Case Study on Maths Class 10 Chapter 13 ‘Statistics’ with Questions and Answers

This collection of case-study questions for Chapter 13: Statistics (Class 10) follows CBSE and NCERT guidelines and presents short, exam-style scenarios designed to strengthen data interpretation, statistical reasoning, and step-by-step problem solving. The set uses real-world contexts and classroom-style prompts to help students practice grouping raw data, constructing and interpreting frequency distributions, finding measures of central tendency for both ungrouped and grouped data, and applying the cumulative frequency method to locate medians and quartiles. Answers are provided with clear reasoning, diagram guidance, and concise steps to help learners develop procedural fluency and confidence for board examinations. The downloadable PDF includes additional practice sets for quick revision and classroom worksheets to improve speed and accuracy.

Solved Statistics Case Study Questions and Answers

CBSE's case-based questions usually give you a short passage describing a real situation, sometimes with a small figure, followed by 4 - 5 questions of increasing difficulty, a couple of MCQs to check basic understanding, a short-answer question that needs a calculation, and a longer question that asks you to interpret or justify your answer.

Key Formulas at a Glance

• Mean: Direct Method

Mean (x̄) = Σfᵢxᵢ / Σfᵢ

where xᵢ = class mark (midpoint of class interval)

• Mean: Step Deviation Method

Mean (x̄) = a + h × (Σfᵢuᵢ / Σfᵢ)

where a = assumed mean, h = class width, uᵢ = (xᵢ − a)/h

• Median

Median = l + [(n/2 − cf) / f] × h

l = lower limit of median class, cf = cumulative frequency of class before median class, f = frequency of median class, h = class width

• Mode

Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h

f₁ = frequency of modal class, f₀ = frequency of class before, f₂ = frequency of class after

• Empirical Relationship

Mode = 3 Median − 2 Mean

OR: 3 Median = Mode + 2 Mean

Case Study 1: Monthly Savings of Families 

Situation: A social welfare organisation surveyed 50 families in a residential colony of Bengaluru to study their monthly savings. The data collected is given in the table below. The organisation wants to find the average monthly savings to plan a financial literacy workshop.

(i) The class mark (mid-point) of the class 1500 - 2000 is:

(a) 1500

(b) 1750

(c) 2000

(d) 1600

(ii) The value of Σfᵢxᵢ for the entire distribution is ________.

(iii) Calculate the mean monthly savings of the families. 

(iv) True or False:

If every families savings increase by ₹500, the mean will remain ₹1810.

(v) Which class interval has the highest frequency?

(a) 1000–1500

(b) 2000–2500

(c) 1500–2000

(d) 2500–3000

Solution: 

(i) (b) 1750

Class mark = (Lower limit + Upper limit) ÷ 2 = (1500 + 2000) ÷ 2 = 1750

(ii) Answer: ₹90,500

Calculation: 

6×750 + 10×1250 + 14×1750 + 12×2250 + 8×2750

= 4500 + 12500 + 24500 + 27000 + 22000

= ₹90,500

(iii) Answer: ₹1810

Mean = Σfᵢxᵢ / Σfᵢ

= 90500 / 50

= ₹1810

The mean monthly savings of the families is ₹1810.

(iv) FALSE

When a constant is added to every observation, the mean also increases by that same constant. So the new mean = ₹1810 + ₹500 = ₹2310.

(v) Answer: (c) 1500–2000

The frequency 14 is the highest, which belongs to the class interval 1500–2000. This is also the modal class for this distribution.

Case Study 2: Shoe Sales in a Store

Situation: A popular footwear store in Jaipur sold 100 pairs of sports shoes in a day. The sizes (in cm) of shoes sold are recorded below. The store manager wants to know the mode shoe size so she can stock inventory smartly.

 

(i) The modal class of the distribution is:

(a) 22–24

(b) 26–28

(c) 24–26

(d) 28–30

(ii) In the mode formula, the value of f₁ is ______, f₀ is _____, and f₂ is ______ .

(iii) Calculate the mode shoe size. 

(iv) True or False:

Mode can only be used for grouped data, not for ungrouped data.

Solutions: 

(i) Answer: (c) 24–26

The highest frequency is 32, which belongs to the class 24 - 26. This is the modal class.

(ii) f₁ = 32, f₀ = 15, f₂ = 28

f₁ = frequency of modal class (24–26) = 32

f₀ = frequency of class before modal class (22–24) = 15

f₂ = frequency of class after modal class (26–28) = 28

(iii) Modal class = 24–26

l = 24, f₁ = 32, f₀ = 15, f₂ = 28, h = 2

Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h

= 24 + [(32 − 15) / (64 − 15 − 28)] × 2

= 24 + [17 / 21] × 2

= 24 + 34/21

= 24 + 1.619...

≈ 25.62 cm

The mode shoe size is approximately 25.62 cm. The manager should stock sizes around 25–26 cm most heavily.

(iv) FALSE

For ungrouped data, mode is simply the value that appears most often. The formula Mode = l + [(f₁−f₀)/(2f₁−f₀−f₂)] × h applies only to grouped data (class intervals with frequencies).

Case Study 3: Hospital Patient Ages

Situation: A government hospital in Delhi recorded the ages of 100 patients who visited the OPD on a particular day. The data, in cumulative form, is given below. The hospital administrator wants to find the median age and also understand how to draw an ogive to represent the data visually.

(i) For drawing a less than ogive, the points plotted are of the form:

(a) (Lower limit, cumulative frequency)

(b) (Upper limit, cumulative frequency)

(c) (Class mark, frequency)

(d) (Upper limit, frequency)

(ii) To find the median graphically from the less than ogive, we draw a horizontal line from _______ on the y-axis.

(iii) Calculate the median age of the patients. 

(iv) True or False:

The less than ogive and more than ogive, when drawn on the same axea, always intersect at a point whose x- coordinate gives the median.

(v) The upper limit of the median class in this data is: 

(a) 30

(b) 50

(c) 40

(d) 35

Solutions:

(i) Answer: (b) (Upper limit, cumulative frequency)

For a less than ogive, we plot (upper class limit, cumulative frequency). For example, the first point is (10, 5), meaning fewer than 10 patients are aged less than 10 years.

(ii) Answer: 50 (i.e., n/2 = 100/2 = 50)

We mark n/2 = 50 on the y-axis, draw a horizontal line to meet the ogive, then drop a perpendicular to the x-axis. The x-value at this point is the median.

(iii) n = 100, n/2 = 50

cf at 20 - 30 = 35 (less than 50)

cf at 30 - 40 = 60 (first to exceed 50) ⇒ Median class = 30 - 40

l = 30, cf = 35, f = 25, h = 10

Median = 30 + [(50 − 35) / 25] × 10

= 30 + [15/25] × 10

= 30 + 6

= 36 years

The median age of patients is 36 years.

(iv) TRUE

This is a direct fact. The x-coordinate of the intersection of the less than and more than ogives gives the median of the distribution. The y-coordinate of that intersection equals n/2.

(v) (c) 40

The median class is 30 - 40. Its upper limit is 40.

Questions PDF with worked-out examples for Class 10 Chapter 13: Statistics, perfect for last-minute CBSE exam revision.

Class 10 Chapter 13: Statistics Case Study PDF