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Statistics – CBSE NCERT Study Resources

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Study Materials

10th

10th - Mathematics

Statistics

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Overview

This chapter introduces students to the fundamental concepts of Statistics, a branch of Mathematics that deals with data collection, organization, analysis, interpretation, and presentation. The CBSE Grade 10 curriculum focuses on measures of central tendency (mean, median, mode) and graphical representation of data (histograms, frequency polygons, etc.).

Key Concepts

Statistics: The science of collecting, organizing, analyzing, interpreting, and presenting data.

Measures of Central Tendency

  • Mean: The average of all observations.
  • Median: The middle value when data is arranged in ascending/descending order.
  • Mode: The most frequently occurring observation in a dataset.

Graphical Representation of Data

  • Histogram: A bar graph representing continuous class intervals.
  • Frequency Polygon: A line graph formed by joining midpoints of histogram bars.
  • Ogive (Cumulative Frequency Curve): A graph showing cumulative frequencies.

Formulas

Mean (Direct Method): \(\bar{x} = \frac{\sum f_i x_i}{\sum f_i}\)

Mean (Assumed Mean Method): \(\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}\)

Median: \(Median = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h\)

Mode: \(Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h\)

Applications

Statistics is widely used in economics, social sciences, business, and daily life for decision-making based on data trends.

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 mean of the first five natural numbers?
Answer:
3
Question 2:
Find the median of 7, 3, 5, 9, 2.
Answer:
5
Question 3:
What is the mode of 2, 3, 4, 2, 5, 2?
Answer:
2
Question 4:
Calculate the range of 10, 15, 8, 20, 12.
Answer:
12
Question 5:
If the mean of 5 numbers is 10, what is their total sum?
Answer:
50
Question 6:
What is the class mark of the class interval 10-20?
Answer:
15
Question 7:
Find the median of the first 10 even numbers.
Answer:
11
Question 8:
What is the empirical relationship between mean, median, and mode?
Answer:

Mode = 3 Median - 2 Mean

Question 9:
If the mean of observations x, x+2, x+4, x+6, x+8 is 11, find x.
Answer:
7
Question 10:
What is the upper limit of the class interval 30-40?
Answer:
40
Question 11:
Find the mean of the first five prime numbers.
Answer:
5.6
Question 12:
What is the cumulative frequency of the last class in a frequency distribution?
Answer:
Total frequency
Question 13:
Define mean in statistics.
Answer:

The mean is the average of all the numbers in a data set.
It is calculated by adding all the values and dividing by the total number of values.
Formula: Mean = (Sum of all observations) / (Number of observations)

Question 14:
What is the class mark of a class interval?
Answer:

The class mark is the midpoint of a class interval.
It is calculated as: (Lower limit + Upper limit) / 2.
For example, for the interval 10-20, the class mark is (10 + 20)/2 = 15.

Question 15:
How is the median of a data set determined?
Answer:

The median is the middle value when data is arranged in ascending or descending order.
For odd number of observations: Median = Middle term.
For even number of observations: Median = Average of two middle terms.

Question 16:
What does the mode represent in a data set?
Answer:

The mode is the value that appears most frequently in a data set.
A data set can have one mode, more than one mode (bimodal), or no mode if all values are unique.

Question 17:
Explain the term frequency in statistics.
Answer:

Frequency refers to the number of times a particular observation occurs in a data set.
For example, if the value '5' appears 3 times in a data set, its frequency is 3.

Question 18:
What is the purpose of a histogram?
Answer:

A histogram is a graphical representation of the distribution of numerical data.
It uses bars to show the frequency of data within equal intervals, helping visualize patterns and trends.

Question 19:
How is the range of a data set calculated?
Answer:

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

Question 20:
What is the difference between discrete and continuous data?
Answer:

Discrete data can only take specific values (e.g., number of students).
Continuous data can take any value within a range (e.g., height, weight).
Discrete data is counted, while continuous data is measured.

Question 21:
Why do we use grouped frequency distribution?
Answer:

Grouped frequency distribution organizes large data sets into intervals (classes) for easier analysis.
It simplifies data presentation and helps identify patterns that might be hidden in raw 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:
Define mean in statistics and write its formula for ungrouped data.
Answer:

The mean is the average of all observations in a dataset.
For ungrouped data, it is calculated as:
Mean = (Sum of all observations) / (Number of observations)

Question 2:
What is the class mark of a class interval? Give the formula.
Answer:

The class mark is the midpoint of a class interval.
Formula: Class Mark = (Lower Limit + Upper Limit) / 2

Question 3:
How is the mode of grouped data determined using the modal class?
Answer:

The mode for grouped data is estimated using the modal class (class with highest frequency).
Formula: Mode = L + [(f₁ - f₀) / (2f₁ - f₀ - f₂)] × h
Where:
L = Lower limit of modal class
f₁ = Frequency of modal class
f₀ = Frequency of class before modal class
f₂ = Frequency of class after modal class
h = Class width

Question 4:
Differentiate between discrete and continuous data with examples.
Answer:

Discrete data can only take specific values (e.g., number of students in a class).
Continuous data can take any value within a range (e.g., height of students).

Question 5:
Explain the step-deviation method for finding the mean of grouped data.
Answer:

This method simplifies calculations by assuming a mean (A) and using deviations in class intervals.
Steps:
1. Choose assumed mean (A)
2. Find deviation (d = (x - A)/h) where h is class width
3. Calculate mean using: Mean = A + (∑fd / ∑f) × h

Question 6:
What is the empirical relationship between mean, median and mode?
Answer:

The empirical relationship is:
Mode = 3 Median - 2 Mean

This holds for moderately skewed distributions.

Question 7:
How is the median of grouped data calculated? Write the formula.
Answer:

Median for grouped data is calculated using:
Median = L + [(N/2 - cf) / f] × h
Where:
L = Lower limit of median class
N = Total frequency
cf = Cumulative frequency before median class
f = Frequency of median class
h = Class width

Question 8:
What does the cumulative frequency represent in statistics?
Answer:

Cumulative frequency is the running total of frequencies up to a particular class.
It helps in finding medians and quartiles in grouped data.

Question 9:
Why do we use class intervals in grouped frequency distribution?
Answer:

Class intervals:
1. Simplify large datasets
2. Make patterns more visible
3. Enable calculation of measures like mean, mode for grouped data
4. Provide better data organization

Question 10:
Calculate the mean of first 10 natural numbers using direct method.
Answer:

First 10 natural numbers: 1, 2, 3, ..., 10
Sum = 1+2+3+...+10 = 55
Number of observations = 10
Mean = Sum/Number = 55/10 = 5.5

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 represents the central value of a data set.


Example: For the data set 5, 10, 15, 20:
Mean = (5 + 10 + 15 + 20) / 4 = 50 / 4 = 12.5

Significance: The mean helps in understanding the overall trend of the data and is widely used in real-life scenarios like calculating average marks, temperatures, etc.

