A = 22.5, \( \Sigma fd = 9 \), n = 45, and c = 5.
\( \bar{x}_B = 22.5 + \frac{9}{45} \times 5 \)
\( = 22.5 + \frac{45}{45} \)
\( = 22.5 + 1 \)
\( \bar{x}_B = 23.5 \) thousand km.
Comparing the mean lives:
Mean life of Brand A tyres = 22.33 thousand km
Mean life of Brand B tyres = 23.5 thousand km
Since the mean life of Brand B tyres (23.5 thousand km) is greater than that of Brand A tyres (22.33 thousand km), Brand B tyres are considered better.
In simple words: Brand B tyres have a higher average lifespan than Brand A tyres. Therefore, Brand B is the better choice for durability.

🎯 Exam Tip: When comparing two datasets, calculating and comparing their means is often a good starting point. Ensure consistent application of the mean formula for both sets of data.

Question 3. The distribution of sale of cars of a company on different days is as follows. Find the mode for the number of cars sold using an appropriate formula.

No. of cars	0-10	10-15	15-20	24	26	28
No. of days	8	14	16	11	4	2

Answer:The provided data mixes class intervals (0-10, 10-15, 15-20) with discrete values (24, 26, 28). To find the mode using an empirical formula, it's often more appropriate to consider it as a mixed distribution or a distribution where the formula \( \text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean} \) might be used if the distribution is moderately skewed. Let's first calculate the mean and median from the given data. We'll treat the discrete values as class midpoints for calculation.

No. of cars (Class Interval)	Mid value x	No. of days f	f.x	Cumulative frequency cf
0-10	5	8	40	8
10-15	12.5	14	175	22
15-20	17.5	16	280	38
24 (as 20-28 for mid-value 24)	24	11	264	49
26 (as 20-28 for mid-value 26)	26	4	104	53
28 (as 20-28 for mid-value 28)	28	2	56	55
Total		n = 55	\( \Sigma fx = 919 \)

**1. Calculate the Mean (\( \bar{x} \)):**
\( \bar{x} = \frac{\Sigma fx}{n} \)
\( \bar{x} = \frac{919}{55} \)
\( \bar{x} \approx 16.71 \) cars.
**2. Calculate the Median (M):**
The median class is the class containing the \( \left(\frac{n}{2}\right) \)th observation.
\( \frac{n}{2} = \frac{55}{2} = 27.5 \)th observation.
Referring to the cumulative frequency (cf) column, the 27.5th observation falls within the 15-20 class.
Therefore, the Median class is 15-20.
Using the formula for Median for grouped data:
\( M = L + \frac{\left(\frac{n}{2}\right)-cf}{f} \times c \)
Here, L (lower boundary of Median class) = 15
\( \left(\frac{n}{2}\right) \) = 27.5
cf (cumulative frequency of the class preceding Median class) = 22
f (frequency of Median class) = 16
c (class width) = 5
Substituting these values into the formula:
\( M = 15 + \frac{27.5 - 22}{16} \times 5 \)
\( = 15 + \frac{5.5}{16} \times 5 \)
\( = 15 + \frac{27.5}{16} \)
\( = 15 + 1.71875 \)
\( M \approx 16.72 \) cars.
**3. Calculate the Mode (Mo) using the empirical formula:**
\( \text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean} \)
\( \text{Mode} = 3 \times 16.72 - 2 \times 16.71 \)
\( = 50.16 - 33.42 \)
\( \text{Mode} = 16.74 \) cars.
Hence, the mode of cars sold is approximately 16.74 cars.
In simple words: The typical number of cars sold, or the mode, is found to be about 16.74 cars per day by using an empirical formula that relates the mean and median.

🎯 Exam Tip: When dealing with mixed data (intervals and discrete values), estimate mid-values carefully. The empirical formula for mode is useful when direct calculation is complex or the distribution is moderately skewed, but ensure mean and median are calculated precisely first.

