Chapter 13 : Class 10 : Statistics : Case Study Question Bank

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Chapter 13: Statistics

Class 10 Standard Mathematics (Case Study Bank)

Curated By Akash Srivastva

✨ Interactive TLMs, Worksheets & Digital Study Notes: www.akashmaths.online

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Case Study 1 Climate Analytics 4 Marks

Weather Monitoring Stations

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A meteorological department deploys automated weather sensors across 50 locations to record daily maximum temperatures in summer:

Max Temp (°C)	Stations (\(f_i\))
24 − 28	4
28 − 32	8
32 − 36	18
36 − 40	14
40 − 44	6

Q1. Modal Class & Midpoint 1 Mark

Identify the modal class of this temperature distribution and calculate its class mark (xᵢ).

(A) 28 − 32, Class mark = 30

(B) 32 − 36, Class mark = 34

(C) 36 − 40, Class mark = 38

(D) 32 − 36, Class mark = 32

Q2. Mean via Direct Method 1 Mark

Find the mean daily maximum temperature (°C) of the region using the Direct Method.

(A) 34.8°C

(B) 35.2°C

(C) 33.6°C

(D) 34.0°C

Q3. Assumed Mean & Network Expansion 2 Marks

Calculate mean temperature using Assumed Mean Method with A = 34.

(A) 35.0°C

(B) 34.4°C

(C) 34.8°C (Σfᵢdᵢ = 40)

(D) 35.4°C

OR (Alternative Q3)

If 4 new stations are added to the bracket 40 − 44, calculate the new mean temperature.

(A) 35.80°C

(B) 35.33°C

(C) 36.12°C

(D) 34.95°C

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Highest frequency is 18 in class 32 − 36. Class mark = (32 + 36)/2 = 34.	[0.5 Mark] for modal class. [0.5 Mark] for class mark = 34.
Q2	Σfᵢxᵢ = 4(26) + 8(30) + 18(34) + 14(38) + 6(42) = 104 + 240 + 612 + 532 + 252 = 1740. Mean = 1740 / 50 = 34.8°C.	[0.5 Mark] for Σfᵢxᵢ = 1740. [0.5 Mark] for Mean = 34.8°C.
Q3	dᵢ = xᵢ − 34: Σfᵢdᵢ = 4(−8) + 8(−4) + 18(0) + 14(4) + 6(8) = −32 − 32 + 0 + 56 + 48 = 40. Mean = 34 + 40/50 = 34 + 0.8 = 34.8°C.	[1 Mark] for Σfᵢdᵢ = 40. [1 Mark] for verifying 34.8°C.
Q3 (OR)	New N = 50 + 4 = 54. New Σfᵢxᵢ = 1740 + 4(42) = 1740 + 168 = 1908. New Mean = 1908 / 54 = 35.33°C.	[1 Mark] for new sum 1908. [1 Mark] for 35.33°C.

Case Study 2 Public Health 4 Marks

Community Health Screening

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A non-profit health clinic screens 80 patients admitted for metabolic therapy:

Age Group (Years)	Patients (\(f_i\))
20 − 30	8
30 − 40	12
40 − 50	25
50 − 60	20
60 − 70	15

Q1. Modal Class Identification 1 Mark

Identify the modal class of this age distribution and state its lower limit (l) and class size (h).

(A) 40 − 50, l = 40, h = 10

(B) 50 − 60, l = 50, h = 10

(C) 40 − 50, l = 45, h = 10

(D) 30 − 40, l = 30, h = 10

Q2. Modal Age Calculation 1 Mark

Calculate the exact modal age (in years) of the patients.

(A) 45.50 years

(B) 47.22 years (47 2/9 years)

(C) 48.00 years

(D) 46.80 years

Q3. Median Age & Empirical Formula 2 Marks

Calculate the median age of the patient group using the cumulative frequency method.