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

Discrete data consists of distinct, separate values that can be counted, while continuous data can take any value within a range and is measured.


Examples:
  • Discrete: Number of students in a class (e.g., 30, 31, 32)
  • Continuous: Height of students (e.g., 150.5 cm, 160.2 cm)

Discrete data is finite, whereas continuous data is infinite and can include fractions or decimals.

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

The median is the middle value in a data set arranged in ascending or descending order. For an odd number of observations, it is the value at the (n+1)/2 position.


Example: Data set: 12, 7, 21, 5, 9
Step 1: Arrange in order: 5, 7, 9, 12, 21
Step 2: Number of observations (n) = 5
Step 3: Median position = (5 + 1) / 2 = 3rd value
Step 4: Median = 9

The median divides the data into two equal halves.

Question 4:
What is a frequency distribution table? Create one for the data set: 3, 5, 3, 7, 5, 3, 9, 5, 7, 3.
Answer:

A frequency distribution table summarizes data by showing how often each value occurs. It has two columns: Value and Frequency.


Example:
ValueFrequency
34
53
72
91

This table helps in analyzing the distribution and patterns in the data.

Question 5:
Describe the empirical relationship between mean, median, and mode for a moderately skewed distribution.
Answer:

For a moderately skewed distribution, the empirical relationship is given by:
Mode = 3 Median - 2 Mean


This relationship helps estimate one measure if the other two are known. For example, if Mean = 30 and Median = 28:
Mode = 3(28) - 2(30) = 84 - 60 = 24


This formula is useful when the data is not perfectly symmetrical but follows a rough pattern.

Question 6:
Explain how to calculate the median for an odd and even number of observations.
Answer:

For odd number of observations, arrange data in ascending order and pick the middle value.


For even number of observations, arrange data and take the average of the two middle values.


Example (Odd): Data: 3, 1, 5 → Sorted: 1, 3, 5 → Median = 3.


Example (Even): Data: 4, 2, 6, 8 → Sorted: 2, 4, 6, 8 → Median = (4+6)/2 = 5.

Question 7:
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.


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


Example: Marks of students (0-10, 10-20, etc.) with frequencies (5, 10, etc.).

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:
The mean of the following frequency distribution is 18. Find the missing frequency f:
Class Interval0-1010-2020-3030-4040-50
Frequency5f15105
Answer:
Introduction

We studied how to find missing frequencies using the mean formula. Here, the total frequency is needed.


Argument 1
  • Given mean = 18, total frequency = 5 + f + 15 + 10 + 5 = 35 + f.
  • Mid-points: 5, 15, 25, 35, 45.
  • Sum of (mid-point × frequency) = (5×5) + (15×f) + (25×15) + (35×10) + (45×5) = 25 + 15f + 375 + 350 + 225 = 975 + 15f.

Argument 2
  • Mean = Sum / Total frequency ⇒ 18 = (975 + 15f) / (35 + f).
  • Solving: 630 + 18f = 975 + 15f ⇒ 3f = 345 ⇒ f = 115.

Conclusion

Our textbook shows similar problems. The missing frequency f is 115.

Question 2:
The following table shows the daily wages of 50 workers. Find the median wage:
Daily Wage (₹)100-120120-140140-160160-180180-200
Number of Workers12148610
Answer:
Introduction

We learned to find median for grouped data using cumulative frequency. Here, N = 50.


Argument 1
  • Median class is where cumulative frequency ≥ N/2 (25).
  • Cumulative frequencies: 12, 26, 34, 40, 50.
  • Median class is 120-140 (cf = 26).

Argument 2
  • l = 120, h = 20, f = 14, cf = 12.
  • Median = l + [(N/2 - cf)/f] × h = 120 + [(25 - 12)/14] × 20 = 120 + (13/14) × 20 ≈ 120 + 18.57 = ₹138.57.

Conclusion

Our textbook explains this method. The median daily wage is ₹138.57.

Question 3:
The mean of the following frequency distribution is 18. Find the missing frequency f:
Class Interval0-1010-2020-3030-4040-50
Frequency5f1078
Answer:
Introduction

We studied mean calculation for grouped data using the formula: Mean = Σ(fixi)/Σfi.


Argument 1
  • Mid-points (xi): 5,15,25,35,45
  • Σfi = 30 + f
  • Σ(fixi) = 25 + 15f + 250 + 245 + 360 = 880 + 15f

Argument 2

Using mean formula: (880 + 15f)/(30 + f) = 18 → 880 + 15f = 540 + 18f → 340 = 3f


Conclusion

Solving gives f = 113.33 ≈ 113 (frequency must be whole number). Our textbook shows similar problems in Chapter 14.

Question 4:
Explain the step-deviation method to find mean with an example from daily life. Why is it better than direct method when class intervals are equal?
Answer:
Introduction

We studied three mean calculation methods. Step-deviation simplifies calculations when intervals are equal.


Argument 1
  • Formula: Mean = a + (Σfiui/Σfi) × h
  • Example: Finding average daily screen time from
    Hours0-22-44-6
    Students10155

Argument 2

It reduces large numbers by choosing assumed mean (a) and scaling factor (h). Our textbook shows this in Example 5 of Statistics.


Conclusion

This method minimizes calculation errors for equal intervals, making it efficient for surveys like population studies.

Question 5:
The following table shows the daily wages of 50 workers. Find the median wage:
Wages (₹)100-120120-140140-160160-180180-200
Workers12148610
Answer:
Introduction

We learned to calculate median for grouped data using cumulative frequency. NCERT Example 6 in Chapter 14 explains this.


Argument 1
  • Total workers = 50 ⇒ Median class = 25th term.
  • Cumulative frequencies: 12, 26, 34, 40, 50.
  • Median class = 120-140 (since 25th term lies here).

Argument 2
  • Using formula: Median = L + [(N/2 − cf)/f] × h.
  • L = 120, cf = 12, f = 14, h = 20 ⇒ Median = 120 + [(25 − 12)/14] × 20 ≈ 138.57.

Conclusion

The median daily wage is ₹138.57, showing central tendency of the data.