Question 4. The wheat crop grown per acre by farmers in different parts of a state is given below:

Wheat crop per acre (quintals)	20-25	25-30	30-40	40-50	50-60
No. of farmers	12	23	45	29	7

Find the mean and median for the wheat crop per acre.
Answer:Notice that the class lengths are unequal (5 for the first two, then 10). This needs to be considered when calculating mid-values or if a direct mean/median formula for unequal classes is preferred. For mean, mid-values are straightforward. For median, class boundaries are critical. Let's calculate mean and median:

Crop per acre (Quintal)	No. of farmers f	Mid value x	f.x	Cumulative frequency cf
20-25	12	22.5	270.0	12
25-30	23	27.5	632.5	35
30-40	45	35.0	1575.0	80
40-50	29	45.0	1305.0	109
50-60	7	55.0	385.0	116
Total	n = 116		\( \Sigma fx = 4167.5 \)

**1. Calculate the Mean (\( \bar{x} \)):**
\( \bar{x} = \frac{\Sigma fx}{n} \)
\( \bar{x} = \frac{4167.5}{116} \)
\( \bar{x} \approx 35.93 \) quintals.
**2. Calculate the Median (M):**
The median class is the class containing the \( \left(\frac{n}{2}\right) \)th observation.
\( \frac{n}{2} = \frac{116}{2} = 58 \)th observation.
Referring to the cumulative frequency (cf) column, the 58th observation falls within the 30-40 class.
Therefore, the Median class is 30-40.
Using the formula for Median for grouped data:
\( M = L + \frac{\left(\frac{n}{2}\right)-cf}{f} \times c \)
Here, L (lower boundary of Median class) = 30
\( \left(\frac{n}{2}\right) \) = 58
cf (cumulative frequency of the class preceding Median class) = 35
f (frequency of Median class) = 45
c (class width) = 10
Substituting these values into the formula:
\( M = 30 + \frac{58 - 35}{45} \times 10 \)
\( = 30 + \frac{23}{45} \times 10 \)
\( = 30 + \frac{230}{45} \)
\( = 30 + 5.111... \)
\( M \approx 35.11 \) quintals.
Hence, the mean wheat crop per acre is 35.93 quintals, and the median wheat crop per acre is 35.11 quintals.
In simple words: On average, farmers produce about 35.93 quintals of wheat per acre. The median value of 35.11 quintals means that half of the farmers produce less than this amount, and half produce more.

🎯 Exam Tip: When class intervals are unequal, carefully calculate the mid-values and ensure the correct class width (c) is used for the median class in the formula. Double-check cumulative frequencies.

Question 5. The distribution of age of 150 spectators In a theatre is as follows:

Age (years)	15-20	20-25	25-30	30-40	40-50	50-60	60-80
No. of spectators	6	13	19	52	34	18	8

Find the mode for age of spectators using graphical method.
Answer:The given frequency distribution has unequal class lengths. To find the mode graphically using a histogram, we first need to adjust the frequencies to create proportionate frequencies, making the class lengths uniform. We choose the minimum class length (which is 5 years) as the basis for proportionate frequency.

Age (years)	No. of spectators f	Class length	Proportionate frequency = \(\frac{\text{frequency of a class}}{\text{class length of a class}} \times \text{minimum class length}\)
15-20	6	5	\(\frac{6}{5} \times 5 = 6\)
20-25	13	5	\(\frac{13}{5} \times 5 = 13\)
25-30	19	5	\(\frac{19}{5} \times 5 = 19\)
30-40	52	10	\(\frac{52}{10} \times 5 = 26\)
40-50	34	10	\(\frac{34}{10} \times 5 = 17\)
50-60	18	10	\(\frac{18}{10} \times 5 = 9\)
60-80	8	20	\(\frac{8}{20} \times 5 = 2\)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक हिस्टोग्राम है जो दर्शकों की आयु (वर्षों में) और उनकी आवृत्ति को दर्शाता है। X-अक्ष पर आयु वर्ग (वर्ष) और Y-अक्ष पर दर्शकों की संख्या (आवृत्ति) है। हिस्टोग्राम के दंडों की ऊँचाई आवृत्ति के समानुपातिक है, और दंडों के शीर्ष बिंदुओं को जोड़कर मोड वर्ग का पता लगाया गया है। सबसे ऊँचे दंड के शीर्ष से खींची गई तिरछी रेखाओं के प्रतिच्छेद बिंदु से X-अक्ष पर एक लंब खींचकर मोड का मान 34 वर्ष निर्धारित किया गया है।
From the histogram (drawn using proportionate frequencies), the mode is found by identifying the highest rectangle (modal class), drawing lines from its top corners to the adjacent inner corners of the rectangles, and then drawing a perpendicular from the intersection point of these lines to the X-axis.
From the histogram, the mode (OQ) is 34 years.
Hence, the mode of the age of spectators is 34 years.
In simple words: Most spectators in the theatre are around 34 years old. This value was found by drawing a histogram and locating the peak frequency, then making adjustments for unequal class sizes.

