Chapter 10 of 12

Updated 6 August 2026

NCERT Solutions

Descriptive Statistics — NCERT Solutions

CBSE · Class 11 · Applied Mathematics

NCERT Solutions for Descriptive Statistics, CBSE Class 11 Applied Mathematics: 9 textbook questions solved step by step.

44 questions24 flashcards5 concepts

Interactive on Super Tutor

Studying Descriptive Statistics? Get the full interactive chapter.

Quizzes, flashcards, AI doubt-solver and a step-by-step study plan — built for NCERT solutions and more.

Try Super Tutor — Free

Free trial, no card needed.

9 Questions Solved · 1 Section

The first 5 solutions are open to read. The other 4 are free with a Super Tutor account.

Check your Progress / Exercise Questions — Descriptive Statistics (Class 11 Applied Mathematics)

1Discuss the difference between bar graph and histogram from charts given in examples of respective concepts.Show solution

Given: Two types of graphical representations — Bar Graph and Histogram.

Differences between Bar Graph and Histogram:

Feature	Bar Graph	Histogram
Type of data	Categorical (discrete) data	Continuous (grouped) data
Bars	Bars are separated by gaps	Bars are adjacent (no gaps)
X-axis	Represents categories or discrete values	Represents class intervals (continuous)
Width of bars	All bars have equal width; width has no meaning	Width represents the class interval size; width is meaningful
Rearrangement	Bars can be rearranged in any order	Bars cannot be rearranged; order is fixed
Purpose	Compares different categories	Shows frequency distribution of continuous data
Base	Each bar stands on its own base	Bars share a common continuous base

Conclusion: A bar graph is used for discrete/categorical data where gaps between bars indicate distinct categories, while a histogram is used for continuous frequency distributions where the area of each bar represents the frequency of that class interval.

2Collect multiple sets of data and discuss which type of graph representation is most suitable for a given set of data. Is it possible to represent the data graphically in two or more ways?Show solution

Given: Various types of data sets.

Concept: Different types of graphs are suitable for different types of data.

Discussion of suitable graph types:

Type of Data	Most Suitable Graph	Reason
Categorical comparison (e.g., population of different states)	Bar Graph	Compares discrete categories clearly
Parts of a whole (e.g., budget allocation)	Pie Chart	Shows proportion/percentage of each part
Continuous frequency distribution (e.g., marks of students)	Histogram / Frequency Polygon	Represents class intervals and frequencies
Trend over time (e.g., GDP growth over years)	Line Graph	Shows change/trend over a continuous period
Two variables relationship (e.g., height vs weight)	Scatter Plot	Shows correlation between two variables

Can data be represented in two or more ways?

Yes, it is possible to represent the same data graphically in two or more ways. For example:

A continuous frequency distribution can be shown using a Histogram, a Frequency Polygon, or an Ogive (cumulative frequency curve).
Categorical data can be shown using both a Bar Graph and a Pie Chart.

Conclusion: While multiple representations are possible, the choice of graph should be guided by the nature of the data and the purpose of analysis. The most appropriate graph makes the data easier to interpret and compare.

3India's union budget 2020-21 proposes to change the following amendment in tax rate. Represent the information using a suitable graph. Justify your choice.

| Taxable Income Slab (Rs.) | Current Tax Rates | New Tax Rates |
|---|---|---|
| 0–2.5 Lakh | Exempt | Exempt |
| 2.5–5 Lakh | 5% | Exempt |
| 5–7.5 Lakh | 20% | 10% |
| 7.5–10 Lakh | 20% | 15% |
| 10–12.5 Lakh | 30% | 20% |
| 12.5–15 Lakh | 30% | 25% |
| Above 15 Lakh | 30% | 30% |Show solution

Given: Tax rate data for different income slabs under current and new tax regimes.

Most Suitable Graph: Double/Grouped Bar Graph

Justification: The data involves comparison of two sets of values (Current Tax Rates vs New Tax Rates) across different income slabs (categories). A grouped (double) bar graph is most suitable because:

It allows direct visual comparison between two data series side by side.
The income slabs are discrete categories, making a bar graph appropriate.
It clearly shows the reduction in tax rates across slabs.

