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Statistics – CBSE NCERT Study Resources

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Study Materials

9th

9th - Mathematics

Statistics

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Chapter Overview: Statistics

Statistics is a branch of Mathematics that deals with the collection, organization, analysis, interpretation, and presentation of data. In this chapter, students will learn about the fundamental concepts of statistics, including measures of central tendency and graphical representation of data.

Statistics: The science of collecting, organizing, analyzing, interpreting, and presenting data to make informed decisions.

Key Topics Covered

  • Collection of Data
  • Presentation of Data
  • Graphical Representation of Data
  • Measures of Central Tendency

Collection of Data

Data can be collected through primary or secondary sources. Primary data is collected firsthand, while secondary data is obtained from existing sources.

Primary Data: Data collected directly by the investigator for a specific purpose.

Secondary Data: Data obtained from existing sources like books, journals, or websites.

Presentation of Data

Data can be presented in various forms such as:

  • Ungrouped Frequency Distribution
  • Grouped Frequency Distribution
  • Histograms
  • Frequency Polygons

Graphical Representation of Data

Graphs provide a visual representation of data, making it easier to understand patterns and trends. Common types include:

  • Bar Graphs
  • Pie Charts
  • Line Graphs

Measures of Central Tendency

These are statistical measures that represent the center of a data set. The three main measures are:

  • Mean
  • Median
  • Mode

Mean: The average of all the numbers in a data set.

Median: The middle value when the data is arranged in ascending or descending order.

Mode: The value that appears most frequently in a data set.

Conclusion

This chapter provides a foundational understanding of statistics, equipping students with the skills to collect, present, and analyze data effectively. Mastery of these concepts is essential for higher-level studies in mathematics and real-world applications.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
What is the range of the data set: 12, 15, 10, 8, 20?
Answer:
12
Question 2:
Define class mark in statistics.
Answer:
Midpoint of a class interval
Question 3:
Find the mean of first five natural numbers.
Answer:
3
Question 4:
What is the mode of the data: 5, 7, 5, 3, 2, 5?
Answer:
5
Question 5:
If the mean of 6, 8, 5, x, 4 is 7, find x.
Answer:
12
Question 6:
What is the median of 9, 3, 5, 7, 6?
Answer:
6
Question 7:
Give one example of primary data.
Answer:
Survey responses
Question 8:
What is the upper limit of class 20-30?
Answer:
30
Question 9:
Find the class mark of class 15-25.
Answer:
20
Question 10:
If the mean of 5 numbers is 10, what is their total?
Answer:
50
Question 11:
What is the frequency of an observation?
Answer:
Number of times it occurs
Question 12:
Find the range of first 10 whole numbers.
Answer:
9
Question 13:
Define statistics in the context of Mathematics.
Answer:

Statistics is the branch of mathematics that deals with the collection, organization, analysis, interpretation, and presentation of data. It helps in making informed decisions based on data.

Question 14:
What is the range of a data set?
Answer:

The range is the difference between the highest and lowest values in a data set.
Formula: Range = Maximum value - Minimum value.

Question 15:
What is the mean of the data set: 5, 8, 10, 12, 15?
Answer:

To find the mean:
Step 1: Add all values → 5 + 8 + 10 + 12 + 15 = 50
Step 2: Divide by number of values → 50 ÷ 5 = 10
Mean = 10.

Question 16:
How is the median calculated for an odd number of observations?
Answer:

For an odd number of observations:
Step 1: Arrange data in ascending order.
Step 2: The median is the middle value of the ordered data set.

Question 17:
What is the mode of the data set: 3, 5, 7, 5, 9, 5, 7?
Answer:

The mode is the value that appears most frequently.
Here, 5 occurs three times → Mode = 5.

Question 18:
Differentiate between primary and secondary data.
Answer:

Primary data: Collected directly by the researcher (e.g., surveys).
Secondary data: Collected by someone else and used by the researcher (e.g., census reports).

Question 19:
What is a frequency distribution table?
Answer:

A frequency distribution table organizes data into classes/intervals and shows the number of observations (frequency) in each class. It helps in summarizing large data sets.

Question 20:
Calculate the mean of first five natural numbers.
Answer:

First five natural numbers: 1, 2, 3, 4, 5
Step 1: Sum = 1 + 2 + 3 + 4 + 5 = 15
Step 2: Mean = 15 ÷ 5 = 3
Mean = 3.

Question 21:
What is the class mark of the class interval 10-20?
Answer:

Class mark is the midpoint of a class interval.
Formula: (Lower limit + Upper limit) ÷ 2
For 10-20 → (10 + 20) ÷ 2 = 15
Class mark = 15.

Question 22:
Why is grouped data preferred over ungrouped data in statistics?
Answer:

Grouped data simplifies large data sets by organizing them into classes, making analysis easier. Ungrouped data lists all individual values, which can be cumbersome for large data.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
What is the difference between primary data and secondary data?
Answer:

Primary data is collected firsthand by the researcher for a specific purpose.
Secondary data is obtained from existing sources like books, journals, or reports.
Example: Survey responses (primary) vs. census data (secondary).

Question 2:
Calculate the mean of the numbers 5, 8, 12, 15, and 20.
Answer:

Step 1: Add all numbers: 5 + 8 + 12 + 15 + 20 = 60
Step 2: Divide by count (5): 60 ÷ 5 = 12
Mean = 12.

Question 3:
What is the range of the data set: 10, 3, 7, 15, 22?
Answer:

Step 1: Identify maximum (22) and minimum (3) values.
Step 2: Subtract: 22 - 3 = 19
Range = 19.

Question 4:
Explain the term frequency in statistics.
Answer:

Frequency refers to the number of times a particular observation or value occurs in a data set.
Example: In {2, 3, 3, 5}, the frequency of 3 is 2.

Question 5:
Find the median of the data: 14, 18, 12, 10, 16.
Answer:

Step 1: Arrange in order: 10, 12, 14, 16, 18
Step 2: Middle value = 14
Median = 14.

