RBSE Solutions Class 11 Economics Chapter 9 Median

RBSE Class 11 Economics Chapter 9 Text Book Questions

RBSE Class 11 Economics Chapter 9 Multiple Choice Questions

 

Question 1. For those facts that cannot be expressed in numbers, the best mean is:
(a) Arithmetic Mean
(b) Median
(c) Mode
(d) Harmonic Mean
Answer: (b) Median
In simple words: When we have information that cannot be counted or measured with numbers, the median is the most suitable way to find the middle value. It works well for things like preferences or qualities.

🎯 Exam Tip: Remember that median is a positional average, which makes it ideal for qualitative data or situations where extreme values might skew other averages.

 

Question 3. The value dividing a series in four equal parts is called:
(a) Average
(b) Median
(c) Quartile
(d) Pentant
Answer: (c) Quartile
In simple words: When a list of numbers is divided into four equal parts, each dividing point is called a quartile. These points help us understand how the data is spread out.

🎯 Exam Tip: Don't confuse quartiles with median; the median divides data into two equal halves, while quartiles divide it into four equal segments.

 

Question 4. The second quadrant of a series is called:
(a) Lower quadrant
(b) Higher quadrant
(c) Mean
(d) Median
Answer: (d) Median
In simple words: The second quartile is the same as the median. It is the middle value that splits the data into two equal halves.

🎯 Exam Tip: The second quartile (Q2) is always equal to the median, as both represent the 50th percentile of a dataset.

 

Question 5. If the mode is 18 and arithmetic mean is 20, then the median will be:
(a) 29.33
(b) 19.33
(c) 18.66
(d) 9.33
Answer: (b) 19.33
In simple words: Using the empirical relationship between mean, median, and mode (Mode \( \approx \) 3 Median - 2 Mean), if the mode is 18 and the mean is 20, the median will be approximately 19.33. This formula is useful for moderately skewed distributions.

🎯 Exam Tip: Always remember the empirical relationship: Mode \( \approx \) 3 Median - 2 Mean. This formula is a quick way to estimate one measure of central tendency if the other two are known.

RBSE Class 11 Economics Chapter 9 Very Short Answer Type Questions

 

Question 2. Write the formula for determining median, if the number of terms in an individual series is even.
Answer: If the number of items in a series is even, the median is not a single value. To find the median, you take the average of the two middle serial numbers. The formula for median (M) in an even individual series is:
\[ M = \frac { \text{Value of } \left( \frac { N }{ 2 } \right)^{th} \text{ item } + \text{ Value of } \left( \frac { N }{ 2 } + 1 \right)^{th} \text{ item } }{ 2 } \]
In simple words: If you have an even number of items, find the two items exactly in the middle. Add them up and then divide by two. That average is your median.

🎯 Exam Tip: For an even number of observations, the median is the average of the two middle values, whereas for an odd number, it's simply the middle value.

 

Question 3. When is the use of median the most suitable?
Answer: The median is most suitable when the data involves qualitative facts. This means for information that describes qualities or categories rather than numbers, such as opinions, preferences, or different ranks. The median works well because it focuses on the middle position, which is still meaningful even without exact numerical values. For example, when asking people to rate something as "good," "average," or "bad," the median can show the most common central rating. This is also why it's suitable for open-ended class intervals, where the exact upper or lower limits are unknown.
In simple words: Median is best when you are dealing with qualities or categories, not just numbers. It helps find the middle point even when numbers are not clear, like in surveys.

🎯 Exam Tip: The median is particularly useful for qualitative data or when a distribution has open-ended classes, as it is less affected by extreme values compared to the mean.

 

Question 4. Which means is suitable for open-ended class- intervals?
Answer: Median
In simple words: The median is the best choice for data that has open-ended parts, where you don't know the exact highest or lowest values. It still helps find the middle point without needing those exact numbers.

🎯 Exam Tip: The median is preferred for open-ended distributions because its calculation does not depend on the exact values of the extreme classes.

 

Question 5. What do you understand by partition values?
Answer: Partition values are specific points that divide a data series into many equal parts. These values help to break down a dataset so it can be analyzed more easily. For instance, a series can be split into four equal parts (called quartiles), five equal parts (quintiles), eight equal parts (octiles), ten equal parts (deciles), or even one hundred equal parts (percentiles). The median itself is a type of partition value, specifically one that divides the series into two equal halves. Understanding partition values helps in grasping the spread and distribution of data.
In simple words: Partition values are like markers that cut a list of numbers into several equal sections. They help you see how the data is spread out across different parts.

🎯 Exam Tip: Common partition values include quartiles (dividing into 4 parts), deciles (10 parts), and percentiles (100 parts), all of which help describe data distribution.

RBSE Class 11 Economics Chapter 9 Short Answer Type Questions

 

Question 1. If the median of four observations 3,4, C and 8 is 5, Find the value of C.
Answer: First, arrange the given observations in ascending order: 3, 4, C, 8. Since there are four observations (an even number), the median is the average of the two middle terms. The two middle terms are 4 and C. The median is given as 5. Therefore, we can write the equation: \( M = \frac { 4 + C }{ 2 } \)
\( \implies 5 = \frac { 4 + C }{ 2 } \)
\( \implies 5 \times 2 = 4 + C \)
\( \implies 10 = 4 + C \)
\( \implies C = 10 - 4 \)
\( \implies C = 6 \) So, the value of C is 6. This helps complete the ordered series and confirms the median.
In simple words: If you have four numbers and the median is 5, and the numbers are 3, 4, C, and 8, then C must be 6. This is because the median for an even set of numbers is the average of the two middle ones.

🎯 Exam Tip: When finding the median for an even number of observations, always average the two middle values after arranging them in order. For an odd number, it's just the middle value itself.

 

Question 2. In order of find the median in a discrete series, (N + 1)/2 is used, while in continuous series, (N/2) is used, Why?
Answer: The way we find the median position differs between discrete and continuous series because of how their data is structured.
**For Discrete Series:** We use \( \frac { N+1 }{ 2 } \) to find the median item. This is because discrete series deal with individual, distinct values. Adding 1 helps to pinpoint the exact item's position, especially if N is even, by effectively creating a "middle" position between two values. For example, if N=4, \( \frac{4+1}{2} = 2.5 \), meaning the median is between the 2nd and 3rd items.
**For Continuous Series:** We use \( \frac { N }{ 2 } \) to identify the median class. Continuous series represent data spread across intervals, not individual points. When we divide N by 2, we are looking for the point that divides the *total frequency* into two equal halves. This point helps us locate the class interval where the median lies. Once the median class is identified, a specific interpolation formula is used to calculate the exact median value within that class. The median is found by looking at the cumulative frequency distribution.
Therefore, the slight difference in formulas accounts for whether we are locating a specific item in a list or a class interval in a range.
In simple words: For simple lists of items (discrete series), we add 1 to N before dividing by 2 to find the exact middle spot. For data grouped into ranges (continuous series), we just divide N by 2 to find which range the middle value falls into, because values are spread out, not distinct.

🎯 Exam Tip: Remember that \( \frac { N+1 }{ 2 } \) is for finding the median's *position* in individual or discrete series, while \( \frac { N }{ 2 } \) is for finding the *median class* in a continuous series before applying the interpolation formula.

 

Question 3. If the arithmetic mean is 75 and mode is 60, find the value of median.
Answer: To find the median when the mode and arithmetic mean are known, we use the empirical formula relating these three measures of central tendency:
\( \text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean} \) Given values: Mode (Z) = 60 Arithmetic Mean \( (\overline{X}) \) = 75 Let Median = M Substitute the values into the formula:
\( 60 = 3M - (2 \times 75) \)
\( 60 = 3M - 150 \) Now, add 150 to both sides of the equation to isolate 3M:
\( 60 + 150 = 3M \)
\( 210 = 3M \) Finally, divide by 3 to find M:
\( M = \frac { 210 }{ 3 } \)
\( M = 70 \) So, the value of the median is 70. This relationship is a useful approximation for moderately skewed distributions.
In simple words: We use a special formula that links mode, median, and mean. If the mode is 60 and the mean is 75, we can work it out. The median will be 70.

🎯 Exam Tip: The empirical relationship (Mode = 3 Median - 2 Mean) is valid for distributions that are moderately asymmetrical; ensure you apply it correctly by substituting the given values.

 

Question 4. Mention any four advantages of median.
Answer: Here are four advantages of using the median:
1. **Ease of Understanding and Calculation:** The median is very easy to understand and calculate, even by simple inspection in many cases. It simply represents the middle value in an ordered dataset, making it intuitive.
2. **Unaffected by Extreme Values:** Unlike the arithmetic mean, the median is not influenced by extremely large or small values (outliers). This makes it a more reliable measure of central tendency for skewed distributions.
3. **Suitable for Qualitative Data:** The median can be used for qualitative data (data that describes qualities or categories), especially when ranks are given or items are not counted numerically. It is also suitable for open-ended class intervals.
4. **Graphical Determination:** The median can be determined graphically using an ogive (cumulative frequency curve). This visual method provides a clear understanding of its position within the data. It also directly represents the value of the middle item in the distribution.
In simple words: The median is good because it's easy to find, it doesn't get messed up by very big or very small numbers, it works for descriptive information, and you can even find it using a graph.