Steps to draw the Grouped Bar Graph:

Step 1: Mark the income slabs on the X-axis.

Step 2: Mark the tax rates (in %) on the Y-axis.

Step 3: For each income slab, draw two bars side by side — one for the Current Tax Rate (e.g., in blue) and one for the New Tax Rate (e.g., in orange).

Step 4: Add a legend to distinguish the two bars.

Data for plotting:

Income Slab	Current Rate (%)	New Rate (%)
0–2.5 L	0	0
2.5–5 L	5	0
5–7.5 L	20	10
7.5–10 L	20	15
10–12.5 L	30	20
12.5–15 L	30	25
Above 15 L	30	30

Observation from the graph: The new tax regime offers lower tax rates for all income slabs up to Rs. 15 Lakh, while the rate remains the same (30%) for income above Rs. 15 Lakh. This shows the government's intent to reduce the tax burden on middle-income groups.

4India's union budget 2020-21 fixes FRBM targets for deficits (as % of GDP). Represent the information using a suitable graph.

| | Actuals 2018-19 | Revised 2019-20 | Budgeted 2020-21 | Target 2021-22 | Target 2022-23 |
|---|---|---|---|---|---|
| Fiscal Deficit | 3.4% | 3.8% | 3.5% | 3.3% | 3.1% |
| Revenue Deficit | 2.4% | 2.4% | 2.7% | 2.3% | 1.9% |Show solution

Given: Fiscal Deficit and Revenue Deficit data (as % of GDP) over five years/periods.

Most Suitable Graph: Double Line Graph (Multiple Line Graph)

Justification: The data shows the trend of two types of deficits over a time period (from 2018-19 to 2022-23). A line graph is most suitable because:

It effectively shows trends and changes over time.
Two lines can be drawn simultaneously to compare Fiscal Deficit and Revenue Deficit.
It clearly shows whether deficits are increasing or decreasing over the years.

Steps to draw the Double Line Graph:

Step 1: Mark the years/periods on the X-axis: 2018-19, 2019-20, 2020-21, 2021-22, 2022-23.

Step 2: Mark the deficit percentages (0% to 4%) on the Y-axis.

Step 3: Plot the points for Fiscal Deficit: (2018-19, 3.4), (2019-20, 3.8), (2020-21, 3.5), (2021-22, 3.3), (2022-23, 3.1) and join them with a line.

Step 4: Plot the points for Revenue Deficit: (2018-19, 2.4), (2019-20, 2.4), (2020-21, 2.7), (2021-22, 2.3), (2022-23, 1.9) and join them with a different coloured line.

Step 5: Add a legend to distinguish the two lines.

Observation: Both Fiscal Deficit and Revenue Deficit show a peak in 2019-20 and 2020-21 respectively, after which they are targeted to decline. The Fiscal Deficit is consistently higher than the Revenue Deficit throughout the period. The government aims to achieve fiscal consolidation by 2022-23.

5Find the standard deviation of the data: 3, 6, 2, 1, 7, 5.Show solution

Given: Data: $3, 6, 2, 1, 7, 5$ ; $n = 6$

Formula:
$\sigma = \sqrt{\frac{1}{N}\sum_{i=1}^{n}(x_i - \bar{x})^2}$

Step 1: Find the Mean
$\bar{x} = \frac{3 + 6 + 2 + 1 + 7 + 5}{6} = \frac{24}{6} = 4$

Step 2: Find $(x_i - \bar{x})$ and $(x_i - \bar{x})^2$ for each value

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
3	$3 - 4 = -1$	1
6	$6 - 4 = 2$	4
2	$2 - 4 = -2$	4
1	$1 - 4 = -3$	9
7	$7 - 4 = 3$	9
5	$5 - 4 = 1$	1
Total		28

Step 3: Calculate Standard Deviation
$\sigma = \sqrt{\frac{\sum(x_i - \bar{x})^2}{N}} = \sqrt{\frac{28}{6}} = \sqrt{4.\overline{6}} = \sqrt{\frac{14}{3}}$

$\sigma = \sqrt{4.667} \approx 2.16$

Therefore, the Standard Deviation $\sigma \approx 2.16$

6Calculate standard deviation for the following set of scores: 40, 38, 42, 60, 72, 54.