Question 6:
What is a histogram? How is it different from a bar graph?
Answer:

A histogram represents continuous data using adjacent bars with no gaps.
A bar graph displays discrete data with separate bars.
Key difference: Histograms show frequency distributions, bar graphs compare categories.

Question 7:
If the mode of a data set is 7 and the data set is 4, 7, 7, 9, 12, what does this indicate?
Answer:

The mode being 7 indicates it is the most frequently occurring value in the data set.
Here, 7 appears twice, more than any other number.

Question 8:
Why is the mean affected by extreme values in a data set?
Answer:

The mean is calculated using all values, so extreme values (very high/low) skew the result.
Example: In {1, 2, 3, 100}, the mean (26.5) doesn’t represent the majority due to 100.

Question 9:
How is a pie chart useful in representing data?
Answer:

A pie chart visually represents proportional data as slices of a circle.
It helps compare parts of a whole, like budget allocation or survey percentages, at a glance.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
Define mean in statistics and explain its significance with an example.
Answer:

The mean is the average of a set of numbers, calculated by dividing the sum of all values by the number of values. It is a measure of central tendency that helps summarize data.


Example: If the marks of 5 students are 10, 20, 30, 40, and 50, the mean is calculated as:
(10 + 20 + 30 + 40 + 50) / 5 = 150 / 5 = 30.

The mean is widely used because it considers all data points, making it useful for comparisons.

Question 2:
Differentiate between discrete and continuous data with examples.
Answer:

Discrete data consists of distinct, separate values (often whole numbers), while continuous data can take any value within a range.


Example of discrete data: Number of students in a class (e.g., 25, 30).
Example of continuous data: Height of students (e.g., 150.5 cm, 160.2 cm).

Discrete data is countable, whereas continuous data is measurable.

Question 3:
Explain how to calculate the median for an odd and even number of observations.
Answer:

The median is the middle value in a sorted dataset.


For odd number of observations:
Arrange data in ascending order.
Median = Middle value.
Example: 5, 7, 9 → Median = 7.

For even number of observations:
Arrange data in ascending order.
Median = Average of the two middle values.
Example: 4, 6, 8, 10 → Median = (6 + 8)/2 = 7.
Question 4:
What is a frequency distribution table? How is it useful in statistics?
Answer:

A frequency distribution table organizes data into classes/intervals and shows how often each class occurs.


Steps to create one:
1. Choose class intervals.
2. Tally the data points falling in each interval.
3. Record frequencies.

Usefulness: It simplifies large datasets, highlights patterns, and aids in visualization (e.g., histograms).

Question 5:
Calculate the mode of the dataset: 2, 3, 4, 2, 5, 2, 6. Explain its importance.
Answer:

The mode is the most frequently occurring value in a dataset.


Calculation:
Given data: 2, 3, 4, 2, 5, 2, 6.
Frequency of 2 = 3 (highest).
Mode = 2.

Importance: Mode helps identify popular or common values, useful in categorical data (e.g., most sold product).

Question 6:
Define mean in statistics and explain how it is calculated with an example.
Answer:

The mean is the average of a set of numbers. It is calculated by adding all the values and dividing by the number of values.


For example, to find the mean of 5, 10, 15:
Step 1: Add the numbers → 5 + 10 + 15 = 30
Step 2: Divide by the count → 30 ÷ 3 = 10
Thus, the mean is 10.

Mean is useful for understanding the central tendency of data.

Question 7:
What is the difference between discrete and continuous data? Provide one example for each.
Answer:

Discrete data can only take specific values (usually whole numbers), while continuous data can take any value within a range.

  • Example of discrete data: Number of students in a class (e.g., 25, 30).
  • Example of continuous data: Height of students (e.g., 150.5 cm, 160.2 cm).

Discrete data is counted, while continuous data is measured.

Question 8:
Explain how to construct a frequency distribution table for the data: 3, 5, 2, 3, 4, 5, 5, 2, 3.
Answer:

A frequency distribution table organizes data by showing how often each value occurs.


Steps:
1. List all unique values → 2, 3, 4, 5
2. Count their occurrences:
- 2 appears 2 times
- 3 appears 3 times
- 4 appears 1 time
- 5 appears 3 times
3. Create the table:
| Value | Frequency |
|-------|-----------|
| 2 | 2 |
| 3 | 3 |
| 4 | 1 |
| 5 | 3 |
Question 9:
What is the range of a dataset? Calculate it for the numbers: 12, 8, 15, 20, 6.
Answer:

The range is the difference between the highest and lowest values in a dataset.


Given data: 12, 8, 15, 20, 6
Step 1: Identify the highest value → 20
Step 2: Identify the lowest value → 6
Step 3: Calculate range → 20 - 6 = 14

Range helps measure the spread of data.

Question 10:
Describe the mode in statistics. Find the mode for the data: 7, 8, 7, 10, 8, 7, 9.
Answer:

The mode is the value that appears most frequently in a dataset.


Given data: 7, 8, 7, 10, 8, 7, 9
Step 1: Count occurrences:
- 7 appears 3 times
- 8 appears 2 times
- 9 appears 1 time
- 10 appears 1 time
Step 2: The highest frequency is 7.

Thus, the mode is 7. Mode helps identify the most common value.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
Explain the mean, median, and mode with examples from our NCERT textbook. How are these measures useful in real-life situations?
Answer:
Introduction

We studied three measures of central tendency: mean, median, and mode. These help summarize data.


Argument 1
  • Mean: Average of data. Example: NCERT shows mean rainfall calculation.
  • Median: Middle value. Example: Finding median height of students.

Argument 2
  • Mode: Most frequent value. Example: Mode of shoe sizes in class.
  • Real-life use: Mean for budgets, median for property prices, mode for popular products.

Conclusion

These measures simplify data analysis, as shown in NCERT examples.

Question 2:
Derive the formula for mean of grouped data using the direct method. Illustrate with an NCERT example.
Answer:
Introduction

Grouped data mean is calculated using class marks and frequencies.


Argument 1
  • Formula: Mean = Σ(fixi)/Σfi, where xi is class mark.
  • Steps: Find midpoints, multiply by frequency, sum up, divide by total frequency.