Question 6:
The mean of the following frequency distribution is 18. Find the missing frequency f:
Class Interval0-1010-2020-3030-4040-50
Frequency5f1078
Answer:
Introduction
We need to find the missing frequency f using the given mean of 18.
Argument 1
  • First, we calculate mid-points (x): 5, 15, 25, 35, 45.
  • Sum of frequencies = 5 + f + 10 + 7 + 8 = 30 + f.
Argument 2
  • Sum of (f × x) = (5×5) + (15×f) + (10×25) + (7×35) + (8×45) = 25 + 15f + 250 + 245 + 360 = 880 + 15f.
  • Mean = Sum of (f × x) / Sum of frequencies ⇒ 18 = (880 + 15f) / (30 + f). Solving, we get f = 20.
Conclusion
Thus, the missing frequency f is 20, as derived step-by-step.
Question 7:
The following table shows the daily wages of workers. Find the median wage:
Daily Wage (₹)100-120120-140140-160160-180180-200
Number of Workers12148610
Answer:
Introduction
We studied how to find the median for grouped data using cumulative frequency.
Argument 1
  • Total workers = 12 + 14 + 8 + 6 + 10 = 50. Median class is where N/2 = 25 lies, i.e., 120-140.
  • Cumulative frequency before median class = 12.
Argument 2
  • Using formula: Median = L + [(N/2 - cf)/f] × h. Here, L = 120, cf = 12, f = 14, h = 20.
  • Median = 120 + [(25 - 12)/14] × 20 = 120 + 18.57 = ₹138.57.
Conclusion
The median daily wage is ₹138.57, calculated using the median formula for grouped data.
Question 8:
The mode of the following data is 55. Find the missing frequency x if the total frequency is 100:
Class Interval0-2020-4040-6060-8080-100
Frequency1520x1817
Answer:
Introduction

We know mode is the value with the highest frequency. Here, the modal class is 40-60 as mode is given as 55.


Argument 1
  • Total frequency = 15 + 20 + x + 18 + 17 = 70 + x = 100 ⇒ x = 30
  • Modal class: 40-60 with frequency f1 = x = 30
  • f0 = 20 (preceding class), f2 = 18 (succeeding class), h = 20

Argument 2

Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] × h ⇒ 55 = 40 + [(30-20)/(60-20-18)] × 20 ⇒ 15 = (10/22) × 20 ≈ 9.09. This confirms x = 30 is correct.


Conclusion

The missing frequency x is 30, ensuring the mode aligns with the given data.

Question 9:
The following table shows the daily wages of 50 workers. Find the median wage:
Daily Wage (₹)100-120120-140140-160160-180180-200
Number of Workers12148610
Answer:
Introduction

We learned to calculate median for grouped data using cumulative frequency. NCERT Example 6 in Chapter 14 explains this.


Argument 1
  • Total workers = 50. Median class is where cumulative frequency ≥ 25.
  • Cumulative frequencies: 12, 26, 34, 40, 50. Median class is 120-140.

Argument 2
  • Using formula: Median = L + [(N/2 - cf) / f] × h.
  • L = 120, cf = 12, f = 14, h = 20 ⇒ Median = 120 + [(25 - 12)/14] × 20 = 120 + 18.57 = ₹138.57.

Conclusion

The median daily wage is ₹138.57, calculated using the median formula for grouped data.

Question 10:
The mean of the following frequency distribution is 18. Find the missing frequency f:
Class Interval0-1010-2020-3030-4040-50
Frequency5f15105
Answer:
Introduction
We studied how to find missing frequencies using the mean formula. Here, the total frequency is 35 + f.

Argument 1
Using the formula: Mean = Σ(fixi)/Σfi, we calculate midpoints (5, 15, 25, 35, 45) and products (25, 15f, 375, 350, 225).

Argument 2
Sum of products = 975 + 15f. Given mean = 18, we solve (975 + 15f)/(35 + f) = 18 → f = 25.

Conclusion
The missing frequency is 25, verified by substituting back into the equation.
Question 11:
Explain the step-deviation method to find mean with an example of daily wages (in ₹) of 30 workers: 100, 110, 120, 130, 140, 150.
Answer:
Introduction
Our textbook shows three methods to calculate mean. The step-deviation method simplifies calculations for large data.

Argument 1
Steps:
  • Choose assumed mean (A = 125)
  • Find deviations (di = xi - A)
  • Divide by class width (h = 10) to get ui

Argument 2
Example: For wages 100-150, Σfiui = -15, Σfi = 30. Mean = 125 + (-15/30)×10 = 120.

Conclusion
This method reduces arithmetic operations, as shown in the NCERT example.
Question 12:
The mean of the following frequency distribution is 18. Find the missing frequency f:
Class Interval0-1010-2020-3030-4040-50
Frequency5f1078
Answer:
Introduction
We need to find the missing frequency f using the given mean (18) and frequency distribution.
Argument 1
  • Mid-points: 5, 15, 25, 35, 45
  • Sum of frequencies = 5 + f + 10 + 7 + 8 = 30 + f

Argument 2
  • Sum of (mid-point × frequency) = (5×5) + (15×f) + (25×10) + (35×7) + (45×8) = 25 + 15f + 250 + 245 + 360 = 880 + 15f
  • Mean = Sum / Total frequency ⇒ 18 = (880 + 15f) / (30 + f)
  • Solving: 540 + 18f = 880 + 15f ⇒ 3f = 340 ⇒ f = 113.33 (approx 113)

Conclusion
The missing frequency f is approximately 113.
Question 13:
The following table shows the daily wages of 50 workers. Calculate the median wage:
Daily Wage (₹)100-120120-140140-160160-180180-200
Number of Workers12148610
Answer:
Introduction
We studied how to find the median for grouped data using cumulative frequency.
Argument 1
  • Total workers = 50 ⇒ Median class = (N/2)th term = 25th term
  • Cumulative frequencies: 12, 26, 34, 40, 50
  • Median class is 120-140 (since 25 falls in 26)

Argument 2
  • l = 120, cf = 12, f = 14, h = 20
  • Median = l + [(N/2 - cf)/f] × h = 120 + [(25 - 12)/14] × 20
  • Median = 120 + (13/14) × 20 ≈ 120 + 18.57 = ₹138.57

Conclusion
The median daily wage is approximately ₹138.57, useful for analyzing income distribution.
Question 14:
The following table shows the daily wages of 50 workers in a factory. Calculate the mean daily wage using the step-deviation method. Explain each step clearly.

Daily Wages (in ₹) | Number of Workers
100-120 | 12
120-140 | 14
140-160 | 8
160-180 | 6
180-200 | 10
Answer:

To calculate the mean daily wage using the step-deviation method, follow these steps:



Step 1: Prepare the frequency distribution table with assumed mean (A) and class interval (h)
Let A = 150 (assumed mean from the class 140-160)
Let h = 20 (class width)

Daily Wages (₹) | Number of Workers (fi) | Mid-point (xi) | di = (xi - A)/h | fidi
100-120 | 12 | 110 | (110-150)/20 = -2 | 12 × -2 = -24
120-140 | 14 | 130 | (130-150)/20 = -1 | 14 × -1 = -14
140-160 | 8 | 150 | (150-150)/20 = 0 | 8 × 0 = 0
160-180 | 6 | 170 | (170-150)/20 = 1 | 6 × 1 = 6
180-200 | 10 | 190 | (190-150)/20 = 2 | 10 × 2 = 20

Step 2: Calculate Σfi and Σfidi
Σfi = 12 + 14 + 8 + 6 + 10 = 50
Σfidi = -24 + (-14) + 0 + 6 + 20 = -12

Step 3: Apply the step-deviation formula for mean
Mean = A + (Σfidi / Σfi) × h
Mean = 150 + (-12/50) × 20
Mean = 150 + (-0.24) × 20
Mean = 150 - 4.8
Mean = ₹145.20

The mean daily wage of the workers is ₹145.20. The step-deviation method simplifies calculations by reducing large numbers to smaller deviations.