Given: Data: $40, 38, 42, 60, 72, 54$ ; $n = 6$

Formula:
$\sigma = \sqrt{\frac{1}{N}\sum_{i=1}^{n}(x_i - \bar{x})^2}$

Step 1: Find the Mean
$\bar{x} = \frac{40 + 38 + 42 + 60 + 72 + 54}{6} = \frac{306}{6} = 51$

Step 2: Find $(x_i - \bar{x})$ and $(x_i - \bar{x})^2$

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
40	$40 - 51 = -11$	121
38	$38 - 51 = -13$	169
42	$42 - 51 = -9$	81
60	$60 - 51 = 9$	81
72	$72 - 51 = 21$	441
54	$54 - 51 = 3$	9
Total		902

Step 3: Calculate Standard Deviation
$\sigma = \sqrt{\frac{902}{6}} = \sqrt{150.33} \approx 12.26$

Therefore, the Standard Deviation $\sigma \approx 12.26$

Free with a Super Tutor account

7By multiplying each of the numbers 3, 6, 2, 1, 7, and 5 by 2 and then adding 5, we obtain the set 11, 17, 9, 7, 19, 15. What is the relationship between the standard deviations and the means for the two sets?

Given:

Set 1: $3, 6, 2, 1, 7, 5$
Set 2 (obtained by $y = 2x + 5$ ): $11, 17, 9, 7, 19, 15$

From Question 5, for Set 1:
$\bar{x}_1 = 4, \quad \sigma_1 \approx 2.16$

For Set 2 — Step 1: Find the Mean
$\bar{x}_2 = \frac{11 + 17 + 9 + 7 + 19 + 15}{6} = \frac{78}{6} = 13$

Step 2: Find Standard Deviation of Set 2

$y_i$	$y_i - \bar{y}$	$(y_i - \bar{y})^2$
11	$-2$	4
17	$4$	16
9	$-4$	16
7	$-6$	36
19	$6$	36
15	$2$	4
Total		112

$\sigma_2 = \sqrt{\frac{112}{6}} = \sqrt{18.\overline{6}} = \sqrt{\frac{56}{3}} \approx 4.32$

Step 3: Establish the Relationship

The transformation used is $y = 2x + 5$ .

Relationship between Means:
$\bar{y} = 2\bar{x} + 5 \implies 13 = 2(4) + 5 = 13 \checkmark$

Relationship between Standard Deviations:
$\sigma_y = |2| \times \sigma_x \implies \sigma_2 = 2 \times \sigma_1$
$4.32 = 2 \times 2.16 \checkmark$

Conclusion:

The mean of Set 2 = $2 \times$ (mean of Set 1) $+ 5$ , i.e., $\bar{y} = 2\bar{x} + 5$ .
The standard deviation of Set 2 = $2 \times$ (standard deviation of Set 1), i.e., $\sigma_y = 2\sigma_x$ .
Key Principle: When each value is multiplied by a constant $k$ and a constant $c$ is added (i.e., $y = kx + c$ ), the mean transforms as $\bar{y} = k\bar{x} + c$ , but the standard deviation transforms as $\sigma_y = |k|\sigma_x$ (the additive constant $c$ does not affect the standard deviation).

Free with a Super Tutor account

8Differentiate between percentile and Percentile rank.

Given: Two statistical measures — Percentile and Percentile Rank.

Differences between Percentile and Percentile Rank:

Feature	Percentile	Percentile Rank
Definition	A value below which a given percentage of observations fall	The percentage of scores in a distribution that fall at or below a given score
What it gives	A score/value corresponding to a given percentage	A percentage/rank corresponding to a given score
Direction	Percentage → Score	Score → Percentage
Example	$P_{75}$ = 80 means 75% of students scored below 80	PR of score 80 = 75 means the student scored better than 75% of students
Use	Used to find the cut-off score for a given percentage	Used to find the relative standing of a particular score
Formula	$P_k = l + \left(\dfrac{\frac{kN}{100} - cf}{f}\right) \times h$	$PR = \dfrac{100}{N}\left(cf + \dfrac{x}{h} \times f\right)$

In summary: Percentile converts a percentage into a score, while Percentile Rank converts a score into a percentage. They are inverse operations of each other.