Argument 2
  • NCERT example: Table of daily wages with frequencies.
  • Real-life use: Calculating average income or test scores.

Conclusion

The direct method simplifies mean calculation for large datasets.

Question 3:
What is a histogram? How is it different from a bar graph? Use an NCERT example to explain.
Answer:
Introduction

Histograms and bar graphs display data visually but differ in usage.


Argument 1
  • Histogram: Shows continuous data (e.g., NCERT height intervals).
  • Bar graph: Compares discrete categories (e.g., favorite subjects).

Argument 2
  • Histogram bars touch; bar graphs have gaps.
  • NCERT example: Histogram of class intervals vs. bar graph of book sales.

Conclusion

Choosing the right graph depends on data type, as shown in NCERT.

Question 4:
Explain the mean, median, and mode with examples from our NCERT textbook. How are these measures useful in real life?
Answer:
Introduction

We studied three measures of central tendency: mean, median, and mode. These help summarize data.


Argument 1
  • Mean: Average of data. Example: NCERT shows mean of 5, 10, 15 is (5+10+15)/3 = 10.
  • Median: Middle value. For 7, 3, 5, arranged as 3,5,7, median is 5.

Argument 2
  • Mode: Most frequent value. In 2,3,3,4, mode is 3.
  • Real-life use: Mean for average income, median for property prices, mode for popular shoe size.

Conclusion

These measures simplify data analysis, as shown in NCERT and daily life.

Question 5:
Derive the formula for mean of grouped data using the direct method. Compare it with the NCERT example of class intervals 0-10, 10-20.
Answer:
Introduction

Grouped data mean is calculated using class marks and frequencies.


Argument 1
  • Formula: Mean = Σ(fi×xi)/Σfi, where xi = class mark, fi = frequency.
  • Derivation: Multiply mid-points by frequencies, sum both, then divide.

Argument 2
  • NCERT example (0-10,10-20): Class marks=5,15. If frequencies=7,3, mean=(5×7+15×3)/(7+3)=8.
  • Units: Always include (e.g., kg, cm) if data has units.

Conclusion

The direct method simplifies mean calculation for grouped data, as in NCERT.

Question 6:
How is a histogram different from a bar graph? Use the NCERT example of students' heights (150-160 cm, etc.) to explain.
Answer:
Introduction

Histograms and bar graphs represent data visually but differ in usage.


Argument 1
  • Bar graph: Spaces between bars, for discrete data (e.g., favorite subjects).
  • Histogram: No gaps, for continuous data (e.g., height groups).

Argument 2
  • NCERT example: Heights 150-160 cm, 160-170 cm plotted as adjacent bars.
  • Real-life: Histograms for temperature ranges, bar graphs for survey results.

Conclusion

As NCERT shows, histograms suit grouped data, while bar graphs compare categories.

Question 7:
Explain the mean, median, and mode with an example from our NCERT textbook. How are these measures useful in real life?
Answer:
Introduction

We studied three measures of central tendency: mean, median, and mode. These help summarize data.


Argument 1
  • Mean: Average of data. Example: NCERT shows heights of 5 students (150, 152, 148, 151, 149 cm). Mean = (150+152+148+151+149)/5 = 150 cm.
  • Median: Middle value when data is ordered. For the same heights, median = 150 cm.
  • Mode: Most frequent value. If heights were 150, 150, 152, 148, 150 cm, mode = 150 cm.

Conclusion

These measures simplify data analysis, like comparing class performance or average temperatures.

Question 8:
Define class interval, frequency, and class mark using the NCERT example of daily wages (in ₹) of 30 workers. Construct a frequency table.
Answer:
Introduction

We learned about organizing data into groups for clarity. Key terms are class interval, frequency, and class mark.


Argument 1
  • Class interval: Groups like 100-120, 120-140. NCERT uses wages: 100-120, 120-140, ..., 180-200.
  • Frequency: Workers in each interval. Example: 6 workers earn ₹100-120.
  • Class mark: Midpoint. For 100-120, mark = (100+120)/2 = 110.

Conclusion

Frequency tables help visualize wage distributions, useful for policymakers.

Question 9:
Derive the formula for mean of grouped data using assumed mean method. Use NCERT’s example of student marks (0-10, 10-20, etc.).
Answer:
Introduction

We studied the assumed mean method to simplify mean calculations for grouped data.


Argument 1
  • Step 1: Choose assumed mean (A) like 25 for marks 0-10, 10-20, ..., 40-50.
  • Step 2: Find deviations (di = xi - A). For 20-30, di = 25 - 25 = 0.
  • Step 3: Multiply di by frequency (fi) and sum (Σfidi).

Conclusion

Mean = A + (Σfidi/Σfi). This method reduces large-number errors.

Question 10:
The median of 7, 12, 5, 9, 14, 6, and x is 9. Find x and justify using step notation.
Answer:
Introduction

Median is the middle value in ordered data. We arrange terms and solve for x.


Argument 1
  • Ordered data: 5, 6, 7, 9, 12, 14, x
  • Median (4th term) = 9, so x must be ≥9.

Argument 2

If x = 9, ordered data becomes 5,6,7,9,9,12,14. Median remains 9.


Conclusion

Our textbook confirms median is unaffected if new values maintain the middle position.

Question 11:
A survey records the mode of transport (Bus/Cycle/Walk) for 30 students. Represent this categorical data in a frequency table and bar graph (description only).
Answer:
Introduction

We studied that mode is the most frequent category. A table and graph help visualize such data.


Argument 1
TransportFrequency
Bus12
Cycle10
Walk8

Argument 2

[Diagram: Bar graph with ‘Transport’ on x-axis and ‘Frequency’ on y-axis, bars for Bus (highest), Cycle, Walk].


Conclusion

Our textbook uses similar examples to show categorical data representation.

Question 12:
The mean of 10 observations is 25. If one observation (35) is replaced by 15, find the new mean. Explain the steps with a real-life example.
Answer:
Introduction

We studied that mean is the average of observations. Here, we adjust the mean after modifying a data point.