(A) 46.5 years

(B) 47.5 years

(C) 48.0 years

(D) 49.2 years

OR (Alternative Q3)

Using Mean = 46.5 and Median = 48, compute empirical Mode and find difference from calculated Mode.

(A) Empirical Mode = 51 years; Difference = 3.78 years

(B) Empirical Mode = 49 years; Difference = 1.78 years

(C) Empirical Mode = 50 years; Difference = 2.78 years

(D) Empirical Mode = 52 years; Difference = 4.78 years

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Highest frequency is 25 in class 40 − 50 ⇒ Modal class = 40 − 50, l = 40, h = 10.	[0.5 Mark] for modal class. [0.5 Mark] for l = 40, h = 10.
Q2	Mode = 40 + [(25 − 12) / (50 − 12 − 20)] × 10 = 40 + (13/18) × 10 = 40 + 7.22 = 47.22 years.	[0.5 Mark] for formula substitution. [0.5 Mark] for 47.22 years.
Q3	N/2 = 40. Median class is 40 − 50 (cf = 20, f = 25). Median = 40 + [(40 − 20)/25] × 10 = 40 + (20/25) × 10 = 40 + 8 = 48 years.	[1 Mark] for identifying median class and cf. [1 Mark] for Median = 48 years.
Q3 (OR)	Mode = 3(Median) − 2(Mean) = 3(48) − 2(46.5) = 144 − 93 = 51 years. Difference = 51 − 47.22 = 3.78 years.	[1 Mark] for Empirical Mode = 51. [1 Mark] for difference = 3.78 years.

Case Study 3 Microfinance 4 Marks

Women Micro-Entrepreneurs Savings

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A regional rural bank analyzes monthly Recurring Deposit contributions (INR) of 100 women entrepreneurs:

Monthly Deposit (INR)	Women (\(f_i\))
500 − 1000	15
1000 − 1500	22
1500 − 2000	30
2000 − 2500	18
2500 − 3000	15

Q1. Median Class 1 Mark

Create cumulative frequencies and identify the median class interval.

(A) 1000 − 1500

(B) 1500 − 2000 (cf = 67)

(C) 2000 − 2500

(D) 500 − 1000

Q2. Median Monthly Deposit 1 Mark

Calculate the median monthly investment (in INR) of these 100 micro-entrepreneurs.

(A) ₹ 1716.67

(B) ₹ 1650.00

(C) ₹ 1750.00

(D) ₹ 1800.00

Q3. Mean vs Modal Investment 2 Marks

Calculate the mean monthly investment (in INR) using the direct method.

(A) ₹ 1680

(B) ₹ 1710

(C) ₹ 1750

(D) ₹ 1730 (Σfᵢxᵢ = 1,73,000)

OR (Alternative Q3)

Calculate the modal monthly investment (in INR) using the mode formula.

(A) ₹ 1700

(B) ₹ 1650

(C) ₹ 1720

(D) ₹ 1680

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	cf: 15, 37, 67, 85, 100. N/2 = 50 ⇒ cf just > 50 is 67, corresponding to class 1500 − 2000.	[0.5 Mark] for cf table. [0.5 Mark] for median class 1500 − 2000.
Q2	Median = 1500 + [(50 − 37)/30] × 500 = 1500 + (13/30) × 500 = 1500 + 216.67 = ₹ 1716.67.	[1 Mark] for Median = ₹ 1716.67.
Q3	Σfᵢxᵢ = 15(750) + 22(1250) + 30(1750) + 18(2250) + 15(2750) = 1,73,000. Mean = 1,73,000 / 100 = ₹ 1730.	[1 Mark] for Σfᵢxᵢ = 173,000. [1 Mark] for Mean = ₹ 1730.
Q3 (OR)	Modal class = 1500 − 2000 (f₁ = 30, f₀ = 22, f₂ = 18). Mode = 1500 + [(30 − 22) / (60 − 22 − 18)] × 500 = 1500 + (8/20) × 500 = 1500 + 200 = ₹ 1700.	[1 Mark] for mode formula substitution. [1 Mark] for Mode = ₹ 1700.