🎯 Exam Tip: Highlight that the median's robustness to extreme values makes it superior to the mean for skewed data, and its applicability to qualitative data sets it apart from other averages.

 

Question 5. Write the formula for finding the first quadrant(Q₁) and third quadrant (Q3) in a continuous series.
Answer: To find the first quartile (Q₁) and third quartile (Q3) in a continuous series, we first need to identify the quartile class for each. This is done by calculating \( \frac{N}{4} \) for Q₁ and \( \frac{3N}{4} \) for Q3, where N is the total frequency.
Once the respective quartile classes are found using cumulative frequency, the formulas for calculation are:
**Formula for First Quartile (Q₁):** \[ Q_1 = l_1 + \left( \frac { \frac { N }{ 4 } - c }{ f } \right) \times i \] **Formula for Third Quartile (Q₃):** \[ Q_3 = l_1 + \left( \frac { \frac { 3N }{ 4 } - c }{ f } \right) \times i \] Where: \( l_1 \) = lower limit of the quartile class (the class containing Q₁ or Q₃) \( \frac { N }{ 4 } \) = position of the first quartile \( \frac { 3N }{ 4 } \) = position of the third quartile \( c \) = cumulative frequency of the class preceding the quartile class \( f \) = frequency of the quartile class \( i \) = class interval (magnitude) of the quartile class These formulas help to precisely locate the quartile values within their respective class intervals.
In simple words: To find Q1 and Q3 in data grouped into ranges, you first find which range they fall into using N/4 and 3N/4. Then, you use a special formula with the lower limit of that range, its frequency, and the cumulative frequency before it, along with the class size, to get the exact value.

🎯 Exam Tip: Clearly define all variables in your formula and ensure you correctly identify the 'c' (cumulative frequency of the *preceding* class) to avoid common calculation errors.

RBSE Class 11 Economics Chapter 9 Long Answer Type Questions

 

Question 1. Find the mode and median from the following table :
Class Interval
0-10
10-20
20-30
30-40
40-50
Frequency
10
3
7
15
5
Answer: To find the mode and median from the given data, we first need to calculate the cumulative frequency (cf) and identify the modal class.

**Calculation for Median:** Total frequency (N) = 40 Median class position \( = \frac { N }{ 2 } = \frac { 40 }{ 2 } = 20^{th} \) item. The 20th item falls in the cumulative frequency of 20, which corresponds to the class interval 20-30. So, the median class is 20-30.
From the median class 20-30: Lower limit \( l_1 = 20 \) Frequency of median class \( f = 7 \) Cumulative frequency of preceding class \( c = 13 \) (cf of 10-20 class) Class interval \( i = 10 \) (30 - 20) Formula for Median: \( M = l_1 + \left( \frac { \frac { N }{ 2 } - c }{ f } \right) \times i \)
\( M = 20 + \left( \frac { 20 - 13 }{ 7 } \right) \times 10 \)
\( M = 20 + \left( \frac { 7 }{ 7 } \right) \times 10 \)
\( M = 20 + 1 \times 10 \)
\( M = 20 + 10 \)
\( M = 30 \) Therefore, the Median (M) = 30. **Calculation for Mode:** The modal class is the class with the highest frequency. From the table, the highest frequency is 15, which corresponds to the class interval 30-40. So, the modal class is 30-40.
From the modal class 30-40: Lower limit \( l_1 = 30 \) Frequency of modal class \( f_1 = 15 \) Frequency of preceding class \( f_0 = 7 \) (frequency of 20-30 class) Frequency of succeeding class \( f_2 = 5 \) (frequency of 40-50 class) Class interval \( i = 10 \) (40 - 30) Formula for Mode: \( \text{Mode} = l_1 + \left( \frac { f_1 - f_0 }{ 2f_1 - f_0 - f_2 } \right) \times i \)
\( \text{Mode} = 30 + \left( \frac { 15 - 7 }{ (2 \times 15) - 7 - 5 } \right) \times 10 \)
\( \text{Mode} = 30 + \left( \frac { 8 }{ 30 - 7 - 5 } \right) \times 10 \)
\( \text{Mode} = 30 + \left( \frac { 8 }{ 18 } \right) \times 10 \)
\( \text{Mode} = 30 + \frac { 80 }{ 18 } \)
\( \text{Mode} = 30 + 4.44 \)
\( \text{Mode} = 34.44 \) Thus, the Mode is 34.44. The median and mode offer different insights into the central tendency of the data.
In simple words: First, we add up the frequencies to get cumulative frequencies. For the median, we find the middle point (20th item) which falls in the 20-30 class and calculate it to be 30. For the mode, we find the class with the most frequency (30-40 class) and calculate it to be 34.44.

🎯 Exam Tip: Always construct a cumulative frequency column to find the median class, and identify the highest frequency to determine the modal class correctly before applying the formulas.

 

Question 2. In the following data item series, calculate the first quadrant (Q₁), third quadrant (Q3) and median (M) :
Size
11-15
16-20
21-25
26-30
31-35
36-40
Frequency
7
10
13
26
35
22
Answer: To calculate Q₁, Q₃, and M, we first need to create a cumulative frequency (cf) table.

Total frequency N = 129. **Calculation for First Quartile (Q₁):** Position of \( Q_1 = \frac { N }{ 4 } = \frac { 129 }{ 4 } = 32.25^{th} \) item. The 32.25th item falls in the cf of 56, so the \( Q_1 \) class is 26-30.
From the \( Q_1 \) class 26-30: \( l_1 = 25.5 \) (lower limit, adjusted for continuous series: 26 - 0.5) \( f = 26 \) \( c = 30 \) (cf of preceding class 21-25) \( i = 5 \) (class interval: 30-26 = 4, but 25.5-20.5 = 5. Assume consistent interval from table rows) \( Q_1 = l_1 + \left( \frac { \frac { N }{ 4 } - c }{ f } \right) \times i \)
\( Q_1 = 25.5 + \left( \frac { 32.25 - 30 }{ 26 } \right) \times 5 \)
\( Q_1 = 25.5 + \left( \frac { 2.25 }{ 26 } \right) \times 5 \)
\( Q_1 = 25.5 + \frac { 11.25 }{ 26 } \)
\( Q_1 = 25.5 + 0.43 \)
\( Q_1 = 25.93 \) **Calculation for Third Quartile (Q₃):** Position of \( Q_3 = \frac { 3N }{ 4 } = \frac { 3 \times 129 }{ 4 } = \frac { 387 }{ 4 } = 96.75^{th} \) item. The 96.75th item falls in the cf of 113, so the \( Q_3 \) class is 36-40.
From the \( Q_3 \) class 36-40: \( l_1 = 35.5 \) (lower limit, adjusted: 36 - 0.5) \( f = 22 \) \( c = 91 \) (cf of preceding class 31-35) \( i = 5 \) \( Q_3 = l_1 + \left( \frac { \frac { 3N }{ 4 } - c }{ f } \right) \times i \)
\( Q_3 = 35.5 + \left( \frac { 96.75 - 91 }{ 22 } \right) \times 5 \)
\( Q_3 = 35.5 + \left( \frac { 5.75 }{ 22 } \right) \times 5 \)
\( Q_3 = 35.5 + \frac { 28.75 }{ 22 } \)
\( Q_3 = 35.5 + 1.31 \)
\( Q_3 = 36.81 \) **Calculation for Median (M):** Position of Median \( = \frac { N }{ 2 } = \frac { 129 }{ 2 } = 64.5^{th} \) item. The 64.5th item falls in the cf of 91, so the Median class is 31-35.
From the Median class 31-35: \( l_1 = 30.5 \) (lower limit, adjusted: 31 - 0.5) \( f = 35 \) \( c = 56 \) (cf of preceding class 26-30) \( i = 5 \) \( M = l_1 + \left( \frac { \frac { N }{ 2 } - c }{ f } \right) \times i \)
\( M = 30.5 + \left( \frac { 64.5 - 56 }{ 35 } \right) \times 5 \)
\( M = 30.5 + \left( \frac { 8.5 }{ 35 } \right) \times 5 \)
\( M = 30.5 + \frac { 42.5 }{ 35 } \)
\( M = 30.5 + 1.21 \)
\( M = 31.71 \) The calculations show Q₁ = 25.93, Q₃ = 36.81, and M = 31.71. These values help in understanding the distribution of sizes in the series.
In simple words: We first make a cumulative frequency table. Then, for Q1, Q3, and Median, we find their positions by dividing N by 4, 3N by 4, and N by 2 respectively. Using these positions, we find their class intervals and then use a formula to calculate their exact values: Q1 is 25.93, Q3 is 36.81, and the Median is 31.71.