Argument 1
  • Original sum = 10 × 25 = 250
  • New sum = 250 − 35 + 15 = 230

Argument 2

New mean = 230 ÷ 10 = 23. In real life, if a student’s score drops in a test, the class average decreases similarly.


Conclusion

Our textbook shows such recalculations help analyze data changes efficiently.

Question 13:
Define mean, median, and mode with examples. Explain how each measure of central tendency is useful in real-life scenarios.
Answer:

Mean is the average of all numbers in a dataset. It is calculated by summing all values and dividing by the number of values.
Example: For the dataset [5, 10, 15], the mean is (5 + 10 + 15) / 3 = 10.
Real-life use: Calculating average monthly expenses.

Median is the middle value when data is arranged in ascending order.
Example: For [7, 3, 5], the median is 5 after arranging as [3, 5, 7].
Real-life use: Finding the median income to understand typical earnings.

Mode is the most frequently occurring value in a dataset.
Example: In [2, 4, 4, 6], the mode is 4.
Real-life use: Identifying the most popular shoe size in a store.

Question 14:
The following data represents the heights (in cm) of 10 students: [150, 152, 148, 155, 160, 152, 153, 150, 151, 154]. Calculate the mean, median, and mode of the data. Show all steps clearly.
Answer:

Step 1: Calculate the Mean
Sum of heights = 150 + 152 + 148 + 155 + 160 + 152 + 153 + 150 + 151 + 154 = 1525
Number of students = 10
Mean = 1525 / 10 = 152.5 cm

Step 2: Calculate the Median
Arrange data in ascending order: [148, 150, 150, 151, 152, 152, 153, 154, 155, 160]
Since there are 10 values (even), median is the average of the 5th and 6th values.
Median = (152 + 152) / 2 = 152 cm

Step 3: Calculate the Mode
The most frequent values are 150 and 152 (each appears twice).
Mode = 150 cm and 152 cm (bimodal)

Question 15:
Explain the concept of range in statistics with an example. How is it useful in analyzing data?
Answer:

Range is the difference between the highest and lowest values in a dataset. It gives a measure of how spread out the data is.
Example: For the dataset [12, 18, 22, 10, 15], the range = 22 (highest) - 10 (lowest) = 12.

Usefulness:

  • Helps understand the spread or variability in data.
  • Used in quality control to check consistency (e.g., temperature ranges in a lab).
  • Simple to calculate but sensitive to extreme values.

Question 16:
The following data represents the marks obtained by 30 students in a Mathematics test (out of 50):
32, 28, 45, 36, 42, 25, 38, 44, 30, 27,
35, 41, 39, 29, 33, 37, 40, 34, 26, 31,
43, 47, 48, 49, 50, 46, 24, 23, 22, 21.
Construct a frequency distribution table with class intervals 20-25, 25-30, and so on. Also, find the mean of the data using the direct method.
Answer:

To construct the frequency distribution table, we first organize the given data into the specified class intervals and count the frequency of each interval.


Frequency Distribution Table:
Class Interval | Tally Marks | Frequency (f)
20-25 | |||| | 4
25-30 | |||| | | 6
30-35 | |||| | | 6
35-40 | |||| | 5
40-45 | ||| | 4
45-50 | |||| | | 5

To find the mean using the direct method, we calculate the class mark (midpoint) for each interval and multiply it by the frequency. The formula for mean is:


Mean = Σ(f × x) / Σf

Where:
- x = Class mark = (Lower limit + Upper limit) / 2
- f = Frequency

Calculations:
Class Interval | Class Mark (x) | Frequency (f) | f × x
20-25 | 22.5 | 4 | 90
25-30 | 27.5 | 6 | 165
30-35 | 32.5 | 6 | 195
35-40 | 37.5 | 5 | 187.5
40-45 | 42.5 | 4 | 170
45-50 | 47.5 | 5 | 237.5

Σf = 30
Σ(f × x) = 90 + 165 + 195 + 187.5 + 170 + 237.5 = 1045

Mean = 1045 / 30 ≈ 34.83

Thus, the mean marks obtained by the students is approximately 34.83.

Question 17:
The heights (in cm) of 15 plants in a garden are as follows:
45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100, 105, 110, 115.
Calculate the median and mode of the data. Also, explain why mode is not an appropriate measure of central tendency in this case.
Answer:

To find the median, we first arrange the data in ascending order (already given) and locate the middle value.


Step 1: Find Median
Number of observations (n) = 15 (odd)
Median = [(n + 1) / 2]th observation
Median = (15 + 1) / 2 = 8th observation

The 8th observation in the data is 80 cm.

Step 2: Find Mode
Mode is the observation that occurs most frequently. Here, each height appears only once, so there is no mode.

Explanation: The mode is not appropriate here because all observations are unique, and no value repeats. Mode is useful only when data has repeating values or clusters. In such cases, median or mean better represent central tendency.

Question 18:
The following data represents the marks obtained by 30 students in a Mathematics test (out of 50):
32, 28, 45, 36, 42, 25, 38, 40, 29, 33,
35, 41, 27, 34, 39, 44, 31, 37, 43, 30,
26, 47, 48, 49, 50, 46, 24, 23, 22, 21.

Construct a frequency distribution table with class intervals 20-25, 25-30, and so on. Also, find the mean of the data using the step-deviation method.

Answer:

To solve the problem, we will first construct the frequency distribution table and then calculate the mean using the step-deviation method.


Step 1: Frequency Distribution Table


We group the data into class intervals of width 5, starting from 20-25:


  • Class Interval (Marks) | Frequency (Number of Students)
  • 20-25 | 4 (21, 22, 23, 24)
  • 25-30 | 5 (25, 26, 27, 28, 29)
  • 30-35 | 6 (30, 31, 32, 33, 34, 35)
  • 35-40 | 6 (36, 37, 38, 39, 40, 41)
  • 40-45 | 4 (42, 43, 44, 45)
  • 45-50 | 5 (46, 47, 48, 49, 50)

Step 2: Mean Calculation (Step-Deviation Method)


Let’s assume the assumed mean (A) = 32.5 (midpoint of 30-35).