🎯 Exam Tip: When constructing a histogram for unequal class intervals, always calculate and use proportionate frequencies. The modal class is the one with the highest proportionate frequency, and the mode is found by interpolation within that class from the graph.

Question 6. A producer believes that the mode of his daily production is 70. The distribution of production from the data obtained after making some changes in the design of the produced units is as follows :

No. of units	60-64	65-69	70-74	75-79	80-84	85-89	90-94
No. of days	5	7	10	8	5	3	2

Is there any change in the mode of the number of items produced?
Answer:The producer initially believes the mode of daily production is 70 units. We need to find the mode of the new production distribution after design changes and compare it. The given data is in inclusive form, so we convert it to exclusive form by adjusting the class boundaries (subtract 0.5 from the lower limit and add 0.5 to the upper limit).
The modal class is the class with the maximum frequency. From the table, the maximum frequency is 10, which corresponds to the 70-74 class.
In exclusive form, the modal class is 69.5 - 74.5.
Using the formula for mode for grouped data:
\( \text{Mode} = L + \frac{f_m - f_1}{2f_m - f_1 - f_2} \times c \)
Here, L (lower boundary of modal class) = 69.5
\( f_m \) (frequency of modal class) = 10
\( f_1 \) (frequency of the class preceding modal class) = 7 (for 65-69 class)
\( f_2 \) (frequency of the class succeeding modal class) = 8 (for 75-79 class)
c (class width) = 5 (e.g., 74.5 - 69.5 = 5)
Substituting these values into the formula:
\( \text{Mode} = 69.5 + \frac{10 - 7}{2(10) - 7 - 8} \times 5 \)
\( = 69.5 + \frac{3}{20 - 15} \times 5 \)
\( = 69.5 + \frac{3}{5} \times 5 \)
\( = 69.5 + 3 \)
\( \text{Mode} = 72.5 \) units.
The new mode of production is 72.5 units.
Since the initial belief was a mode of 70 units and the calculated mode after changes is 72.5 units, there is a change in the mode of the number of items produced. The mode has increased.
In simple words: The production design changes have shifted the most frequent production quantity from 70 units to 72.5 units, indicating an increase in typical daily output.

🎯 Exam Tip: When working with inclusive class intervals, always convert them to exclusive intervals before applying formulas for mode or median to ensure accuracy. Carefully identify \( f_m, f_1, f_2 \), L, and c.

Question 7. The data for sales of oil tins of two companies sold in a shop are as follows, which show the sales of 40 days:

No. of oil tins	2-5	6-9	10-13	14-17	18-21	22-25
No. of days Company X	1	3	17	9	6	4
No. of days Company Y	5	9	20	3	2	1

If median is used to compare the sales, which company can be said to have higher sale?
Answer:To compare the sales of Company X and Company Y using the median, we first need to create cumulative frequency tables for both companies and then calculate their respective medians. The classes are inclusive, so we'll adjust boundaries for calculation. The class width is 4 (e.g., 5.5 - 1.5). **For Company X:**

No. of oil tins (Exclusive)	No. of days f	Cumulative frequency cf
1.5-5.5	1	1
5.5-9.5	3	4
9.5-13.5	17	21
13.5-17.5	9	30
17.5-21.5	6	36
21.5-25.5	4	40
Total	n = 40