🎯 Exam Tip: Remember to adjust class limits for continuous series (e.g., 15 to 16 becomes 15.5) if the classes are inclusive to ensure accurate calculations for quartiles and median.

Arithmetic Mean:

The arithmetic mean is a very popular and important type of average among mathematical means, commonly used in daily life. It is calculated by dividing the total sum of all values in a series by the total number of items in that series. For example, if you want to find the average height of students, you add all their heights and divide by the number of students. This measure gives a sense of the central value of a dataset.

According to W.I. King, the arithmetic average can be described as the total sum of items in a series divided by their count.

H. Secrist further explains that the arithmetic mean is the amount obtained by dividing the sum of values of items in a series by their number.

Therefore, the arithmetic mean is found by adding up all the values in a general category and then dividing by the total number of values.

For example, if the monthly incomes of 5 families are Rs 2000, Rs 3000, Rs 4000, Rs 5000, and Rs 6000, to find their average income, all these incomes are added together. The total income is Rs 20000. This total income is then divided by the total number of families, which is 5. The average monthly income, or the arithmetic mean, will be Rs 4000. The arithmetic mean is commonly used and gives a clear picture of the typical value in a dataset.

There are two types of arithmetic mean:

1. Simple Arithmetic Mean
2. Weighted Arithmetic Mean

Merits of Arithmetic Mean:

Here are the benefits of using the arithmetic mean:

• **Easy to compute and understand:** It is the simplest average to understand and calculate. Even someone without advanced mathematical knowledge can easily grasp it.
• **Based on all items of the series:** The arithmetic mean considers every single item in the series when it's calculated. This makes it a good representative of the entire dataset because no information is left out.
• **Stable in comparison to other averages:** The mean is generally quite stable. If you take multiple samples from the same group, the mean won't change drastically compared to other types of statistical descriptions.
• **Suitable for algebraic treatment:** Because it is determined by a strict mathematical formula, the arithmetic mean can be used easily in further algebraic calculations, making it useful in more complex statistical analyses.
• **No need for arranging data:** You do not need to arrange the data values in any specific order (like ascending or descending) to calculate the arithmetic mean. You can simply sum the values and divide.

Demerits of Arithmetic Mean:

Here are the disadvantages of using the arithmetic mean:

• **Effect of extreme values:** Since the arithmetic mean takes every single item into account, very small or very large values (outliers) can unfairly affect the average. This can make the average figure appear misleading.
• **Unrealistic representation:** Sometimes, the calculated arithmetic mean might be a value that does not actually exist in the data series and might seem unrealistic or impossible in a practical sense.
• **Graphical representation not possible:** The arithmetic mean cannot be found using graphical methods, unlike the median and mode.
• **Calculation difficulties:** Compared to positional averages, calculating the arithmetic mean can be harder. It cannot be found by simply looking at the data, and if even one value is missing, it cannot be calculated because it requires all items. It is also not suitable for qualitative facts (non-numerical data).
• **Fallacious conclusions:** Sometimes, using the arithmetic mean can lead to conclusions that are misleading or inconsistent, especially when the data distribution is very uneven.
• **Not suitable for certain studies:** It is not always the best measure for studying rates, ratios, and percentages, where other types of averages might be more appropriate.

Mode:

The mode is an important measure of central tendency. It is simply the value that appears most often in a series. This means the number or item with the highest frequency is considered the 'mode'. For instance, if most people wear a shoe size '7', then '7' is the mode for shoe sizes in that group. The mode helps quickly identify the most common item or value in a dataset.

Merits of Mode:

Here are the benefits of using the mode:

• **Simple and Popular:** The mode is straightforward and popular. In some situations, it can be determined by just looking at the data. It is often used in daily life, such as finding the most common size for clothes or shoes.
• **Best Representative:** The mode is the value that occurs most frequently in a series, making it the best representative of what is most common. Its value is directly taken from the series itself.
• **Minimum impact of extreme values:** A key feature of the mode is that it is not affected by very large or very small values in the dataset (extreme values). This is different from the arithmetic mean, which can be heavily influenced by outliers.
• **Determination by graphical method:** The mode can also be found graphically, usually with the help of a histogram (a rectangular diagram). This visual method provides another way to understand the most frequent value.
• **Possible to find the mode of qualitative facts:** The mode can be found for qualitative data, which includes facts that can be classified and graded, such as favorite colors or types of items.
• **Unaffected by deviations:** The mode is not affected by changes or deviations in class intervals. This means its value remains stable even if the class boundaries are adjusted.
• **Calculation of all frequencies not required:** You do not need all the frequencies to calculate the mode. Often, just the frequencies of the modal class, its preceding class, and its succeeding class are enough for the calculation.

Demerits of Mode:

Here are the disadvantages of using the mode:

• **Uncertain and ambiguous:** The mode can often be unclear or confusing. It's hard to determine when all observations appear the same number of times. Also, sometimes a series can have more than one mode (bimodal or multimodal), which makes it less clear.
• **Lack of algebraic treatment:** The mode is not suitable for further algebraic calculations. For instance, you cannot combine the modes of two different datasets to find an overall mode for the combined data.
• **Complexity in the computation process:** While simple by inspection, if the mode needs to be calculated by grouping or interpolation for more complex data, it can become very difficult for ordinary people to understand and compute.
• **Illusory mean:** In some cases, the mode might not accurately represent the series. For example, if the highest frequency is for a very small or very large value, it might not be a good central indicator for the rest of the data.

One drawback of mode is that when the size of the classes changes, its value can also change.

Median:

The median is the middle value in a dataset that has been arranged in order. It divides the ordered data into two equal parts: one part contains all values greater than the median, and the other part contains all values lesser than the median. Essentially, it's the 50th percentile of the data. The median helps in understanding the central position without being affected by extreme values.

Following are the four advantages of median:

1. It is especially useful for open-ended classes, because only the position of items is needed, not their exact values. The median is also suggested if the data has unequal classes, as it is easier to calculate than the mean.
2. The median is not affected by very large or very small values (extreme deviations), making it a stable measure.
3. It is the most appropriate average for qualitative data, such as ranks or other types of items that are scored but not directly counted or measured.
4. Perhaps its biggest benefit is that the median truly indicates what many people mistakenly think the arithmetic mean represents—the absolute middle value.
5. The median shows the value of the middle item in the distribution. This is a clear and simple meaning, and it makes the median an easy-to-explain measure.

Following are the disadvantages of median:

• **Lack of Representation:** The median may not truly represent the average of a group if there are big differences in the values of different items.
• **No algebraic treatment:** It cannot be used in algebraic calculations. For example, you cannot combine the medians of two or more groups to find a combined median, which is possible with the mean.
• **Sorting class problem:** To find the median, you must arrange the data in either ascending (smallest to largest) or descending (largest to smallest) order. This process can take a lot of time for large datasets.
• **Unrealistic:** If the median falls between two actual values, it represents a possible value rather than a real, existing one in the dataset.

RBSE Class 11 Economics Chapter 9 Other Important Questions

RBSE Class 11 Economics Chapter 9 Objective Type Questions

 

Question 1. Which one is the appropriate measurement for qualitative measurement-
(a) Arithmetic mean
(b) Median
(c) Mode
(d) Geometric Mean
Answer: (b) Median
In simple words: The median is the most suitable measure for qualitative data, which describes qualities rather than numbers. It helps find the central point of such information.

🎯 Exam Tip: Remember that median, being a positional average, is highly suitable for qualitative data or data where exact numerical values are not available.

 

Question 2. The median value of the following item-value is
(a) 16
(b) 24
(c) 15
(d) 13
Answer: (a) 16
In simple words: The median value, which is the middle number in an ordered list, is 16. To find this, you would usually list all the given numbers from smallest to largest and pick the one in the very middle.

🎯 Exam Tip: Always ensure you have the complete data series to calculate the median accurately; otherwise, it must be provided directly.

 

Question 3. In how many parts does the median divide the data series?
(a) 2
(b) 4
(c) 10
(d) None of the options
Answer: (a) 2
In simple words: The median splits a list of numbers right down the middle into two equal halves. One half has numbers smaller than the median, and the other half has numbers larger than it.

🎯 Exam Tip: Understand that the fundamental role of the median is to divide a dataset into two equal parts, making it the 50th percentile.

 

Question 5. Which one of the following is a place-related mean?
(a) Median
(b) Arithmetic Mean
(c) Geometric Mean
(d) Harmonic Mean
Answer: (a) Median
In simple words: The median is called a "place-related mean" because it depends on its position in the data, rather than the actual numerical values of all the items. It finds the middle spot.

🎯 Exam Tip: Positional averages like the median and mode are less affected by extreme values compared to mathematical averages such as the arithmetic, geometric, and harmonic means.