We calculate the deviation (di) as (xi - A)/h, where h = class width = 5.


  • Class Interval | Midpoint (xi) | Frequency (fi) | di = (xi - A)/h | fidi
  • 20-25 | 22.5 | 4 | (22.5 - 32.5)/5 = -2 | -8
  • 25-30 | 27.5 | 5 | (27.5 - 32.5)/5 = -1 | -5
  • 30-35 | 32.5 | 6 | (32.5 - 32.5)/5 = 0 | 0
  • 35-40 | 37.5 | 6 | (37.5 - 32.5)/5 = 1 | 6
  • 40-45 | 42.5 | 4 | (42.5 - 32.5)/5 = 2 | 8
  • 45-50 | 47.5 | 5 | (47.5 - 32.5)/5 = 3 | 15

Total frequency (Σfi) = 30


Sum of fidi (Σfidi) = -8 + (-5) + 0 + 6 + 8 + 15 = 16


Mean = A + (Σfidi / Σfi) × h


Mean = 32.5 + (16/30) × 5


Mean = 32.5 + (80/30) ≈ 32.5 + 2.67 = 35.17


Thus, the mean marks obtained by the students is approximately 35.17.

Question 19:
The following data represents the marks obtained by 30 students in a Mathematics test (out of 50):
32, 28, 45, 36, 42, 25, 38, 40, 29, 33,
35, 41, 27, 34, 39, 44, 30, 37, 43, 31,
26, 47, 48, 49, 50, 46, 24, 23, 22, 21.

Construct a frequency distribution table with class intervals 20-25, 25-30, and so on. Also, find the class mark for each class interval.

Answer:

To construct the frequency distribution table, we first organize the given data into the specified class intervals and count the frequency (number of students) in each interval. The class mark is calculated as the average of the lower and upper limits of each class interval.


Step 1: Arrange the data in ascending order:
21, 22, 23, 24, 25, 26, 27, 28, 29, 30,
31, 32, 33, 34, 35, 36, 37, 38, 39, 40,
41, 42, 43, 44, 45, 46, 47, 48, 49, 50.

Step 2: Create the frequency distribution table:
Class IntervalTally MarksFrequencyClass Mark
20-25||||4(20+25)/2 = 22.5
25-30|||| |6(25+30)/2 = 27.5
30-35|||| |6(30+35)/2 = 32.5
35-40|||| |6(35+40)/2 = 37.5
40-45||||4(40+45)/2 = 42.5
45-50||||4(45+50)/2 = 47.5

Key Points:

  • The class mark represents the midpoint of a class interval and is used for further calculations like mean, median, etc.
  • Ensure that class intervals are continuous and non-overlapping.
  • Tally marks help in counting frequencies accurately.
Question 20:
The following data represents the marks obtained by 30 students in a Mathematics test (out of 50):
32, 28, 45, 36, 42, 25, 38, 44, 30, 27,
35, 41, 39, 29, 33, 40, 37, 34, 26, 31,
43, 46, 48, 47, 49, 50, 24, 23, 22, 21.
Construct a frequency distribution table with class intervals 20-25, 25-30, and so on. Also, find the mean of the data using the direct method.
Answer:

To construct the frequency distribution table, we first organize the given data into class intervals and count the frequency of each interval.


Step 1: Frequency Distribution Table
Class Interval | Tally Marks | Frequency (f)
20-25 | |||| | 4
25-30 | |||| | | 5
30-35 | |||| || | 6
35-40 | |||| | | 5
40-45 | |||| | 4
45-50 | |||| || | 6

Step 2: Calculate Mean Using Direct Method
To find the mean, we use the formula:
Mean = (Σfixi) / Σfi
where xi = midpoint of each class interval.

Class Interval | Midpoint (xi) | Frequency (fi) | fixi
20-25 | 22.5 | 4 | 90
25-30 | 27.5 | 5 | 137.5
30-35 | 32.5 | 6 | 195
35-40 | 37.5 | 5 | 187.5
40-45 | 42.5 | 4 | 170
45-50 | 47.5 | 6 | 285

Σfi = 30
Σfixi = 90 + 137.5 + 195 + 187.5 + 170 + 285 = 1065

Mean = 1065 / 30 = 35.5

Thus, the mean marks obtained by the students is 35.5.

Question 21:
The following data represents the marks obtained by 30 students in a Mathematics test (out of 50):
32, 45, 28, 36, 42, 25, 39, 48, 29, 35,
41, 27, 33, 46, 30, 38, 44, 26, 34, 40,
47, 31, 37, 43, 24, 49, 50, 23, 22, 45.

Construct a frequency distribution table with class intervals 20-25, 25-30, and so on. Also, find the mean of the data using the assumed mean method (assume a = 37).

Answer:

To construct the frequency distribution table, we first organize the data into the given class intervals and count the frequency of each interval:

  • Class Intervals | Tally Marks | Frequency (f)
  • 20-25 | ||| | 3
  • 25-30 | |||| ||| | 8
  • 30-35 | |||| || | 7
  • 35-40 | |||| | 5
  • 40-45 | |||| | 5
  • 45-50 | || | 2

Now, to find the mean using the assumed mean method (a = 37), we follow these steps:


1. Class Mark (xi): Calculate the midpoint of each class interval.
- 20-25: (20+25)/2 = 22.5
- 25-30: (25+30)/2 = 27.5
- 30-35: (30+35)/2 = 32.5
- 35-40: (35+40)/2 = 37.5
- 40-45: (40+45)/2 = 42.5
- 45-50: (45+50)/2 = 47.5

2. Deviation (di = xi - a): Subtract the assumed mean (a = 37) from each class mark.
- 22.5 - 37 = -14.5
- 27.5 - 37 = -9.5
- 32.5 - 37 = -4.5
- 37.5 - 37 = 0.5
- 42.5 - 37 = 5.5
- 47.5 - 37 = 10.5

3. fi × di: Multiply frequency (f) by deviation (d).
- 3 × (-14.5) = -43.5
- 8 × (-9.5) = -76
- 7 × (-4.5) = -31.5
- 5 × 0.5 = 2.5
- 5 × 5.5 = 27.5
- 2 × 10.5 = 21

4. Sum of fi × di: Add all the products.
- (-43.5) + (-76) + (-31.5) + 2.5 + 27.5 + 21 = -100

5. Mean Formula: Mean = a + (Σfidi / Σfi)
- Σfi = 30 (total frequency)
- Mean = 37 + (-100 / 30) = 37 - 3.33 = 33.67 (approx).