**Median for Company X:**
Position of Median (\( \frac{n}{2} \)) = \( \frac{40}{2} = 20 \)th observation.
Referring to the cf column, the 20th observation falls within the 9.5-13.5 class.
Therefore, the Median class for Company X is 9.5-13.5.
Using the formula for Median for grouped data:
\( M_X = L + \frac{\left(\frac{n}{2}\right)-cf}{f} \times c \)
Here, L = 9.5, \( \frac{n}{2} = 20 \), cf = 4, f = 17, c = 4.
\( M_X = 9.5 + \frac{20 - 4}{17} \times 4 \)
\( = 9.5 + \frac{16}{17} \times 4 \)
\( = 9.5 + \frac{64}{17} \)
\( = 9.5 + 3.764... \)
\( M_X \approx 13.26 \) tins.
**For Company Y:**

No. of oil tins (Exclusive)	No. of days f	Cumulative frequency cf
1.5-5.5	5	5
5.5-9.5	9	14
9.5-13.5	20	34
13.5-17.5	3	37
17.5-21.5	2	39
21.5-25.5	1	40
Total	n = 40

**Median for Company Y:**
Position of Median (\( \frac{n}{2} \)) = \( \frac{40}{2} = 20 \)th observation.
Referring to the cf column, the 20th observation falls within the 9.5-13.5 class.
Therefore, the Median class for Company Y is 9.5-13.5.
Using the formula for Median for grouped data:
\( M_Y = L + \frac{\left(\frac{n}{2}\right)-cf}{f} \times c \)
Here, L = 9.5, \( \frac{n}{2} = 20 \), cf = 14, f = 20, c = 4.
\( M_Y = 9.5 + \frac{20 - 14}{20} \times 4 \)
\( = 9.5 + \frac{6}{20} \times 4 \)
\( = 9.5 + \frac{24}{20} \)
\( = 9.5 + 1.2 \)
\( M_Y = 10.7 \) tins.
**Comparison:**
Median sales for Company X = 13.26 tins
Median sales for Company Y = 10.7 tins
Since the median sales of Company X (13.26 tins) are higher than that of Company Y (10.7 tins), Company X can be said to have higher sales.
In simple words: Company X generally sells more oil tins per day than Company Y, as indicated by its higher median sales figure.

🎯 Exam Tip: When comparing distributions using median for grouped data, ensure consistency in converting inclusive to exclusive classes and accurately identifying the median class and its parameters (L, cf, f, c).

Question 8. The distribution of age (in complete years) at the time of marriage of 50 married men is as follows:

Age (years)	24-26	27-29	30-32	33-35
No. of males f	21	15	6	2

Find the mode of their age at the time of marriage using graphical method.
Answer:To find the mode graphically, we first convert the inclusive continuous frequency distribution into an exclusive form.
The conversion involves subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class.

Age (years) (Exclusive Form)	No. of males f
23.5-26.5	21
26.5-29.5	15
29.5-32.5	6
32.5-35.5	2

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक हिस्टोग्राम है जो विवाह के समय पुरुषों की आयु (वर्षों में) और उनकी संख्या को दर्शाता है। X-अक्ष पर आयु वर्ग (वर्ष) और Y-अक्ष पर पुरुषों की संख्या (आवृत्ति) है। हिस्टोग्राम के दंडों की ऊँचाई आवृत्ति के समानुपातिक है, और सबसे ऊँचे दंड के शीर्ष से खींची गई तिरछी रेखाओं के प्रतिच्छेद बिंदु से X-अक्ष पर एक लंब खींचकर मोड का मान 25.5 वर्ष निर्धारित किया गया है।
From the histogram (drawn using the exclusive class intervals and frequencies), the mode (OQ) is determined by finding the modal class (the class with the highest frequency, 23.5-26.5 years), drawing lines from the top corners of this modal rectangle to the adjacent inner corners of the neighboring rectangles, and then dropping a perpendicular from the intersection point of these lines to the X-axis.
From the histogram, the mode (OQ) is 25.5 years.
Hence, the mode of the age at marriage is 25.5 years.
In simple words: The most common age for men to marry in this dataset is 25.5 years. This was found by visualizing the data in a histogram and identifying the peak frequency.

🎯 Exam Tip: For graphical determination of mode in continuous data, always ensure class intervals are converted to exclusive form. The mode is found by interpolation within the modal class by connecting diagonal lines from the modal bar to adjacent bars on the histogram.