 

Question 6. Median in continuous series is
(a) Value of \( \left[ \frac { N+1 }{ 2 } \right] \) th item
(b) Value of \( \left[ \frac { N }{ 2 } \right] \) th item
(c) \( \left[ \frac {N}{4}\ \) th item's value
(d) None of the options
Answer: (b) Value of \( \left[ \frac { N }{ 2 } \right] \) th item
In simple words: In a continuous series (where data is grouped into ranges), the median is found by first locating the median class using the \( \frac { N }{ 2 } \) item. This tells you which group the middle value falls into, before a more precise calculation.

🎯 Exam Tip: For continuous series, \( \frac { N }{ 2 } \) helps identify the *median class*, not the median itself; further calculation is needed to find the exact median value within that class.

 

Question 7. Low quartile (Q₁) in continuous series is
(a) Value of \( \left[ \frac { N+1 }{4} \right] \) th item
(b) Value of \( \left[ \frac { N }{4} \right] \) th item
(c) \( \left[ \frac {N}{2} \right] \) th item's Value
(d) None of the options
Answer: (b) Value of \( \left[ \frac { N }{4} \right] \) th item
In simple words: For data grouped in ranges (continuous series), the first quartile (Q₁) is found by looking for the \( \frac { N }{ 4 } \) th item. This helps find the class where the first quarter of the data lies.

🎯 Exam Tip: Remember that for continuous series, \( \frac { N }{ 4 } \) indicates the position of the first quartile class, which is then used in a formula to find the exact Q₁ value.

RBSE Class 11 Economics Chapter 9 Very Short Answer Type Questions

 

Question 1. What is meant by median?
Answer: The median is a middle value in a list of numbers. When you put all the numbers in order from smallest to largest, the median is the number right in the middle. Half the numbers are smaller than it, and half are bigger. This helps us find the center point of data.
In simple words: The median is the middle value in an ordered list of numbers, with half the values above it and half below.

🎯 Exam Tip: Define median clearly and mention its role in dividing data into two equal halves.

 

Question 2. What is ascending order?
Answer: Ascending order means arranging numbers or items from the smallest to the largest. It is like climbing stairs, going up one step at a time.
In simple words: Ascending order means listing numbers from the smallest to the largest.

🎯 Exam Tip: Always give an example with numbers or letters to illustrate ascending order.

 

Question 3. What is descending order?
Answer: Descending order means arranging numbers or items from the largest to the smallest. It is like coming down stairs, going down one step at a time.
In simple words: Descending order means listing numbers from the largest to the smallest.

🎯 Exam Tip: Ensure your example clearly shows numbers going from big to small.

 

Question 4. Write down the formula of median in individual series.
Answer: The formula to find the median in a simple list of numbers is to first arrange them in order. Then, you find the value of the item at the position calculated by adding 1 to the total number of items (\(N\)) and dividing by 2.
`\( M = \text{Value of } \left[ \frac { N+1}{2} \right] \text{ th item} \)`
This formula helps locate the exact middle value.
In simple words: First, order the numbers. The median is the value at the position \((N+1)/2\).

🎯 Exam Tip: Remember to state that the data must first be arranged in ascending or descending order before applying this formula.

 

Question 5. Mention one characteristic of median.
Answer: One good thing about the median is that very high or very low numbers, which are far from the others, do not affect it much. This makes it a stable measure.
In simple words: The median is not much affected by extremely high or low values in the data.

🎯 Exam Tip: Point out that the median's resistance to extreme values makes it robust for skewed data.

 

Question 7. What kind of 'quartile' is Q₁?
Answer: \(Q_1\) is called the lower quartile. It represents the value below which 25% of the data falls.
In simple words: \(Q_1\) is the lower quartile, meaning 25% of the data is below this value.

🎯 Exam Tip: Remember that \(Q_1\) marks the first quarter of the data.

 

Question 8. What is Q2 called?
Answer: \(Q_2\) is known as the second quartile. It is also the same as the median, dividing the data into two equal halves.
In simple words: \(Q_2\) is the second quartile, which is the same as the median.

🎯 Exam Tip: Highlight that the median and the second quartile are the same concept, showing the exact middle of the data.

 

Question 9. What do you call the 4 equal parts of a series?
Answer: The four equal parts into which a data series can be divided are called quartiles. These divisions help in understanding data spread.
In simple words: The four equal parts of a data series are called quartiles.

🎯 Exam Tip: Explain that quartiles divide data into quarters, making it easy to see how data is distributed.

 

Question 10. In the analysis of social problems, which means is used?
Answer: When studying social problems, the median is often used to find the middle value. This is because median is not much affected by very high or very low numbers, which can be common in social data.
In simple words: The median is often used in social problems because extreme values do not affect it much.

🎯 Exam Tip: Mention that median is preferred in social studies because it provides a more realistic 'average' when data might have extreme outliers, unlike the mean.

 

Question 11. Write the formula of median in continuous series.
Answer: For a continuous series, the median is found using a specific formula. It involves the lower limit of the median class (\(L_1\)), the total frequency (\(N\)), the cumulative frequency of the class before the median class (\(c\)), the frequency of the median class (\(f\)), and the class interval (\(i\)). This formula helps to pinpoint the median within a range.
`\( M = L_1 + \frac{(\frac{N}{2} - c)}{f} \times i \)`
In simple words: The median for a continuous series is found using a formula that includes the lower limit of the median class, total frequency, cumulative frequency before the median class, median class frequency, and class interval.

🎯 Exam Tip: Ensure all terms in the formula (\(L_1\), \(N\), \(c\), \(f\), \(i\)) are correctly identified from the given data to avoid errors.

 

Question 12. Write 3, 1,7, 5, 4, 2, 8 in ascending order.
Answer: To arrange these numbers in ascending order means to list them from the smallest to the largest. The correct order is 1, 2, 3, 4, 5, 7, 8.
In simple words: Arranging the numbers from smallest to largest gives: 1, 2, 3, 4, 5, 7, 8.

🎯 Exam Tip: Double-check that all numbers from the original list are included and correctly ordered.

 

RBSE Class 11 Economics Chapter 9 Short Answer Type Questions

 

Question 1. State the four flaws of median.
Answer: The four flaws of median are:
1. The median might not show a good average for a group if the numbers in that group are very different from each other.
2. You cannot use the median for certain math calculations, unlike the mean. For example, you cannot easily combine medians from two different groups.
3. If the median falls between two actual numbers in the data, it is a calculated value, not a real one from the data itself.
4. To find the median, you always have to put the data in order (from smallest to largest or vice versa), which can take a lot of time for big datasets.
These limitations mean that the median is not always the best choice for every type of data analysis.
In simple words: The median doesn't always show a true average, can't be used for all math, might not be an actual data point, and takes time to sort data.

🎯 Exam Tip: When asked for flaws, provide concise points and a brief explanation for each, focusing on why it's a limitation.

 

Question 2. Explain the uses of median.
Answer: The median is very useful because it's easy to calculate and understand, especially in real-world situations. It helps when looking at things like how wealth or property is shared among people. It's also great for understanding social issues and for measuring qualities like health or intelligence, where exact numbers might not be available. We use it when we don't need to give different importance (weights) to different values, and it helps in seeing how data spreads out. The median is especially helpful when extreme values in the data could mislead the average. This makes the median a robust measure for many economic and social indicators.
In simple words: Median is simple to find and good for understanding how wealth is shared or for social problems. It works well when numbers are not exact or when very high/low numbers would skew the average.

🎯 Exam Tip: Focus on the median's practicality, its use with qualitative data, and its robustness against extreme values.

 

Question 3. What do you mean by partition value?
Answer: Partition values are numbers that split a list of data into many equal sections. Imagine cutting a cake into equal slices; each cut point is a partition value. For example, data can be divided into 2 equal parts (by the median), 4 equal parts (by quartiles), 5 parts (pentants), 8 parts (octants), 10 parts (decatants), or 100 parts (centiles). These values help to understand the spread and distribution of data in more detail. They are essential for percentile and quartile analysis, which is common in statistical reports.
In simple words: Partition values are points that divide a set of data into many equal smaller parts, such as the median (2 parts) or quartiles (4 parts).

🎯 Exam Tip: Clearly define partition values and list a few examples like median, quartiles, and percentiles to show a complete understanding.

 

RBSE Class 11 Economics Chapter 9 Long Answer Type Questions

 

Question 1. Find out the median value from the following data - 50,42,48,52,47,58,60,40,51
Answer: The first step is to arrange the given numbers in ascending order (from smallest to largest). The data becomes: 40, 42, 47, 48, 50, 51, 52, 58, 60.
Next, count the total number of items (\(N\)). Here, \(N = 9\).
Since \(N\) is an odd number, the median is found using the formula:
`\( M = \text{Value of } \left[ \frac { N+1}{2} \right] \text{ th item} \)`
Substitute \(N=9\):
`\( M = \text{Value of } \left[ \frac { 9+1}{2} \right] \text{ th item} \)`
`\( M = \text{Value of } \left[ \frac { 10}{2} \right] \text{ th item} \)`
`\( M = \text{Value of 5th item} \)`
Looking at the ordered list, the 5th item is 50.
Therefore, the median is 50. This calculation is clearly shown below:

`\( M = \text{Value of } \left[ \frac { N+1}{2} \right] \text{ th item where } N = \text{Number of items, } M = \text{Median} \)`
`\( \text{Value of } \left[ \frac { 9+1}{2} \right] \text{ th item = Value of 5th item = 50} \)`
The median helps in understanding the central tendency of the data.
In simple words: First, arrange the numbers from smallest to largest. Then, use the formula (N+1)/2 to find which position the median is in. The number at that position is the median.