Thus, the mean marks obtained by the students is approximately 33.67.

Question 22:
The following data represents the marks obtained by 30 students in a Mathematics test (out of 50):
32, 45, 28, 36, 42, 25, 39, 48, 30, 22,
35, 41, 27, 33, 47, 29, 38, 44, 31, 26,
40, 37, 43, 34, 46, 24, 49, 23, 50, 21.

Construct a frequency distribution table with class intervals 20-25, 25-30, and so on. Also, find the class mark of each class interval and identify the modal class.

Answer:

To construct the frequency distribution table, we first organize the given data into the specified class intervals and count the frequency of each class. Then, we calculate the class mark and identify the modal class (the class with the highest frequency).


Step 1: Frequency Distribution Table
Class Interval | Tally Marks | Frequency
20-25 | |||| | 4
25-30 | |||| ||| | 7
30-35 | |||| || | 6
35-40 | |||| | | 5
40-45 | |||| | 4
45-50 | |||| | 4

Step 2: Class Mark Calculation
The class mark is calculated as:
(Lower Limit + Upper Limit) / 2

Class Interval | Class Mark
20-25 | (20 + 25)/2 = 22.5
25-30 | (25 + 30)/2 = 27.5
30-35 | (30 + 35)/2 = 32.5
35-40 | (35 + 40)/2 = 37.5
40-45 | (40 + 45)/2 = 42.5
45-50 | (45 + 50)/2 = 47.5

Step 3: Modal Class Identification
The modal class is the class with the highest frequency. Here, the class 25-30 has the highest frequency of 7.

Note: The tally marks are used for quick counting, and the class mark represents the midpoint of each interval. The modal class helps identify where most data points lie.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
A survey recorded the number of siblings for 20 students: 1, 0, 2, 1, 3, 2, 1, 0, 1, 2, 4, 1, 2, 1, 0, 3, 2, 1, 1, 2.
Problem Interpretation: Organize the data in a frequency distribution table and find the mean number of siblings.
Answer:
Problem Interpretation: We need to tabulate the data and compute the mean.
Mathematical Modeling:
Number of Siblings (x)Frequency (f)f×x
030
177
2612
326
414

Solution: Sum of f×x = 29, Total frequency = 20. Mean = 29/20 = 1.45 siblings.
Question 2:
The weights (in kg) of 10 students are: 42, 38, 45, 50, 48, 52, 40, 44, 46, 49.
Problem Interpretation: Calculate the range and median weight. Explain why median is a better measure here.
Answer:
Problem Interpretation: We need to find range and median.
Mathematical Modeling:
  • Range = Maximum − Minimum = 52 − 38 = 14 kg
  • Arranged data: 38, 40, 42, 44, 45, 46, 48, 49, 50, 52

Solution: Median = (45+46)/2 = 45.5 kg. Median is better as it is unaffected by extreme values, unlike mean.
Question 3:
A survey recorded the number of siblings for 20 students: 1, 0, 2, 1, 3, 2, 1, 0, 1, 2, 1, 3, 4, 1, 2, 1, 0, 1, 2, 3.
(i) Organize the data in a frequency distribution table.
(ii) Find the mean number of siblings per student.
Answer:
Problem Interpretation

We need to create a frequency table and calculate the mean for the given data.


Mathematical Modeling
  • Frequency table:
    No. of Siblings (x)Frequency (f)
    03
    18
    25
    33
    41
  • Mean = Σ(f×x)/Σf

Solution

Mean = (0×3 + 1×8 + 2×5 + 3×3 + 4×1)/20 = 25/20 = 1.25 siblings.

Question 4:
The daily wages (in ₹) of 10 workers are: 150, 200, 180, 150, 200, 180, 200, 150, 180, 200.
(i) Find the mode of the data.
(ii) If one worker earning ₹150 is replaced by ₹250, how does the mode change?
Answer:
Problem Interpretation

We studied that mode is the most frequent value. We need to find it before and after data modification.


Mathematical Modeling
  • Original data: 150 appears 3 times (mode)
  • Modified data: Replace one 150 with 250

Solution

(i) Original mode = ₹200 (appears 4 times). (ii) After replacement, ₹200 still appears most frequently (4 times), so mode remains unchanged.

Question 5:
A survey was conducted in a class of 30 students to find their favorite subject. The data collected is: Maths (10), Science (8), English (6), Social Science (4), and Others (2). Represent this data in a pie chart and find the angle for each sector.
Answer:
Problem Interpretation

We studied how to represent data using pie charts. Here, we need to show the favorite subjects of 30 students.


Mathematical Modeling
  • Total students = 30
  • Angle for Maths = (10/30) × 360° = 120°
  • Science = (8/30) × 360° = 96°
  • English = (6/30) × 360° = 72°
  • Social Science = (4/30) × 360° = 48°
  • Others = (2/30) × 360° = 24°

Solution

The angles are: Maths (120°), Science (96°), English (72°), Social Science (48°), Others (24°). [Diagram: Pie chart with labeled sectors]

Question 6:
The mean of 5 numbers is 18. If one number is excluded, the mean becomes 16. Find the excluded number using the sum of observations method.
Answer:
Problem Interpretation

Our textbook shows how mean changes when data is modified. Here, we need to find the excluded number.


Mathematical Modeling
  • Total sum of 5 numbers = 18 × 5 = 90
  • Sum after exclusion = 16 × 4 = 64

Solution

The excluded number = 90 - 64 = 26. We verified this by recalculating the mean of the remaining numbers (64 ÷ 4 = 16).