Question 15:
The following data gives the marks obtained by 100 students in a mathematics test. Draw a less than type ogive and hence find the median marks. Show all steps and the graph.

Marks | Number of Students
0-10 | 5
10-20 | 10
20-30 | 18
30-40 | 30
40-50 | 20
50-60 | 12
60-70 | 5
Answer:

To find the median marks using a less than type ogive, follow these steps:



Step 1: Prepare the cumulative frequency table (less than type)
Marks (Less than) | Cumulative Frequency
10 | 5
20 | 5 + 10 = 15
30 | 15 + 18 = 33
40 | 33 + 30 = 63
50 | 63 + 20 = 83
60 | 83 + 12 = 95
70 | 95 + 5 = 100

Step 2: Plot the ogive
On graph paper:
1. X-axis: Upper limits of class intervals (10, 20, ..., 70)
2. Y-axis: Cumulative frequencies (5, 15, ..., 100)
3. Plot points (10,5), (20,15), ..., (70,100)
4. Join points with a smooth freehand curve

Step 3: Find median using N/2
N = 100 ⇒ N/2 = 50
Locate 50 on Y-axis, draw horizontal line to meet ogive, then vertical line to X-axis

Step 4: Determine median value
The point where the horizontal line meets the ogive corresponds to approximately 35 on X-axis

Thus, the median marks are approximately 35. The ogive provides a graphical representation of cumulative frequencies, making it easier to find percentiles.

Question 16:
The following table shows the daily wages of 50 workers in a factory. Calculate the mean daily wage using the step-deviation method. Explain each step clearly.

Daily Wages (in ₹): 100-120, 120-140, 140-160, 160-180, 180-200
Number of Workers: 12, 14, 8, 6, 10
Answer:

To calculate the mean daily wage using the step-deviation method, follow these steps:



Step 1: Prepare the frequency distribution table with assumed mean (A) and class interval (h)
Let A = 150 (mid-value of 140-160)
h = 20 (class width)

Step 2: Find the deviation (di) and step-deviation (ui)
ui = (xi - A)/h

Class Interval | Mid-value (xi) | Frequency (fi) | di = xi - A | ui = di/h | fiui
100-120 | 110 | 12 | -40 | -2 | -24
120-140 | 130 | 14 | -20 | -1 | -14
140-160 | 150 | 8 | 0 | 0 | 0
160-180 | 170 | 6 | 20 | 1 | 6
180-200 | 190 | 10 | 40 | 2 | 20

Step 3: Calculate Σfi and Σfiui
Σfi = 50
Σfiui = -12

Step 4: Apply the step-deviation formula
Mean = A + (Σfiui/Σfi) × h
= 150 + (-12/50) × 20
= 150 - 4.8
= ₹145.20

The mean daily wage is ₹145.20. This method simplifies calculations by reducing large numbers to smaller deviations.

Question 17:
The following table shows the daily wages of 50 workers in a factory. Calculate the mean daily wage using the step-deviation method. Justify why this method is suitable here.

Daily Wages (in ₹): 100-120, 120-140, 140-160, 160-180, 180-200
Number of Workers: 12, 14, 8, 6, 10
Answer:

To calculate the mean daily wage using the step-deviation method, follow these steps:



Step 1: Create a table with necessary columns
Class Interval | Frequency (fi) | Mid-point (xi) | Assumed Mean (A = 150) | di = xi - A | ui = di/h (h = 20) | fiui
100-120 | 12 | 110 | 150 | -40 | -2 | -24
120-140 | 14 | 130 | 150 | -20 | -1 | -14
140-160 | 8 | 150 | 150 | 0 | 0 | 0
160-180 | 6 | 170 | 150 | 20 | 1 | 6
180-200 | 10 | 190 | 150 | 40 | 2 | 20

Step 2: Calculate Σfi and Σfiui
Σfi = 50
Σfiui = -24 + (-14) + 0 + 6 + 20 = -12

Step 3: Apply the step-deviation formula
Mean = A + (Σfiui / Σfi) × h
Mean = 150 + (-12/50) × 20
Mean = 150 - 4.8
Mean = ₹145.2

Why step-deviation method is suitable: This method simplifies calculations when class intervals are equal (here, h = 20). It reduces large numbers (like xi) to smaller values (ui), minimizing errors. It is efficient for grouped data with uniform class widths.

Question 18:
The following table shows the daily wages of workers in a factory. Calculate the mean daily wage using the step-deviation method. Explain each step clearly.

Daily Wages (in ₹): 100-120, 120-140, 140-160, 160-180, 180-200
Number of Workers: 10, 15, 20, 12, 8
Answer:

To calculate the mean daily wage using the step-deviation method, follow these steps:



Step 1: Prepare the frequency distribution table
Class Interval | Frequency (fi) | Midpoint (xi) | Assumed Mean (A = 150) | di = xi - A | ui = di/h (h = 20) | fiui
100-120 | 10 | 110 | 150 | -40 | -2 | -20
120-140 | 15 | 130 | 150 | -20 | -1 | -15
140-160 | 20 | 150 | 150 | 0 | 0 | 0
160-180 | 12 | 170 | 150 | 20 | 1 | 12
180-200 | 8 | 190 | 150 | 40 | 2 | 16

Step 2: Calculate Σfi and Σfiui
Σfi = 10 + 15 + 20 + 12 + 8 = 65
Σfiui = -20 + (-15) + 0 + 12 + 16 = -7

Step 3: Apply the step-deviation formula
Mean = A + (Σfiui / Σfi) × h
Mean = 150 + (-7 / 65) × 20
Mean = 150 - (140 / 65)
Mean = 150 - 2.15
Mean = ₹147.85

Thus, the mean daily wage of the workers is ₹147.85. The step-deviation method simplifies calculations by reducing large numbers to smaller values (ui), making it efficient for grouped data.

Question 19:
The following table shows the daily wages of 50 workers in a factory. Calculate the mean daily wage using the step-deviation method. Explain each step clearly.

Daily Wages (in ₹) | Number of Workers
200-250 | 12
250-300 | 14
300-350 | 8
350-400 | 10
400-450 | 6
Answer:

To calculate the mean daily wage using the step-deviation method, follow these steps:


Step 1: Identify the assumed mean (A) and class width (h)
Let us assume A = 325 (mid-point of the class 300-350) and h = 50 (class width).