🎯 Exam Tip: Always order the data first, then count the total number of items (N), and apply the correct median formula for odd or even N.

 

Example: Median Calculation (Discrete Series)
Answer: First, arrange the data by serial number, ensuring monthly incomes are in ascending order. This creates a clear progression in the dataset.

Count the total number of items (\(N\)). Here, \(N = 10\).
Since \(N\) is an even number, the median is the average of the two middle items. The median position is found by taking the \(N/2\)th item and the \((N/2) + 1\)th item.
`\( N/2 = 10/2 = 5\text{th item} \)`
`\( (N/2) + 1 = 6\text{th item} \)`
From the table, the 5th item (income) is Rs 1800, and the 6th item (income) is Rs 2000.
So, Median = `\( \frac{\text{Value of 5th item} + \text{Value of 6th item}}{2} \)`
`\( = \frac{\text{Rs } 1800 + \text{Rs } 2000}{2} \)`
`\( = \frac{\text{Rs } 3800}{2} \)`
`\( \text{Median = Rs } 1900 \)`
This calculation shows how to find the central point when there are an even number of data points.
In simple words: If there are an even number of items, find the two middle ones. Add them up and divide by two to get the median.

🎯 Exam Tip: When N is even, calculate the average of the two middle terms after arranging the data.

 

Question 2. How is the median calculated in discrete series? Explain it with an example.
Answer: To find the median in a discrete series, follow these steps:
1. First, create a column for cumulative frequencies. This means adding up the frequencies as you go down the list.
2. Next, calculate the median's position using the formula: `\( M = \text{Value of } \left[ \frac { N+1}{2} \right] \text{ th item} \)` where \(N\) is the total sum of all frequencies (\(\sum f\)).
The median is the value of the item that corresponds to this calculated position in the cumulative frequency column. This method helps to find the central point even when data points have different frequencies.
Example:
Let's find the median for the given discrete series:

Total frequency \(N = \sum f = 63\).
Now, use the median formula for discrete series:
`\( M = \text{Value of } \left[ \frac { N+1}{2} \right] \text{ th item} \)`
`\( M = \text{Value of } \left[ \frac { 63+1 }{2} \right] \text{ th item} \)`
`\( M = \text{Value of } \left[ \frac { 64 }{2} \right] \text{ th item} \)`
`\( M = \text{Value of 32nd item} \)`
Look at the cumulative frequency column. The 32nd item falls within the cumulative frequency of 42 (which corresponds to the item-value 8).
Therefore, the median value is 8.
In simple words: First, make a running total of frequencies. Then, use (Total Frequencies + 1) / 2 to find the position of the median. The item value that matches this position in the running total is your median.

🎯 Exam Tip: Always create the cumulative frequency column carefully and accurately before locating the median item.

 

Question 3. Explain the process of determining median in continuous series with example.
Answer: To find the median in a continuous series, follow these steps:
1. First, calculate the cumulative frequencies for all classes.
2. Then, find the median position using the formula: `\( \text{Median Number (M.No)} = \text{Value of } \left[ \frac { N}{2} \right] \text{ th term} \)` , where \(N\) is the total frequency.
3. The class where this median number first appears in the cumulative frequency column is called the median class.
4. Finally, use the following formula to calculate the exact median value from the median class:
`\( M = L_1 + \frac{(\frac{N}{2} - c)}{f} \times i \)`
Here, \(L_1\) is the lower limit of the median class, \(N\) is the total frequency, \(c\) is the cumulative frequency of the class just before the median class, \(f\) is the frequency of the median class, and \(i\) is the class interval (width) of the median class. This detailed process allows us to find a precise median even with data grouped into ranges.
In simple words: First, make a running total of frequencies. Find the median's position by dividing total frequencies by 2. Locate the class where this position falls. Then use a special formula with the class limits and frequencies to calculate the exact median.

🎯 Exam Tip: Always remember to calculate cumulative frequencies and find the median class before applying the main median formula for continuous series.

 

Example 1. Find out the median from the following item series
Answer: First, let's create a cumulative frequency column for the given data:

Total frequency \(N = \sum f = 31\).
Next, find the median position:
`\( \text{Median Number (M.No)} = \text{Value of } \frac{N}{2} \text{ th term} \)`
`\( = \text{Value of } \frac{31}{2} \text{ th term} = 15.5\text{th item} \)`
The 15.5th item falls in the class 20-30 (because its cumulative frequency is 19). So, this is our median class.
From the median class (20-30):
Lower limit (\(L_1\)) = 20
Frequency of median class (\(f\)) = 9
Cumulative frequency of preceding class (\(c\)) = 10 (for class 10-20)
Class interval (\(i\)) = \(30 - 20 = 10\)
Now, apply the median formula:
`\( M = L_1 + \frac{(\frac{N}{2} - c)}{f} \times i \)`
`\( = 20 + \frac{(15.5 - 10)}{9} \times 10 \)`
`\( = 20 + \frac{5.5}{9} \times 10 \)`
`\( = 20 + \frac{55}{9} \)`
`\( = 20 + 6.11 \)`
`\( M = 26.11 \text{ marks} \)`
This median indicates the central score within the student group.
In simple words: First, make a total of running frequencies. Find the median's place by dividing total students by 2. This helps find the "median class". Then use the special formula with the median class's numbers to get the exact median value.

🎯 Exam Tip: Double-check the values for \(L_1\), \(N/2\), \(c\), \(f\), and \(i\) before plugging them into the formula. A common mistake is using the wrong cumulative frequency.

 

Example 2. Find out the median of the following series
Answer: To find the median, first, arrange the class intervals and their frequencies in ascending order and calculate the cumulative frequencies:

Total number of workers \(N = \sum f = 161\).
Next, find the median position:
`\( \text{Median Number (M.No)} = \text{Value of } \frac{N}{2} \text{ th term} \)`
`\( = \text{Value of } \frac{161}{2} \text{ th term} = 80.5\text{th item} \)`
The 80.5th item falls within the cumulative frequency of 106, which corresponds to the class 20-30. So, 20-30 is our median class.
From the median class (20-30):
Lower limit (\(L_1\)) = 20
Frequency of median class (\(f\)) = 46
Cumulative frequency of preceding class (\(c\)) = 60 (for class 10-20)
Class interval (\(i\)) = \(30 - 20 = 10\)
Now, apply the median formula for ascending series:
`\( M = L_1 + \frac{(\frac{N}{2} - c)}{f} \times i \)`
`\( = 20 + \frac{(80.5 - 60)}{46} \times 10 \)`
`\( = 20 + \frac{20.5}{46} \times 10 \)`
`\( = 20 + \frac{205}{46} \)`
`\( = 20 + 4.46 \)`
`\( M = 24.46 \)`
This value represents the median wage among the workers.
In simple words: First, reorder the wage groups from smallest to largest and add up the frequencies step-by-step. Find the middle position by dividing the total number of workers by 2. Find which wage group this middle position falls into. Then, use a specific formula to calculate the exact median wage.

🎯 Exam Tip: Be careful when data is given in descending order; always reorder it to ascending before creating cumulative frequencies and applying the standard formula.

 

Determination of Median in Inclusive Series:

Answer: When class intervals are inclusive (like 1-5, 6-10, where the upper limit of one class is not the lower limit of the next), we must first change them to exclusive series (like 0.5-5.5, 5.5-10.5). This means adjusting the class limits so that the upper limit of one class matches the lower limit of the next, by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit. After this conversion, the median is calculated using the same method as for a continuous series. This step ensures that there are no gaps between classes.
In simple words: For data groups like 1-5, 6-10 (where numbers don't touch), we first change them to 0.5-5.5, 5.5-10.5 (where numbers touch). After that, we find the median the normal way.

🎯 Exam Tip: Remember that for inclusive series, the first step is always to convert them into exclusive series before any calculations.