Case Study 4 Smart Grids 4 Marks

Smart Power Grid & Households

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A smart grid monitors weekly consumption (kWh) of 100 households with Median = 46 kWh. Frequencies x and y were temporarily lost:

Consumption (kWh)	Households (\(f_i\))
0 − 20	12
20 − 40	x
40 − 60	30
60 − 80	y
80 − 100	14

Q1. Frequency Sum Equation 1 Mark

Formulate a linear algebraic equation in terms of x and y using total households N = 100.

(A) x + y = 40

(B) x + y = 44

(C) x + y = 46

(D) x + y = 50

Q2. Preceding Cumulative Frequency 1 Mark

State the cumulative frequency (cf) of the class preceding median class 40 − 60.

(A) 12

(B) x

(C) 12 + x

(D) 42 + x

Q3. Finding Missing Frequencies 2 Marks

Calculate exact values of missing telemetry frequencies x and y using Median = 46 kWh.

(A) x = 29, y = 15

(B) x = 25, y = 19

(C) x = 28, y = 16

(D) x = 30, y = 14

OR (Alternative Q3)

With recovered values x = 29 and y = 15, calculate the exact mean weekly consumption.

(A) 45 kWh

(B) 46.5 kWh

(C) 50 kWh

(D) 48 kWh (Σfᵢxᵢ = 4800)

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	12 + x + 30 + y + 14 = 100 ⇒ 56 + x + y = 100 ⇒ x + y = 44.	[1 Mark] for x + y = 44.
Q2	Median 46 lies in 40 − 60. Preceding class is 20 − 40 with cf = 12 + x.	[1 Mark] for cf = 12 + x.
Q3	Median = 40 + [(50 − (12 + x))/30] × 20 = 46 ⇒ 6 = 2(38 − x)/3 ⇒ 9 = 38 − x ⇒ x = 29. y = 44 − 29 = 15.	[1 Mark] for solving x = 29. [1 Mark] for y = 15.
Q3 (OR)	Σfᵢxᵢ = 12(10) + 29(30) + 30(50) + 15(70) + 14(90) = 120 + 870 + 1500 + 1050 + 1260 = 4800. Mean = 4800 / 100 = 48 kWh.	[1 Mark] for Σfᵢxᵢ = 4800. [1 Mark] for Mean = 48 kWh.

Case Study 5 Agri-Analytics 4 Marks

Wheat Fields – Precision Agriculture

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A farming cooperative records wheat yield (kg/hectare) across 100 experimental plots:

Yield (kg/ha)	Plots (\(f_i\))
50 − 55	8
55 − 60	12
60 − 65	20
65 − 70	25
70 − 75	15
75 − 80	20

Q1. Modal Class & Class Mark 1 Mark

Identify the modal class of the wheat yield distribution and calculate its class mark (xᵢ).

(A) 60 − 65, Class mark = 62.5

(B) 65 − 70, Class mark = 67.5

(C) 75 − 80, Class mark = 77.5

(D) 65 − 70, Class mark = 65.0

Q2. Mean via Direct Method 1 Mark

Compute the mean wheat yield using the Direct Method (in kg/hectare).

(A) 66.85 kg/ha

(B) 67.25 kg/ha

(C) 65.50 kg/ha

(D) 68.00 kg/ha

Q3. Assumed Mean & Shift Property 2 Marks

Calculate mean wheat yield using Assumed Mean Method with A = 67.5.

(A) 67.50 kg/ha

(B) 66.20 kg/ha

(C) 66.85 kg/ha (Σfᵢdᵢ = −65)

(D) 67.15 kg/ha

OR (Alternative Q3)

If a bio-fertilizer increases yield of every plot by 5 kg/ha, calculate the new mean yield.