Step 2: Create a table with necessary columns
Daily Wages | Mid-Value (xi) | fi | di = (xi - A)/h | fidi
200-250 | 225 | 12 | (225-325)/50 = -2 | -24
250-300 | 275 | 14 | (275-325)/50 = -1 | -14
300-350 | 325 | 8 | (325-325)/50 = 0 | 0
350-400 | 375 | 10 | (375-325)/50 = 1 | 10
400-450 | 425 | 6 | (425-325)/50 = 2 | 12

Step 3: Calculate Σfi and Σfidi
Σfi = 50
Σfidi = -24 -14 +0 +10 +12 = -16

Step 4: Apply the step-deviation formula
Mean = A + (Σfidi/Σfi) × h
Mean = 325 + (-16/50) × 50
Mean = 325 - 16 = ₹309

Thus, the mean daily wage is ₹309. The step-deviation method simplifies calculations by reducing large numbers.

Question 20:
The following table shows the daily wages of 50 workers in a factory. Calculate the mean daily wage using the step-deviation method. Explain each step clearly.

Daily Wages (in ₹) | Number of Workers
------------------|------------------
200 - 250 | 12
250 - 300 | 15
300 - 350 | 10
350 - 400 | 8
400 - 450 | 5
Answer:

To calculate the mean daily wage using the step-deviation method, follow these steps:


Step 1: Identify the class intervals and frequencies
The given data is already grouped into class intervals with corresponding frequencies.

Step 2: Find the mid-point (xi) of each class interval
The mid-point is calculated as (Lower Limit + Upper Limit) / 2.

For the first class (200 - 250): (200 + 250) / 2 = 225
For the second class (250 - 300): (250 + 300) / 2 = 275
For the third class (300 - 350): (300 + 350) / 2 = 325
For the fourth class (350 - 400): (350 + 400) / 2 = 375
For the fifth class (400 - 450): (400 + 450) / 2 = 425

Step 3: Choose an assumed mean (A)
Let’s assume A = 325 (mid-point of the third class).

Step 4: Calculate the deviation (di) of each mid-point from the assumed mean
di = xi - A

For the first class: 225 - 325 = -100
For the second class: 275 - 325 = -50
For the third class: 325 - 325 = 0
For the fourth class: 375 - 325 = 50
For the fifth class: 425 - 325 = 100

Step 5: Calculate the step-deviation (ui)
ui = di / h, where h is the class width (50 in this case).

For the first class: -100 / 50 = -2
For the second class: -50 / 50 = -1
For the third class: 0 / 50 = 0
For the fourth class: 50 / 50 = 1
For the fifth class: 100 / 50 = 2

Step 6: Multiply ui by the frequency (fi) and calculate Σfiui
For the first class: 12 × (-2) = -24
For the second class: 15 × (-1) = -15
For the third class: 10 × 0 = 0
For the fourth class: 8 × 1 = 8
For the fifth class: 5 × 2 = 10

Σfiui = -24 + (-15) + 0 + 8 + 10 = -21

Step 7: Calculate the mean using the formula:
Mean = A + (Σfiui / Σfi) × h
Σfi = 50 (total number of workers)

Mean = 325 + (-21 / 50) × 50
Mean = 325 - 21
Mean = ₹304

Thus, the mean daily wage of the workers is ₹304. The step-deviation method simplifies calculations by reducing large numbers to smaller deviations.

Question 21:
The following data represents the marks obtained by 30 students in a Mathematics test (out of 100). Construct a cumulative frequency table and draw an ogive (less than type). Use it to estimate the median marks.

Marks | Number of Students
0-20 | 3
20-40 | 5
40-60 | 10
60-80 | 8
80-100 | 4
Answer:

To estimate the median marks using the ogive (less than type), follow these steps:


Step 1: Create a cumulative frequency table (less than type)
Marks | Cumulative Frequency (CF)
Less than 20 | 3
Less than 40 | 3 + 5 = 8
Less than 60 | 8 + 10 = 18
Less than 80 | 18 + 8 = 26
Less than 100 | 26 + 4 = 30

Step 2: Plot the ogive (less than type)
  • Take marks on the x-axis and CF on the y-axis.
  • Plot points: (20,3), (40,8), (60,18), (80,26), (100,30).
  • Join these points smoothly to form the ogive.

Step 3: Estimate the median
Median = (n/2)th term = (30/2)th term = 15th term.
Locate 15 on the y-axis, draw a horizontal line to meet the ogive, then drop a perpendicular to the x-axis.
The x-coordinate at this point is the median.

From the ogive, the median marks are approximately 58. The ogive helps visualize cumulative data and estimate central tendencies graphically.

Question 22:
The following data gives the marks obtained by 100 students in a mathematics test. Draw a less than ogive and find the median marks from the graph. Also, verify it using the formula.

Marks: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60
No. of Students: 5, 10, 20, 30, 20, 15
Answer:

To find the median marks using the less than ogive and verify it:



Step 1: Prepare the cumulative frequency table
Marks | Cumulative Frequency (Less than type)
Less than 10 | 5
Less than 20 | 15 (5+10)
Less than 30 | 35 (15+20)
Less than 40 | 65 (35+30)
Less than 50 | 85 (65+20)
Less than 60 | 100 (85+15)

Step 2: Plot the ogive
• X-axis: Upper limits of class intervals (10,20,30,40,50,60)
• Y-axis: Cumulative frequencies (5,15,35,65,85,100)
• Plot points and join them with a smooth curve

Step 3: Find median graphically
• Locate N/2 = 50 on Y-axis
• Draw horizontal line to meet ogive
• Drop perpendicular to X-axis
• The meeting point (~36) is the median

Step 4: Verify using formula
Median class = 30-40 (where cumulative frequency ≥ N/2)
l = 30, cf = 35, f = 30, h = 10
Median = l + [(N/2 - cf)/f] × h
= 30 + [(50-35)/30] × 10
= 30 + 5
= 35

The median marks are 35. The slight difference (36 vs 35) occurs due to graph approximation. Ogives help visualize data distribution and central tendency.

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 daily wages (in ₹) of 30 workers: 100, 120, 150, 100, 120, 180, 200, 150, 120, 100, 180, 200, 150, 120, 100, 150, 180, 200, 120, 150, 100, 120, 150, 180, 200, 150, 120, 100, 150, 200.
Problem Interpretation: Organize the data in a frequency distribution table and find the mean wage.
Answer:
Problem Interpretation: We need to tabulate wages and compute mean.
Mathematical Modeling:
Wage (₹)Frequency (f)f × x
1005500
1206720
15071050
1804720
20051000

Solution: Mean = Σ(f × x)/Σf = (500+720+1050+720+1000)/30 = ₹144.33.
Question 2:
The median of observations 11, 12, 14, 18, x+2, x+4, 30, 32, 35, 41 arranged in ascending order is 24.
Problem Interpretation: Find the value of x and the mode if 18 occurs twice.
Answer:
Problem Interpretation: Median is average of 5th & 6th terms.
Mathematical Modeling:
  • Median = (x+2 + x+4)/2 = 24 → 2x+6=48 → x=21
  • Observations: 11,12,14,18,23,25,30,32,35,41

Solution: Mode = 18 (most frequent).