 

Example 3. Find out the median from the following :
Answer: First, convert the inclusive class intervals into exclusive class intervals and then calculate the cumulative frequencies:

Total frequency \(N = \sum f = 32\).
Next, find the median position:
`\( \text{Median Number (M.No)} = \text{Value of } \frac{N}{2} \text{ th term} \)`
`\( = \text{Value of } \frac{32}{2} \text{ th term} = 16\text{th item} \)`
The 16th item falls within the cumulative frequency of 20, which corresponds to the exclusive class 10.5-15.5. So, this is our median class.
From the median class (10.5-15.5):
Lower limit (\(L_1\)) = 10.5
Frequency of median class (\(f\)) = 8
Cumulative frequency of preceding class (\(c\)) = 12 (for class 5.5-10.5)
Class interval (\(i\)) = \(15.5 - 10.5 = 5\)
Now, apply the median formula:
`\( M = L_1 + \frac{(\frac{N}{2} - c)}{f} \times i \)`
`\( = 10.5 + \frac{(16 - 12)}{8} \times 5 \)`
`\( = 10.5 + \frac{4}{8} \times 5 \)`
`\( = 10.5 + \frac{20}{8} \)`
`\( = 10.5 + 2.5 \)`
`\( M = 13 \)`
The median score is 13.
In simple words: First, change the groups so numbers meet (like 1-5 becomes 0.5-5.5). Then, make a running total of frequencies. Find the median's position, then the "median class". Finally, use the special formula with the class numbers to find the exact median.

🎯 Exam Tip: Clearly show the conversion of inclusive to exclusive classes as the first step, as this is crucial for accurate calculations.

 

Example 4. Find out the median from the following data
Answer: First, we need to create class intervals from the given mid-values. The difference between consecutive mid-values is \(20 - 10 = 10\), which is our class interval (\(i\)).
To find the lower limit (\(L_1\)) of each class, subtract \(i/2\) from the mid-value. To find the upper limit (\(L_2\)), add \(i/2\).
So, for mid-value 10, the class is \(10 - 5 = 5\) to \(10 + 5 = 15\), i.e., 5-15.
Now, let's form the class intervals and calculate the cumulative frequencies:

Total frequency \(N = \sum f = 35\).
Next, find the median position:
`\( \text{Median Number (M.No)} = \text{Value of } \frac{N}{2} \text{ th term} \)`
`\( = \text{Value of } \frac{35}{2} \text{ th term} = 17.5\text{th item} \)`
The 17.5th item falls within the cumulative frequency of 20, which corresponds to the class 25-35. So, 25-35 is our median class.
From the median class (25-35):
Lower limit (\(L_1\)) = 25
Frequency of median class (\(f\)) = 7
Cumulative frequency of preceding class (\(c\)) = 13 (for class 15-25)
Class interval (\(i\)) = \(35 - 25 = 10\)
Now, apply the median formula:
`\( M = L_1 + \frac{(\frac{N}{2} - c)}{f} \times i \)`
`\( = 25 + \frac{(17.5 - 13)}{7} \times 10 \)`
`\( = 25 + \frac{4.5}{7} \times 10 \)`
`\( = 25 + \frac{45}{7} \)`
`\( = 25 + 6.43 \)`
`\( M = 31.43 \)`
The median is 31.43.
In simple words: First, use the middle values to find the start and end of each group. Then, sum up the frequencies and find the median's position. This leads to the "median group". Finally, use the median formula with the numbers from this group to get the exact median.

🎯 Exam Tip: Carefully construct the class intervals from mid-values before proceeding. A mistake here will affect the entire calculation.

 

Question 4. How is the median determined in cumulative frequency? Explain with examples.
Answer: To find the median when data is presented in a cumulative frequency distribution (like "more than" or "less than" type), follow these steps:
1. Convert the cumulative frequency distribution into a simple frequency distribution with proper class intervals. This involves calculating the frequency for each individual class.
2. Then, calculate the new cumulative frequencies for this simple frequency distribution.
3. Follow the standard steps for finding the median in a continuous series: first, calculate \(N/2\); second, identify the median class where this \(N/2\) value falls in the cumulative frequency; and third, apply the median formula: `\( M = L_1 + \frac{(\frac{N}{2} - c)}{f} \times i \)`.
This method allows for calculating the median even when the data is initially presented in a cumulative form.
Example 2:
To find the median from "More Than" cumulative frequency data, first convert it into a standard frequency distribution with class intervals in ascending order and then calculate the cumulative frequencies.
Given "Obtained Marks (More Than)" and "No. of Students" (which are cumulative frequencies):

Now, create the standard frequency distribution:
Total number of students \(N = \sum f = 65\).
Next, find the median position:
`\( \text{Median Number (M.No)} = \text{Value of } \frac{N}{2} \text{ th term} \)`
`\( = \text{Value of } \frac{65}{2} \text{ th term} = 32.5\text{th item} \)`
The 32.5th item falls within the cumulative frequency of 47, which corresponds to the class 50-60. So, 50-60 is our median class.
From the median class (50-60):
Lower limit (\(L_1\)) = 50
Frequency of median class (\(f\)) = 22
Cumulative frequency of preceding class (\(c\)) = 25 (for class 40-50)
Class interval (\(i\)) = \(60 - 50 = 10\)
Now, apply the median formula:
`\( M = L_1 + \frac{(\frac{N}{2} - c)}{f} \times i \)`
`\( = 50 + \frac{(32.5 - 25)}{22} \times 10 \)`
`\( = 50 + \frac{7.5}{22} \times 10 \)`
`\( = 50 + \frac{75}{22} \)`
`\( = 50 + 3.41 \)`
`\( M = 53.41 \text{ Marks} \)`
The median marks obtained by students is 53.41.
In simple words: When data is given as "more than" or "less than", first convert it into normal groups with their actual counts. Then, create a running total of these counts. Find the median's position, pick the matching group, and use the specific formula to find the exact median.

🎯 Exam Tip: Carefully convert "More Than" or "Less Than" data into regular frequency distributions before any other calculations. Pay attention to zero frequencies.

 

Question 5. How is the median determined in unequal distribution? Explain with examples.
Answer: To find the median when class intervals are not of the same size, we generally try to make them equal if it's feasible. However, if that is not possible, or if the formula is applied correctly, the median can still be calculated directly using the standard formula. This is because the class interval (\(i\)) used in the median formula specifically refers to the width of the *median class* itself. The standard formula for continuous series implicitly handles varying class intervals by using the 'i' of the specific median class.
Example 1:
First, let's calculate the cumulative frequencies for the given data:

Total number of boys \(N = \sum f = 22\).
Next, find the median position:
`\( \text{Median Number (M.No)} = \text{Value of } \frac{N}{2} \text{ th term} \)`
`\( = \text{Value of } \frac{22}{2} \text{ th term} = 11\text{th item} \)`
The 11th item falls within the cumulative frequency of 17, which corresponds to the class 6-8. So, 6-8 is our median class.
From the median class (6-8):
Lower limit (\(L_1\)) = 6
Frequency of median class (\(f\)) = 7
Cumulative frequency of preceding class (\(c\)) = 10 (for class 4-6)
Class interval (\(i\)) = \(8 - 6 = 2\)
Now, apply the median formula:
`\( M = L_1 + \frac{(\frac{N}{2} - c)}{f} \times i \)`
`\( = 6 + \frac{(11 - 10)}{7} \times 2 \)`
`\( = 6 + \frac{1}{7} \times 2 \)`
`\( = 6 + \frac{2}{7} \)`
`\( = 6 + 0.29 \)`
`\( M = 6.29 \text{ years} \)`
The median age of boys is 6.29 years.
In simple words: First, add up the frequencies step-by-step. Find the middle position by dividing the total number of boys by 2. This helps find the "median age group". Then, use the special formula with the numbers from this group to find the exact median age.

🎯 Exam Tip: Even if class intervals are not equal, the formula works correctly by using the 'i' (class interval) of the specific median class.

RBSE Class 11 Economics Chapter 9 Long Answer Type Questions

 

Example 2. Find out the median of following series :

Answer:
To find the median, we first create a cumulative frequency table from the given data:
Here, the total number of items \( N = 156 \).
Now, we find the median item:
\( \text{Median No.} = \text{value of } \frac { N }{ 2 } \text{th item } = \frac { 156 }{ 2 } = 78\text{th item} \)
The 78th item falls within the cumulative frequency of 92, which corresponds to the class interval 20-30. So, the median class is 20-30.
\( L_1 = 20 \), \( c = 52 \) (cumulative frequency of the preceding class), \( f = 40 \) (frequency of the median class), \( i = 10 \) (class width).
Using the formula for median in a continuous series:
\( M = L_1 + \frac { N/2 - c }{ f } \times i \)
\( M = 20 + \frac { 78 - 52 }{ 40 } \times 10 \)
\( M = 20 + \frac { 26 }{ 40 } \times 10 \)
\( M = 20 + \frac { 260 }{ 40 } \)
\( M = 20 + 6.5 \)
\( M = 26.5 \)
The median of the given series is 26.5. This value helps understand the central point of the data distribution.
In simple words: First, we arranged the data and found the middle position. Then, we used a formula to pinpoint the exact median value within its group. The median is 26.5.