Question 7:
A survey recorded the number of hours students spent on homework daily. The data is: 2, 3, 4, 2, 5, 1, 6, 4, 3, 2.
Problem Interpretation: Find the mean, median, and mode of the data.
Mathematical Modeling: Use the formulas we studied in class.
Answer:
Problem Interpretation: We need to find central tendencies of the given data.
Mathematical Modeling:
  • Mean = Sum of observations / Total observations
  • Median = Middle value when arranged in order
  • Mode = Most frequent value
Solution:

Mean = (2+3+4+2+5+1+6+4+3+2)/10 = 3.2 hours. Ordered data: 1,2,2,2,3,3,4,4,5,6. Median = (3+3)/2 = 3 hours. Mode = 2 hours (appears 3 times).

Question 8:
The weights (in kg) of 8 students are: 42, 50, 45, 48, 52, 40, 55, 60.
Problem Interpretation: Calculate the range and mean deviation from the median.
Mathematical Modeling: Use the steps from our textbook examples.
Answer:
Problem Interpretation: We need measures of dispersion for the weight data.
Mathematical Modeling:
  • Range = Maximum - Minimum
  • Mean deviation = Σ|X - Median| / n
Solution:

Range = 60 - 40 = 20 kg. Ordered data: 40,42,45,48,50,52,55,60. Median = (48+50)/2 = 49 kg. Mean deviation = (|40-49| + |42-49| + ... + |60-49|)/8 = 6.25 kg.

Question 9:
A survey was conducted in a class of 30 students to find their favorite subject. The data collected is: Math (10), Science (8), English (6), Social Science (4), and Hindi (2).

(i) Represent this data in a bar graph.
(ii) Calculate the central angle for Math in a pie chart.
Answer:
Problem Interpretation

We studied bar graphs and pie charts in our textbook. Here, we need to represent the given data visually.


Mathematical Modeling
  • Bar Graph: X-axis (Subjects), Y-axis (Number of Students)
  • Pie Chart: Central Angle = (Frequency/Total) × 360°

Solution

(i) [Diagram: Bar graph with labeled axes]

(ii) Central Angle for Math = (10/30) × 360° = 120°

Question 10:
The mean of 5 numbers is 18. If one number is excluded, the mean becomes 16.

(i) Find the excluded number.
(ii) If the excluded number is replaced by 12, what is the new mean?
Answer:
Problem Interpretation

Our textbook shows how mean changes when data is modified. We need to find the excluded number and recalculate the mean.


Mathematical Modeling
  • Total of 5 numbers = 18 × 5 = 90
  • Total of 4 numbers = 16 × 4 = 64

Solution

(i) Excluded number = 90 - 64 = 26

(ii) New total = 64 + 12 = 76. New mean = 76/4 = 19

Question 11:

A survey was conducted in a school to find the favorite sports of students. The data collected is represented in the following frequency distribution table:


SportNumber of Students
Cricket45
Football30
Basketball25
Badminton20
Others10

Based on the data, answer the following:


(i) Represent the data using a bar graph.


(ii) Which sport is the most popular among students? Justify your answer.

Answer:

(i) Bar Graph Representation:


To draw the bar graph:


  • Draw two perpendicular lines (X-axis and Y-axis) on graph paper.
  • Label X-axis with sports names: Cricket, Football, Basketball, Badminton, Others.
  • Label Y-axis with the number of students (scale: 1 unit = 5 students).
  • Draw rectangular bars of equal width for each sport with heights corresponding to their frequencies.

Example Diagram: (Visual representation not provided here, but ensure bars are accurately scaled.)


(ii) Most Popular Sport:


Cricket is the most popular sport among students.


Justification: Cricket has the highest frequency (45 students) in the frequency distribution table, indicating it is preferred by the maximum number of students compared to other sports.

Question 12:

The following data represents the marks obtained (out of 50) by 20 students in a Mathematics test:


32, 28, 45, 22, 38, 42, 30, 25, 40, 35, 29, 33, 27, 41, 36, 31, 39, 44, 26, 34


(i) Prepare a grouped frequency distribution table with class intervals 20-25, 25-30, and so on.


(ii) Calculate the mean marks of the students using the assumed mean method (take assumed mean = 35).

Answer:

(i) Grouped Frequency Distribution Table:


Class IntervalFrequency
20-251
25-304
30-356
35-405
40-454

(ii) Mean Calculation (Assumed Mean Method):


Assumed mean (A) = 35


Step 1: Find deviations (d = x - A) for each class midpoint.
Step 2: Multiply deviations by frequencies (f × d).
Step 3: Sum of (f × d) = -15
Step 4: Sum of frequencies (Σf) = 20
Step 5: Mean = A + (Σf × d / Σf) = 35 + (-15 / 20) = 35 - 0.75 = 34.25


Final Answer: The mean marks of the students is 34.25.

Question 13:

A survey was conducted in a school to find the favorite subject of students in Class 9. The data collected is represented in the following pie chart:


Pie Chart Data:
Mathematics: 90°
Science: 72°
Social Science: 60°
English: 54°
Hindi: 84°

If there are 200 students in Class 9, answer the following:


a) How many students prefer Mathematics?
b) Which subject is the least preferred and by how many students?
Answer:

a) To find the number of students who prefer Mathematics, we calculate the proportion of the angle in the pie chart to the total angle (360°).


Number of students = (Angle of Mathematics / Total angle) × Total students
= (90° / 360°) × 200
= 0.25 × 200
= 50 students

b) The least preferred subject is the one with the smallest angle in the pie chart, which is English (54°).


Number of students = (54° / 360°) × 200
= 0.15 × 200
= 30 students
Question 14:

The following frequency distribution table shows the marks obtained by 40 students in a Mathematics test:


Marks Range | Number of Students
0-10 | 5
10-20 | 8
20-30 | 12
30-40 | 10
40-50 | 5

Answer the following:


a) What is the class size of the given data?
b) How many students scored less than 30 marks?
Answer:

a) The class size is the difference between the upper and lower limits of any class interval.


Class size = Upper limit - Lower limit
= 10 - 0
= 10

b) To find the number of students who scored less than 30 marks, we add the frequencies of the classes 0-10, 10-20, and 20-30.