Our textbook shows median calculation for even data.

Question 3:
A class has 20 students with mean height 150 cm. Two students of heights 160 cm and 165 cm leave.
Problem Interpretation: Find the new mean height. (Use step notation)
Answer:
Problem Interpretation: Recalculate mean after data change.
Mathematical Modeling:
  • Total height = 20 × 150 = 3000 cm
  • New total = 3000 - (160+165) = 2675 cm

Solution: New mean = 2675/18 ≈ 148.61 cm.

We studied mean adjustments in NCERT Example 4.

Question 4:
A survey recorded the daily wages (in ₹) of 30 workers: 100, 120, 150, 100, 120, 180, 200, 150, 120, 180, 200, 150, 100, 120, 150, 180, 200, 150, 120, 100, 150, 180, 200, 120, 150, 100, 120, 150, 180, 200.
Problem Interpretation: Organize the data in a frequency distribution table.
Mathematical Modeling: Find the mean wage.
Answer:
Problem Interpretation:
We arrange the data in a table with wages (₹) and frequencies:
Wage (₹)Frequency
1005
1207
1508
1805
2005

Mathematical Modeling:
Mean = (100×5 + 120×7 + 150×8 + 180×5 + 200×5)/30 = ₹146.67.
Question 5:
The median of the following data is 25. Find the missing frequency x:
Class: 0-10, 10-20, 20-30, 30-40, 40-50
Frequency: 5, 8, x, 10, 7
Problem Interpretation: Identify the median class.
Mathematical Modeling: Use the median formula to find x.
Answer:
Problem Interpretation:
Total frequency = 30 + x. Median class is 20-30 since median is 25.
Mathematical Modeling:
Median = L + [(N/2 - cf)/f] × h
25 = 20 + [(15 + x/2 - 13)/x] × 10
Solving, x = 12. Our textbook shows similar problems.
Question 6:
The mode of the data: 5, 8, 6, 5, 8, 9, 5, 6, 8, 5, 6, 5 is 5.
Problem Interpretation: Verify the mode.
Mathematical Modeling: If two more 8s are added, does the mode change? Justify.
Answer:
Problem Interpretation:
Frequency table shows 5 occurs 5 times (highest). Mode is correct.
Mathematical Modeling:
After adding two 8s, frequency of 8 becomes 5 (equal to 5). Now both 5 and 8 are modes. We studied that data can have multiple modes.
Question 7:
A survey recorded the daily wages (in ₹) of 30 workers in a factory: 100, 120, 150, 100, 120, 180, 200, 150, 120, 100, 180, 200, 150, 120, 100, 150, 120, 180, 200, 150, 120, 100, 180, 200, 150, 120, 100, 150, 120, 180.

(i) Prepare a frequency distribution table for the data. (ii) Find the mean daily wage.
Answer:
Problem Interpretation

We need to organize the given wage data and calculate the average wage.


Mathematical Modeling
  • Frequency table lists wages and their counts.
  • Mean = Sum of (wage × frequency) / Total workers.

Solution
Wage (₹)Frequency
1006
1208
1506
1805
2005

Mean = (100×6 + 120×8 + 150×6 + 180×5 + 200×5)/30 = ₹141.33.

Question 8:
The height (in cm) of 10 students in a class are: 155, 160, 145, 150, 157, 159, 161, 165, 162, 158.

(i) Find the median height. (ii) If a new student of height 152 cm joins, how does the median change?
Answer:
Problem Interpretation

We calculate the central value of heights before and after adding a new data point.


Mathematical Modeling
  • Arrange data in ascending order.
  • Median = Middle value (odd count) or average of two middle values (even count).

Solution

Original data (sorted): 145, 150, 155, 157, 158, 159, 160, 161, 162, 165. Median = (158+159)/2 = 158.5 cm.

After adding 152 cm: New median = 158 cm (6th term in 11 data points).

Question 9:
A survey recorded the daily wages (in ₹) of 30 workers in a factory: 100, 120, 150, 100, 120, 180, 200, 150, 120, 100, 180, 200, 150, 120, 100, 150, 180, 200, 120, 100, 150, 120, 180, 200, 150, 120, 100, 150, 180, 200.

(i) Prepare a frequency distribution table. (ii) Find the mean daily wage.
Answer:
Problem Interpretation

We need to organize the given wage data and compute the average wage.


Mathematical Modeling
  • Frequency table lists wages and their counts.
  • Mean = Sum of (wage × frequency) / Total workers.

Solution
Wage (₹)Frequency
1006
1208
1507
1804
2005

Mean = (100×6 + 120×8 + 150×7 + 180×4 + 200×5)/30 = ₹141.

Question 10:
The height (in cm) of 10 students in a class are: 150, 152, 151, 148, 149, 152, 150, 151, 153, 150.

(i) Find the mode. (ii) Calculate the median after arranging data in ascending order.
Answer:
Problem Interpretation

We identify the most frequent height and the middle value of ordered data.


Mathematical Modeling
  • Mode = Most repeated value.
  • Median = Middle value of sorted list.

Solution

Ascending order: 148, 149, 150, 150, 150, 151, 151, 152, 152, 153.

  • Mode = 150 cm (repeats 3 times).
  • Median = (150 + 151)/2 = 150.5 cm.
Question 11:
A survey was conducted to find the number of hours students spent on social media daily. The data collected is as follows: 2, 3, 4, 2, 5, 1, 6, 4, 2, 3.

(i) Find the mean time spent by the students.
(ii) Calculate the median of the given data.
Answer:

(i) Mean = Sum of all observations / Total number of observations
Mean = (2+3+4+2+5+1+6+4+2+3)/10 = 32/10 = 3.2 hours

(ii) To find the median, arrange data in ascending order: 1, 2, 2, 2, 3, 3, 4, 4, 5, 6
Median = Average of 5th and 6th terms = (3+3)/2 = 3 hours

Question 12:
The following table shows the marks obtained by 10 students in a Mathematics test:

Marks: 40, 50, 60, 70, 80, 90
No. of Students: 2, 3, 1, 2, 1, 1

(i) Find the mode of the data.
(ii) Calculate the range of the marks.
Answer:

(i) The mode is the observation with the highest frequency.
Here, 50 marks occur most frequently (3 times).
Mode = 50 marks

(ii) Range = Highest observation - Lowest observation
Range = 90 - 40 = 50 marks

Question 13:
The daily wages (in ₹) of 8 workers are: 300, 350, 400, 450, 500, 550, 600, 650.