🎯 Exam Tip: When working with class intervals, always identify the correct median class based on the cumulative frequency before applying the formula. Double-check your calculation of 'c' (preceding cumulative frequency).

 

Question 6. What do you mean by quartile? How is quartile calculated? Explain with examples.
Answer:
Meaning of Quartile:
A quartile is a measure that divides a series of data into four equal parts. When any group or series of items is divided into four equal parts, the dividing points are called quartiles. There are four quartiles in any series. The fourth quartile represents the upper limit of the data, so it's not usually calculated separately. The second quartile is actually the median, meaning only the first and third quartiles need to be computed for analysis. These measures help in understanding the spread and distribution of data.
The first quartile is known as the lower quartile, and the third quartile is called the upper quartile. These are often shown using symbols like \( Q_1 \) and \( Q_3 \) respectively. In the first quartile, 25% of the data items are less than this value, and 75% are greater. Conversely, for the third quartile, 75% of the data items are less than this value, and 25% are greater.

How to Calculate Quartiles:
(i) Individual and Discrete Series:
The following formulas are used to calculate quartiles in individual and discrete series:
\( Q_1 = \text{Value of } \left[ \frac { N+1 }{ 4 } \right] \text{ th item} \)
\( Q_3 = \text{Value of } \left[ \frac { 3 (N+1) }{ 4 } \right] \text{ th item} \)
Here, \( N \) represents the total number of items in an individual series, and in a discrete series, it is the total of frequencies.

(ii) Uninterrupted and Continuous Series:
For continuous series, \( Q_1 \) and \( Q_3 \) are first identified. The following formulas are used for this:
\( Q_1 = \text{Value of } \left[ \frac { N }{ 4 } \right] \text{ th item} \)
\( Q_3 = \text{Value of } \left[ \frac { 3N }{ 4 } \right] \text{ th item} \)
After finding the item number, we locate which cumulative frequency range it falls into. The class interval corresponding to that cumulative frequency is the quartile class. Then, the specific values of \( Q_1 \) and \( Q_3 \) are calculated using the following formulas:
\( Q_1 = L_1 + \frac { N/4 - c }{ f } \times i \)
\( Q_3 = L_1 + \frac { 3N/4 - c }{ f } \times i \)

Examples:
(i) Calculation in individual series:
First, the values must be arranged in ascending or descending order.

Example 1.
Find out the first and third quartiles from the following production data (in Tons):
10, 12, 13, 14, 15, 17, 18, 20

Here, \( N = 8 \).
For the first quartile (\( Q_1 \)):
\( Q_1 = \text{Value of } \left[ \frac { N+1 }{ 4 } \right] \text{ th item} = \text{value of } \left[ \frac { 8+1 }{ 4 } \right] \text{ th item} = \text{Value of 2.25th item} \)
To find the value of the 2.25th item, we interpolate:
\( \text{Value of 2.25th item} = \text{Value of second item} + 0.25 \times (\text{Value of third item} - \text{Value of second item}) \)
\( = 12 + 0.25 \times (13 - 12) \)
\( = 12 + 0.25 \times 1 \)
\( = 12.25 \)
So, \( Q_1 = 12.25 \text{ tonnes} \).

For the third quartile (\( Q_3 \)):
\( Q_3 = \text{Value of } \left[ \frac { 3N+1 }{ 4 } \right] \text{ th item} = \text{Value of } \left[ \frac { 3(8+1) }{ 4 } \right] \text{ th item} = \text{Value of 6.75th item} \)
To find the value of the 6.75th item, we interpolate:
\( \text{Value of 6.75th item} = \text{Value of sixth item} + 0.75 \times (\text{Value of seventh item} - \text{Value of sixth item}) \)
\( = 17 + 0.75 \times (18 - 17) \)
\( = 17 + 0.75 \times 1 \)
\( = 17.75 \)
So, \( Q_3 = 17.75 \text{ tonnes} \).

Example 2.
Calculate the first and third quartile from the following data:
To find the quartiles, we first arrange the data and calculate the cumulative frequency:
Here, the total number of items \( N = 59 \).
For the third quartile (\( Q_3 \)):
\( Q_3 = \text{Value of } \frac { 3N+1 }{ 4 } \text{ th item } = \text{Value of } \frac { 3(59+1) }{ 4 } \text{ th item } = \text{Value of 45th item.} \)
The 45th item is included in the cumulative frequency 49, which corresponds to the value 13.
So, \( Q_3 = 13 \).

(iii) Calculation of quartile in continuous series:
Example 3.
Calculate the first and third quartile from the following frequency distribution:
To find the quartiles, we construct the cumulative frequency table:
Here, the total frequency \( N = 109 \).
For the first quartile (\( Q_1 \)):
\( Q_1 \text{ item } = \text{value of } \frac { N }{ 4 } \text{ th term } = \frac { 109 }{ 4 } = 27.25\text{th item} \)
The 27.25th item falls in the cumulative frequency 34, so the \( Q_1 \) class is 12-16.
\( L_1 = 12 \), \( c = 16 \), \( f = 18 \), \( i = 4 \).
\( Q_1 = L_1 + \frac { N/4 - c }{ f } \times i \)
\( Q_1 = 12 + \frac { 27.25-16 }{ 18 } \times 4 \)
\( Q_1 = 12 + \frac { 11.25 \times 4 }{ 18 } \)
\( Q_1 = 12 + \frac { 45 }{ 18 } \)
\( Q_1 = 12 + 2.5 \)
\( Q_1 = 14.5 \)

For the third quartile (\( Q_3 \)):
\( Q_3 \text{ item } = \text{value of } \frac { 3N }{ 4 } \text{ th term } = \frac { 3 \times 109 }{ 4 } \text{ th item } = 81.75\text{th item} \)
The 81.75th item falls in the cumulative frequency 91, so the \( Q_3 \) class is 24-28.
\( L_1 = 24 \), \( c = 79 \), \( f = 12 \), \( i = 4 \).
\( Q_3 = L_1 + \frac { 3N/4 - c }{ f } \times i \)
\( Q_3 = 24 + \frac { 81.75-79 }{ 12 } \times 4 \)
\( Q_3 = 24 + \frac { 2.75 \times 4 }{ 12 } \)
\( Q_3 = 24 + \frac { 11 }{ 12 } \)
\( Q_3 = 24 + 0.9167 \)
\( Q_3 = 24.92 \) (rounded to two decimal places).
The quartiles help segment the data, showing where the first 25% and 75% of the values lie, which is useful for detailed statistical analysis.
In simple words: Quartiles divide data into four equal parts. We use different formulas for individual, discrete, and continuous data. For individual data, we find the item number directly. For continuous data, we find the quartile class first, then apply a specific formula to get the exact value.

🎯 Exam Tip: Always arrange data in ascending order before calculating quartiles for individual or discrete series. For continuous series, ensure you correctly identify the quartile class and the values of \( L_1 \), \( c \), \( f \), and \( i \) to avoid errors.

RBSE Class 11 Economics Chapter 9 Numerical Questions

 

Question 1. Find out the median value from the following data-
7, 12, 14, 9, 10, 13, 15, 11, 27, 18, 31
Answer:
To find the median, we first arrange the given data in ascending order:
7, 9, 10, 11, 12, 13, 14, 15, 18, 27, 31
The total number of observations, \( N = 11 \).
For an odd number of observations, the median is calculated using the formula:
\( M = \text{Value of } \left[ \frac { N+1 }{ 2 } \right] \text{ th item} \)
\( M = \text{Value of } \left[ \frac { 11+1 }{ 2 } \right] \text{ th item} \)
\( M = \text{Value of } 6\text{th item} \)
Looking at the arranged series, the 6th item is 13.
So, the Median \( M = 13 \). The median gives us the middle value in a sorted dataset.
In simple words: We sort the numbers from smallest to largest. Since there are 11 numbers, the middle one is the 6th number, which is 13.

🎯 Exam Tip: Always remember to arrange the data in ascending or descending order before calculating the median. For an odd number of observations, the formula \( \frac{N+1}{2} \) directly gives the position of the median.

 

Question 2. Find out the median from the following data :

Answer:
To find the median for this discrete series, we first create a cumulative frequency table:
The total number of observations, \( N = 55 \).
For a discrete series, the median item is found using the formula:
\( M = \text{Value of } \left[ \frac { N+1 }{ 2 } \right] \text{ th item} \)
\( M = \text{Value of } \left[ \frac { 55+1 }{ 2 } \right] \text{ th item} \)
\( M = \text{Value of } 28\text{th item} \)
Looking at the cumulative frequency (cf) column, the 28th item falls within the cf of 31. The value corresponding to this cf is 14.
So, the Median \( M = 14 \). This indicates that half of the workers earn less than or equal to Rs. 14, and half earn more.
In simple words: First, we added up the frequencies to get a running total. Then we found the position of the middle value. The value at that middle position in our list of wages is 14, which is our median.