Free with a Super Tutor account

9Calculate reasonable value for

PR_{70}

and

PR_{50}

for example no. 43 (given in Percentile rank of grouped data).

Note: Example 43 refers to the grouped data table provided in the solution section of the OCR text. The data table is:

Class Interval	f	cf	Percentage of cf
93–97	4	59	100
88–92	7	55	93.22
83–87	5	48	81.36
78–82	8	43	72.88
73–77	3	35	59.32
68–72	6	32	54.24
63–67	7	26	44.07
58–62	10	19	32.30
53–57	5	9	15.25
48–52	4	4	6.78

Here $N = 59$ , class size $h = 5$ .

Formula for Percentile Rank of grouped data:
$PR = \frac{100}{N}\left(cf + \frac{x}{h} \times f\right)$
where $x$ = score $-$ lower true boundary of the class, $cf$ = cumulative frequency below the class, $f$ = frequency of the class.

Calculating $PR_{70}$ (Percentile Rank of score 70):

Score = 70 lies in class interval 68–72.

Lower true boundary $l = 67.5$
$x = 70 - 67.5 = 2.5$
$f = 6$ (frequency of class 68–72)
$cf = 26$ (cumulative frequency below class 68–72)
$h = 5$ , $N = 59$

$PR_{70} = \frac{100}{59}\left(26 + \frac{2.5}{5} \times 6\right) = \frac{100}{59}\left(26 + 3\right) = \frac{100 \times 29}{59}$

$PR_{70} = \frac{2900}{59} \approx 49.15$

$\therefore$ Percentile Rank of score 70 $\approx 49.15$

Calculating $PR_{50}$ (Percentile Rank of score 50):

Score = 50 lies in class interval 48–52.

Lower true boundary $l = 47.5$
$x = 50 - 47.5 = 2.5$
$f = 4$ (frequency of class 48–52)
$cf = 0$ (no class below 48–52)
$h = 5$ , $N = 59$

$PR_{50} = \frac{100}{59}\left(0 + \frac{2.5}{5} \times 4\right) = \frac{100}{59}\left(0 + 2\right) = \frac{200}{59}$

$PR_{50} = \frac{200}{59} \approx 3.39$

$\therefore$ Percentile Rank of score 50 $\approx 3.39$

Interpretation:

A student scoring 70 has performed better than approximately 49.15% of the students.
A student scoring 50 has performed better than approximately 3.39% of the students.

Free with a Super Tutor account

4 more solved questions in Descriptive Statistics

They are free with a Super Tutor account, along with practice quizzes and flashcards for this chapter. Free to start, no card needed.

Frequently Asked Questions

What are the important topics in Descriptive Statistics for CBSE Class 11 Applied Mathematics?

Key topics in Descriptive Statistics include Types of Data and Data Sources, Data Measurement Scales, Data Representation and Visualization, Measures of Central Tendency. Study these first, then practise questions on each for Class 11 exams.

Are these NCERT Solutions for Descriptive Statistics free?

The first 5 of the 9 solutions on this page are open to read. The other 4 are free with a Super Tutor account — signing up is free and needs no card.

How should I revise Descriptive Statistics for Class 11 exams?

Learn the core ideas first, then work through the 44 practice questions on Descriptive Statistics. Revise definitions regularly and use flashcards for quick recall before the exam.

Sources & Official References

Content is aligned to the official syllabus. Refer to the board website for the latest curriculum.

More resources for Descriptive Statistics

All CBSE Class 11 Applied Mathematics NCERT Solutions

For serious students

Get the full Descriptive Statistics chapter — start free.

Quizzes, flashcards, an AI doubt solver and a study plan for CBSE Class 11 Applied Mathematics. Free to start, no card needed.

Try Super Tutor Free

Try this chapter free

Descriptive Statistics — NCERT Solutions

Studying Descriptive Statistics? Get the full interactive chapter.

Check your Progress / Exercise Questions — Descriptive Statistics (Class 11 Applied Mathematics)

Frequently Asked Questions

Sources & Official References

More resources for Descriptive Statistics

Practice Quiz

Important Questions

Revision Notes

Chapter Summary

Flashcards

Syllabus

Get the full Descriptive Statistics chapter — start free.