(i) Find the mean deviation about the median.
(ii) What does this deviation indicate about the data?
Answer:

(i) First, find median: (450+500)/2 = 475
Mean deviation = Σ|Xi - Median| / n
= (175+125+75+25+25+75+125+175)/8 = 100

(ii) The mean deviation of 100 indicates that, on average, wages vary by ₹100 from the median wage. This shows moderate dispersion in the data.

Question 14:
A survey was conducted in a school to find the favorite subject of 50 students. The data collected is as follows:
Mathematics - 15, Science - 12, Social Science - 10, English - 8, Hindi - 5.
Represent this data in a pie chart and justify why a pie chart is suitable for this representation.
Answer:

To represent the given data in a pie chart, follow these steps:


Step 1: Calculate the central angle for each subject using the formula:
Central Angle = (Frequency / Total Frequency) × 360°

Step 2: Compute the angles:
Mathematics: (15/50) × 360° = 108°
Science: (12/50) × 360° = 86.4°
Social Science: (10/50) × 360° = 72°
English: (8/50) × 360° = 57.6°
Hindi: (5/50) × 360° = 36°

Step 3: Draw the pie chart using these angles, labeling each sector with the subject name and percentage.

Justification: A pie chart is suitable because:

  • It visually represents the proportional distribution of categories (subjects).
  • It helps compare the popularity of each subject at a glance.
  • It is effective for displaying categorical data with a small number of categories.
Question 15:
The following table shows the daily wages (in ₹) of 30 workers in a factory:
Wages (₹): 100-120, 120-140, 140-160, 160-180, 180-200
Number of Workers: 5, 8, 6, 7, 4
Calculate the mean daily wage using the assumed mean method. Show all steps.
Answer:

To calculate the mean daily wage using the assumed mean method, follow these steps:


Step 1: Choose an assumed mean (A). Let A = 150 (midpoint of 140-160).

Step 2: Create a table with columns for class intervals, midpoints (xi), deviations (di = xi - A), and frequency (fi):

Class Interval | Midpoint (xi) | di = xi - 150 | Frequency (fi) | fidi
100-120 | 110 | -40 | 5 | -200
120-140 | 130 | -20 | 8 | -160
140-160 | 150 | 0 | 6 | 0
160-180 | 170 | 20 | 7 | 140
180-200 | 190 | 40 | 4 | 160

Step 3: Sum the frequencies (Σfi = 30) and fidi (Σfidi = -60).

Step 4: Apply the formula:
Mean = A + (Σfidi / Σfi)
Mean = 150 + (-60 / 30) = 150 - 2 = ₹148

The mean daily wage is ₹148.

Question 16:
The heights (in cm) of 10 students are: 155, 160, 145, 150, 157, 148, 152, 149, 146, 153. Calculate the median height and explain why the median is a better measure of central tendency than the mean in this case.
Answer:

To calculate the median height:


Step 1: Arrange the data in ascending order:
145, 146, 148, 149, 150, 152, 153, 155, 157, 160

Step 2: Since there are 10 observations (even number), the median is the average of the 5th and 6th terms.

Step 3: Median = (150 + 152) / 2 = 151 cm

Explanation: The median is a better measure of central tendency than the mean here because:

  • It is less affected by extreme values (if present).
  • It accurately represents the middle value of the dataset, ensuring a balanced view.
  • For small datasets, the median provides a more reliable central value compared to the mean, which can be skewed by outliers.
Question 17:

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


Pie Chart Data:
Cricket - 90°
Football - 60°
Basketball - 45°
Badminton - 75°
Others - 90°

If 120 students chose cricket, find the total number of students surveyed and the number of students who chose football.

Answer:

To solve this problem, we use the concept of central angles in a pie chart representing proportions.


Step 1: Calculate the total angle of the pie chart.
Total angle = 90° (Cricket) + 60° (Football) + 45° (Basketball) + 75° (Badminton) + 90° (Others) = 360°.

Step 2: Find the total number of students surveyed.
Since 90° corresponds to 120 students,
360° corresponds to (120 × 360°)/90° = 480 students.

Step 3: Calculate the number of students who chose football.
60° corresponds to (120 × 60°)/90° = 80 students.

Thus, the total number of students surveyed is 480, and the number of students who chose football is 80.

Question 18:

The following frequency distribution shows the marks obtained by 50 students in a Mathematics test:


Marks (Class Interval): 0-10 | 10-20 | 20-30 | 30-40 | 40-50
Number of Students (Frequency): 5 | 12 | 18 | 10 | 5

Calculate the mean marks of the students using the direct method.

Answer:

To find the mean marks, we use the direct method formula: Mean = Σ(fi × xi) / Σfi, where xi is the class mark.


Step 1: Calculate the class marks (xi) for each interval.
Class marks: (0+10)/2 = 5, (10+20)/2 = 15, (20+30)/2 = 25, (30+40)/2 = 35, (40+50)/2 = 45.

Step 2: Multiply each class mark by its frequency (fi × xi).
5 × 5 = 25
12 × 15 = 180
18 × 25 = 450
10 × 35 = 350
5 × 45 = 225

Step 3: Sum the products and frequencies.
Σ(fi × xi) = 25 + 180 + 450 + 350 + 225 = 1230
Σfi = 5 + 12 + 18 + 10 + 5 = 50

Step 4: Compute the mean.
Mean = 1230 / 50 = 24.6

Thus, the mean marks obtained by the students is 24.6.

Question 19:

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


SubjectNumber of Students
Mathematics45
Science60
Social Science30
English40
Hindi25

Based on the data, answer the following:


(i) Represent the data using a bar graph.


(ii) Which subject is the most preferred and by how many more students than the least preferred subject?

Answer:

(i) Bar Graph Representation:


To draw the bar graph:

  • Draw two perpendicular axes (X-axis and Y-axis).
  • Label the X-axis with Subjects (Mathematics, Science, Social Science, English, Hindi).
  • Label the Y-axis with Number of Students (scale: 0 to 70, with intervals of 10).
  • Draw rectangular bars of equal width for each subject, with heights corresponding to their frequencies.

(ii) Most and Least Preferred Subject:


The most preferred subject is Science with 60 students.
The least preferred subject is Hindi with 25 students.
Difference = 60 - 25 = 35 students.


Thus, Science is preferred by 35 more students than Hindi.

Question 20:

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


(i) Calculate the mean savings of the students.


(ii) Find the mode of the data and explain its significance in this context.

Answer:

(i) Calculation of Mean Savings:


Mean = (Sum of all observations) / (Number of observations)
Sum = 200 + 250 + 300 + 150 + 400 + 350 + 200 + 250 + 300 + 350 = 2750
Number of observations = 10
Mean = 2750 / 10 = ₹275.


(ii) Mode and Its Significance:


To find the mode, identify the most frequently occurring value:
Arranged data: 150, 200, 200, 250, 250, 300, 300, 350, 350, 400.
Here, 200, 250, 300, and 350 each occur twice, making the data multimodal.


In this context, the mode(s) indicate the most common savings amounts among students, helping to identify trends or typical behavior.