🎯 Exam Tip: When dealing with discrete frequency distributions, always calculate the cumulative frequency first. The median value corresponds to the item whose cumulative frequency is just greater than or equal to \( \frac{N+1}{2} \).

 

Question 3. Find out median from the following data

Answer:
To find the median for this continuous series, we first prepare a cumulative frequency table:
The total frequency \( N = 24 \).
Now, we determine the median item:
\( \text{Median Class} = \frac { N }{ 2 } \text{ th item} = \frac { 24 }{ 2 } \text{ th item} = 12\text{th item} \)
The 12th item is included in the cumulative frequency 16. This corresponds to the class interval 20-30. So, the median class is 20-30.
From the median class, we have:
\( L_1 = 20 \) (lower limit of median class)
\( c = 8 \) (cumulative frequency of the class preceding the median class)
\( f = 8 \) (frequency of the median class)
\( i = 10 \) (class width)
Using the formula for median in a continuous series:
\( M = L_1 + \frac { N/2 - c }{ f } \times i \)
\( M = 20 + \frac { 12-8 }{ 8 } \times 10 \)
\( M = 20 + \frac { 4 }{ 8 } \times 10 \)
\( M = 20 + 0.5 \times 10 \)
\( M = 20 + 5 \)
\( M = 25 \)
The median of this data is 25. This value helps identify the central point within the frequency distribution.
In simple words: We first make a running total of frequencies. Then we find the middle position. The group where this middle position falls is our median class. Using a formula with values from this group, we calculate the exact median, which is 25.

🎯 Exam Tip: For continuous series, the first step is always to calculate cumulative frequencies. Then, correctly identify the median class by locating \( \frac{N}{2} \) in the cumulative frequency column.

 

Question 4. Calculate Median from the following item series :

Answer:
Since the class intervals are not in ascending order, we first rearrange the series and calculate the cumulative frequency:
The total frequency \( N = 31 \).
Now, we find the median item:
\( \text{Median Class} = \frac { N }{ 2 } \text{ th item} = \frac { 31 }{ 2 } \text{ th item} = 15.5\text{th item} \)
The 15.5th item is included in the cumulative frequency 19. This corresponds to the class interval 20-30. So, the median class is 20-30.
From the median class, we have:
\( L_1 = 20 \), \( c = 10 \), \( f = 9 \), \( i = 10 \).
Using the formula for median in a continuous series:
\( M = L_1 + \frac { N/2 - c }{ f } \times i \)
\( M = 20 + \frac { 15.5-10 }{ 9 } \times 10 \)
\( M = 20 + \frac { 5.5 \times 10 }{ 9 } \)
\( M = 20 + \frac { 55 }{ 9 } \)
\( M = 20 + 6.11 \)
\( M = 26.11 \)
The median marks obtained by students is 26.11. This helps understand the typical performance.
In simple words: We first put the marks in order and found the running total. Then we found the middle position. The group that holds this middle position is our median class. Using its values in a formula, we calculated the median to be 26.11.

🎯 Exam Tip: Always check if the class intervals are in ascending order. If not, rearrange them before creating the cumulative frequency table to ensure correct median calculation.

 

Question 5. Find out median from the following distribution :

Answer:
To find the median for this continuous distribution, we first construct a cumulative frequency table:
The total frequency \( N = 32 \).
Now, we find the median item:
\( \text{Median Class} = \frac { N }{ 2 } \text{ th item} = \frac { 32 }{ 2 } \text{ th item} = 16\text{th item} \)
The 16th item is included in the cumulative frequency 20. This corresponds to the class interval 10.5-15.5. So, the median class is 10.5-15.5.
From the median class, we have:
\( L_1 = 10.5 \), \( c = 12 \), \( f = 8 \), \( i = 5 \).
Using the formula for median in a continuous series:
\( M = L_1 + \frac { N/2 - c }{ f } \times i \)
\( M = 10.5 + \frac { 16-12 }{ 8 } \times 5 \)
\( M = 10.5 + \frac { 4 \times 5 }{ 8 } \)
\( M = 10.5 + \frac { 20 }{ 8 } \)
\( M = 10.5 + 2.5 \)
\( M = 13 \)
The median of this distribution is 13. This gives us the central point of the data distribution.
In simple words: We created a running total of frequencies. Then we located the middle position, which falls into the 10.5-15.5 group. Using a formula with values from this group, we found the exact median to be 13.

🎯 Exam Tip: Pay close attention to inclusive or exclusive class intervals. These intervals are already exclusive, making direct calculation easier. If they were inclusive, an adjustment would be needed first.

 

Question 6. Find out median from the following series :

Answer:
To find the median from mid-values, we first need to determine the class intervals and then calculate the cumulative frequency:
The total frequency \( N = 30 \).
Now, we find the median item:
\( \text{Median Class} = \frac { N }{ 2 } \text{ th item} = \frac { 30 }{ 2 } \text{ th item} = 15\text{th item} \)
The 15th item is included in the cumulative frequency 20. This corresponds to the class interval 25-35. So, the median class is 25-35.
From the median class, we have:
\( L_1 = 25 \), \( c = 13 \), \( f = 7 \), \( i = 10 \).
Using the formula for median in a continuous series:
\( M = L_1 + \frac { N/2 - c }{ f } \times i \)
\( M = 25 + \frac { 15-13 }{ 7 } \times 10 \)
\( M = 25 + \frac { 2 \times 10 }{ 7 } \)
\( M = 25 + \frac { 20 }{ 7 } \)
\( M = 25 + 2.86 \)
\( M = 27.86 \)
The median of this series is 27.86. Converting mid-values to class intervals is key for such problems.
In simple words: First, we turned the mid-values into proper groups and made a running total of frequencies. Then we found the middle position, which falls into the 25-35 group. Using its values in a formula, we calculated the median to be 27.86.

🎯 Exam Tip: When given mid-values, the first step is always to construct the class intervals by adding/subtracting half the class width. Ensure this step is accurate as it forms the base for all subsequent calculations.

 

Question 8. Calculate median from the following series :

Answer:
To calculate the median from a "less than" cumulative frequency distribution, we first convert it into a simple frequency distribution and then create a cumulative frequency table for calculation:
The total frequency \( N = 100 \).
Now, we determine the median item:
\( \text{Median Class} = \frac { N }{ 2 } \text{ th item} = \frac { 100 }{ 2 } \text{ th item} = 50\text{th item} \)
The 50th item is included in the cumulative frequency 70. This corresponds to the class interval 20-30. So, the median class is 20-30.
From the median class, we have:
\( L_1 = 20 \), \( c = 20 \), \( f = 50 \), \( i = 10 \).
Using the formula for median in a continuous series:
\( M = L_1 + \frac { N/2 - c }{ f } \times i \)
\( M = 20 + \frac { 50-20 }{ 50 } \times 10 \)
\( M = 20 + \frac { 30 \times 10 }{ 50 } \)
\( M = 20 + \frac { 300 }{ 50 } \)
\( M = 20 + 6 \)
\( M = 26 \)
The median of this series is 26. Converting the "less than" series to a standard frequency distribution is a critical first step.
In simple words: We first changed the "less than" data into normal frequency groups. Then we made a running total and found the middle position. The group holding this position is our median class. Using the formula with its values, we calculated the median to be 26.

🎯 Exam Tip: When given "less than" or "more than" cumulative frequency, always convert it into a simple frequency distribution first. This allows for correct identification of the median class and its specific frequency.

 

Question 8. Calculate median from the following series :

Answer:
To calculate the median from a "more than" cumulative frequency distribution, we first convert it into a simple frequency distribution and then create an 'actual' cumulative frequency table for calculation:
The total frequency \( N = 100 \).
Now, we determine the median item:
\( \text{Median Class} = \frac { N }{ 2 } \text{ th item} = \frac { 100 }{ 2 } \text{ th item} = 50\text{th item} \)
The 50th item is included in the cumulative frequency 56. This corresponds to the class interval 30-40. So, the median class is 30-40.
From the median class, we have:
\( L_1 = 30 \), \( c = 40 \), \( f = 16 \), \( i = 10 \).
Using the formula for median in a continuous series:
\( M = L_1 + \frac { N/2 - c }{ f } \times i \)
\( M = 30 + \frac { 50-40 }{ 16 } \times 10 \)
\( M = 30 + \frac { 10 \times 10 }{ 16 } \)
\( M = 30 + \frac { 100 }{ 16 } \)
\( M = 30 + 6.25 \)
\( M = 36.25 \)
The median income for these families is Rs. 36.25. This shows the income level that divides the families into two equal halves.
In simple words: We changed the "more than" data into regular frequency groups and then calculated a running total. We found the middle position, which falls into the 30-40 income group. Using the formula with values from this group, we calculated the median income to be Rs. 36.25.

🎯 Exam Tip: Converting "more than" or "less than" cumulative frequencies into simple frequencies is a common point of error. Always subtract successive cumulative frequencies to get the individual class frequencies. Ensure the total frequency \( N \) matches the highest cumulative frequency.

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