Number of students = 5 (0-10) + 8 (10-20) + 12 (20-30)
= 5 + 8 + 12
= 25 students
Question 15:

A survey was conducted in a school to find the favorite sport of students. The data collected is represented in the following pie chart:


Cricket: 90°
Football: 120°
Basketball: 60°
Badminton: 45°
Others: 45°

If 120 students chose Football, find the total number of students surveyed and the number of students who chose Cricket.

Answer:

To find the total number of students surveyed:


1. The angle for Football is 120° and corresponds to 120 students.
2. Total angle in a pie chart = 360°.
3. Let the total number of students be x.
4. Using proportion: (120° / 360°) = (120 / x).
5. Solving: x = (120 × 360) / 120 = 360 students.

To find the number of students who chose Cricket:


1. Angle for Cricket = 90°.
2. Proportion: (90° / 360°) = (Number of Cricket students / 360).
3. Solving: Number of Cricket students = (90 × 360) / 360 = 90 students.

Thus, the total number of students surveyed is 360, and the number of students who chose Cricket is 90.

Question 16:

The following frequency distribution table shows the marks obtained by 40 students in a Mathematics test:


Marks | Number of Students
0-10 | 5
10-20 | 8
20-30 | 12
30-40 | 10
40-50 | 5

Calculate the mean marks obtained by the students using the direct method.

Answer:

To calculate the mean marks using the direct method, follow these steps:


1. Find the class mark (xi) for each class interval:
- 0-10: (0 + 10)/2 = 5
- 10-20: (10 + 20)/2 = 15
- 20-30: (20 + 30)/2 = 25
- 30-40: (30 + 40)/2 = 35
- 40-50: (40 + 50)/2 = 45

2. Multiply each class mark by its corresponding frequency (fi):
- 5 × 5 = 25
- 15 × 8 = 120
- 25 × 12 = 300
- 35 × 10 = 350
- 45 × 5 = 225

3. Sum all the products: 25 + 120 + 300 + 350 + 225 = 1020.

4. Total number of students (Σfi) = 40.

5. Mean = Σ(fixi) / Σfi = 1020 / 40 = 25.5.

Thus, the mean marks obtained by the students is 25.5.

Question 17:
A survey was conducted in a school to find the favorite sport of students. The data collected is as follows:
Cricket: 35 students
Football: 25 students
Basketball: 20 students
Badminton: 15 students
Tennis: 5 students

Represent this data using a bar graph and answer the following:

  • Which sport is the most popular?
  • What is the total number of students surveyed?
Answer:

To represent the given data in a bar graph, follow these steps:


1. Draw two perpendicular lines (x-axis and y-axis) on graph paper.
2. Label the x-axis with the names of the sports: Cricket, Football, Basketball, Badminton, and Tennis.
3. Label the y-axis with the number of students, using an appropriate scale (e.g., 1 unit = 5 students).
4. Draw rectangular bars of equal width for each sport, with heights corresponding to the number of students.

From the graph:

  • The most popular sport is Cricket as it has the highest bar (35 students).
  • The total number of students surveyed is calculated as:
    35 (Cricket) + 25 (Football) + 20 (Basketball) + 15 (Badminton) + 5 (Tennis) = 100 students.

Bar graphs are useful for comparing categorical data visually, making it easy to identify trends and preferences.

Question 18:
The following data represents the monthly savings (in ₹) of 10 students:
200, 250, 300, 150, 400, 350, 200, 250, 300, 450

Calculate the mean, median, and mode of the data. Which measure of central tendency best represents this data and why?

Answer:

To analyze the given data, let's calculate the mean, median, and mode:


Mean:
Sum of all savings = 200 + 250 + 300 + 150 + 400 + 350 + 200 + 250 + 300 + 450 = ₹2,850
Number of students = 10
Mean = Total savings / Number of students = 2850 / 10 = ₹285

Median:
First, arrange the data in ascending order: 150, 200, 200, 250, 250, 300, 300, 350, 400, 450
Since there are 10 observations (even number), the median is the average of the 5th and 6th values.
Median = (250 + 300) / 2 = ₹275

Mode:
The most frequently occurring values are ₹200, ₹250, and ₹300 (each appears twice).
Thus, the data is multimodal with modes: ₹200, ₹250, and ₹300.

The median (₹275) best represents this data because:

  • The mean is affected by the extreme value (₹450), pulling it higher.
  • The mode is not unique, making it less reliable.
  • The median gives a central value that is not skewed by outliers.

In real-life scenarios like savings, the median is often a better measure when data has outliers.

Question 19:
A survey was conducted in a school to find the favorite subject of students. The data collected is as follows:
Mathematics: 45 students
Science: 30 students
Social Science: 25 students
English: 20 students
Hindi: 15 students
Represent this data using a bar graph and find the subject preferred by the majority of students.
Answer:

To represent the given data in a bar graph, follow these steps:


1. Draw two perpendicular lines (X-axis and Y-axis) on graph paper.
2. Label the X-axis with the subjects: Mathematics, Science, Social Science, English, Hindi.
3. Label the Y-axis with the number of students, scaling appropriately (e.g., 5 students per unit).
4. Draw rectangular bars for each subject with heights corresponding to the number of students.

From the data:
- The highest number of students (45) prefer Mathematics.
- Thus, Mathematics is the subject preferred by the majority.

Additional Tip: Always ensure the bars are of equal width and evenly spaced for clarity.

Question 20:
The following data represents the monthly savings (in ₹) of 10 students:
200, 250, 300, 350, 400, 450, 500, 550, 600, 650
Calculate the mean savings and explain how it helps in understanding the overall trend.
Answer:

To calculate the mean savings:


1. Add all the savings:
200 + 250 + 300 + 350 + 400 + 450 + 500 + 550 + 600 + 650 = 4250.
2. Divide the total by the number of students (10):
4250 ÷ 10 = 425.

The mean savings is ₹425.

Explanation: The mean gives the average savings, helping to understand the central tendency of the data. It shows that, on average, each student saves around ₹425 per month.


Additional Insight: The mean is useful for comparing larger datasets or predicting future trends.

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