Question 21:
A survey was conducted in a school to find the favorite sport of students. The data collected is represented in the following table:

SportNumber of Students
Cricket45
Football30
Basketball25
Badminton20
Others10

Based on the data, answer the following:

  • Find the mean number of students who prefer a sport.
  • Represent the data using a pie chart with appropriate labels.
Answer:

To find the mean number of students:


Total number of students = 45 (Cricket) + 30 (Football) + 25 (Basketball) + 20 (Badminton) + 10 (Others) = 130
Number of sports = 5
Mean = Total students / Number of sports = 130 / 5 = 26

To draw the pie chart:


First, calculate the central angle for each sport using the formula: (Number of students / Total students) × 360°
  • Cricket: (45/130) × 360° ≈ 124.62°
  • Football: (30/130) × 360° ≈ 83.08°
  • Basketball: (25/130) × 360° ≈ 69.23°
  • Badminton: (20/130) × 360° ≈ 55.38°
  • Others: (10/130) × 360° ≈ 27.69°

Now, draw a circle and divide it into sectors with the calculated angles. Label each sector with the sport name and the corresponding percentage or number of students.
Question 22:
The following table shows the monthly savings (in ₹) of 10 families in a locality:

FamilySavings (₹)
12500
23000
31800
43500
52700
62200
74000
83200
92900
103100

Answer the following:

  • Calculate the median savings of the families.
  • If one more family with savings of ₹5000 is added, how does it affect the median?
Answer:

To find the median savings:


First, arrange the data in ascending order:
1800, 2200, 2500, 2700, 2900, 3000, 3100, 3200, 3500, 4000

Since there are 10 observations (even number), the median is the average of the 5th and 6th values.
Median = (2900 + 3000) / 2 = ₹2950

When a new family with savings of ₹5000 is added:


New data in ascending order:
1800, 2200, 2500, 2700, 2900, 3000, 3100, 3200, 3500, 4000, 5000

Now, there are 11 observations (odd number), so the median is the 6th value.
New median = ₹3000

The median increases slightly from ₹2950 to ₹3000 because the new value is higher than the previous median but does not drastically alter the central tendency.
Question 23:
A survey was conducted in a school to find the favorite sport of students. The data collected is represented in the following table:

SportNumber of Students
Cricket45
Football30
Basketball25
Badminton20
Others10

Based on the data, answer the following:

  • Represent the data using a pie chart.
  • Which sport is the most popular among students? Justify your answer.
Answer:

To represent the data using a pie chart, follow these steps:


Step 1: Calculate the total number of students surveyed.
Total students = 45 (Cricket) + 30 (Football) + 25 (Basketball) + 20 (Badminton) + 10 (Others) = 130.

Step 2: Find the central angle for each sport using the formula:
Central angle = (Number of students for the sport / Total students) × 360°.
  • Cricket: (45/130) × 360° ≈ 124.62°
  • Football: (30/130) × 360° ≈ 83.08°
  • Basketball: (25/130) × 360° ≈ 69.23°
  • Badminton: (20/130) × 360° ≈ 55.38°
  • Others: (10/130) × 360° ≈ 27.69°

Step 3: Draw the pie chart with the calculated angles and label each sector with the sport name and percentage.

The most popular sport among students is Cricket because it has the highest number of students (45) compared to other sports. This is also reflected in the pie chart, where Cricket occupies the largest sector (≈124.62°).

Question 24:
The following table shows the monthly savings (in ₹) of 10 families in a locality:

FamilySavings (₹)
A2000
B2500
C1800
D3000
E2200
F2700
G1900
H2100
I2300
J2600

Based on the data, answer the following:

  • Calculate the mean savings of the families.
  • How many families have savings above the mean savings?
Answer:

To calculate the mean savings of the families, follow these steps:


Step 1: Add the savings of all families.
Total savings = 2000 + 2500 + 1800 + 3000 + 2200 + 2700 + 1900 + 2100 + 2300 + 2600 = ₹23,100.

Step 2: Divide the total savings by the number of families (10).
Mean savings = 23,100 / 10 = ₹2,310.

The mean savings of the families is ₹2,310.


To find the number of families with savings above the mean (₹2,310), identify the families with savings > ₹2,310:

  • B (₹2,500)
  • D (₹3,000)
  • F (₹2,700)
  • J (₹2,600)
  • I (₹2,300) → Note: ₹2,300 is not greater than ₹2,310.

Only 4 families (B, D, F, J) have savings above the mean savings of ₹2,310.

Question 25:
A survey was conducted in a school to find the favorite sport of students. The data collected is as follows:
Cricket - 45 students
Football - 30 students
Basketball - 25 students
Tennis - 20 students
Others - 10 students.

Represent this data in a pie chart and find the central angle for each sport.

Answer:

To represent the data in a pie chart, we first calculate the total number of students surveyed:


Total students = 45 (Cricket) + 30 (Football) + 25 (Basketball) + 20 (Tennis) + 10 (Others) = 130 students.

The central angle for each sport is calculated using the formula:


Central Angle = (Number of students for the sport / Total students) × 360°

Calculations:
Cricket: (45 / 130) × 360° ≈ 124.62°
Football: (30 / 130) × 360° ≈ 83.08°
Basketball: (25 / 130) × 360° ≈ 69.23°
Tennis: (20 / 130) × 360° ≈ 55.38°
Others: (10 / 130) × 360° ≈ 27.69°

These angles are used to draw the pie chart, where each sector represents a sport proportionally.

Question 26:
The following table shows the marks obtained by 30 students in a Mathematics test:
Marks: 0-10 | 10-20 | 20-30 | 30-40 | 40-50
No. of Students: 2 | 5 | 8 | 10 | 5

Calculate the mean marks of the students using the step-deviation method.

Answer:

To find the mean using the step-deviation method, follow these steps:


1. Identify the class marks (midpoints) for each class interval:
0-10: 5
10-20: 15
20-30: 25
30-40: 35
40-50: 45

2. Assume a mean (A) from the class marks, say A = 25.

3. Calculate the deviation (d) for each class: d = (Class Mark - A) / h, where h is the class width (10).
0-10: (5 - 25)/10 = -2
10-20: (15 - 25)/10 = -1
20-30: (25 - 25)/10 = 0
30-40: (35 - 25)/10 = 1
40-50: (45 - 25)/10 = 2

4. Multiply deviations by frequencies (f × d):
2 × (-2) = -4
5 × (-1) = -5
8 × 0 = 0
10 × 1 = 10
5 × 2 = 10
Sum of (f × d) = -4 -5 + 0 + 10 + 10 = 11

5. Calculate the mean using the formula:
Mean = A + (Sum of (f × d) / Total frequency) × h
Mean = 25 + (11 / 30) × 10 ≈ 25 + 3.67 = 28.67

Thus, the mean marks of the students is approximately 28.67.

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