(A) 71.85 kg/ha (Mean increases by constant k = 5)

(B) 70.85 kg/ha

(C) 72.50 kg/ha

(D) 66.85 kg/ha (Unchanged)

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Highest frequency is 25 in class 65 − 70. Class mark = (65 + 70)/2 = 67.5 kg/ha.	[0.5 Mark] for modal class. [0.5 Mark] for class mark 67.5.
Q2	Σfᵢxᵢ = 8(52.5) + 12(57.5) + 20(62.5) + 25(67.5) + 15(72.5) + 20(77.5) = 420 + 690 + 1250 + 1687.5 + 1087.5 + 1550 = 6685. Mean = 6685 / 100 = 66.85 kg/ha.	[0.5 Mark] for Σfᵢxᵢ = 6685. [0.5 Mark] for Mean = 66.85.
Q3	dᵢ = xᵢ − 67.5: Σfᵢdᵢ = 8(−15) + 12(−10) + 20(−5) + 25(0) + 15(5) + 20(10) = −120 − 120 − 100 + 0 + 75 + 200 = −65. Mean = 67.5 + (−65/100) = 67.5 − 0.65 = 66.85 kg/ha.	[1 Mark] for deviations and Σfᵢdᵢ = −65. [1 Mark] for Mean = 66.85.
Q3 (OR)	When constant k is added to all observations, New Mean = Old Mean + k. New Mean = 66.85 + 5 = 71.85 kg/ha.	[1 Mark] for shift property. [1 Mark] for 71.85 kg/ha.

Case Study 6 Logistics Operations 4 Marks

Express Package Delivery

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An e-commerce dispatch hub records delivery durations (hours) for 120 express packages:

Delivery Time (Hours)	Packages (\(f_i\))
0 − 10	8
10 − 20	12
20 − 30	24
30 − 40	40
40 − 50	16
50 − 60	20

Q1. Modal Class & Limit 1 Mark

Identify the modal class of the delivery distribution and state its upper limit.

(A) 20 − 30, Upper limit = 30

(B) 30 − 40, Upper limit = 40

(C) 40 − 50, Upper limit = 50

(D) 30 − 40, Upper limit = 30

Q2. Modal Delivery Time 1 Mark

Calculate the modal delivery time for these dispatched packages.

(A) 32 hours

(B) 36 hours

(C) 34 hours

(D) 35 hours

Q3. Median & Empirical Relationship 2 Marks

Construct cumulative frequencies and calculate median delivery time.

(A) 34 hours (Median class 30 − 40)

(B) 32 hours

(C) 36 hours

(D) 35.5 hours

OR (Alternative Q3)

Given Mean = 33.67, verify empirical relation Mode = 3 Median − 2 Mean and comment on skewness.

(A) Highly right-skewed

(B) Bimodal

(C) Severely left-skewed

(D) 3(34) − 2(33.67) = 34.66 ≈ 34 hours; approximately symmetrical

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Highest frequency is 40 in class 30 − 40 ⇒ Modal class = 30 − 40, Upper limit = 40 hours.	[0.5 Mark] for modal class. [0.5 Mark] for upper limit = 40.
Q2	Mode = 30 + [(40 − 24) / (80 − 24 − 16)] × 10 = 30 + (16/40) × 10 = 30 + 4 = 34 hours.	[0.5 Mark] for formula substitution. [0.5 Mark] for Mode = 34 hours.
Q3	cf: 8, 20, 44, 84, 100, 120. N/2 = 60 ⇒ Median class is 30 − 40 (cf = 44, f = 40). Median = 30 + [(60 − 44)/40] × 10 = 30 + (16/40) × 10 = 30 + 4 = 34 hours.	[1 Mark] for median class setup. [1 Mark] for Median = 34 hours.
Q3 (OR)	Empirical Mode = 3(34) − 2(33.67) = 102 − 67.34 = 34.66 hours ≈ 34 hours. Since Mean ≈ Median ≈ Mode, the distribution is approximately symmetrical.	[1 Mark] for empirical formula verification. [1 Mark] for symmetry comment.

Case Study 7 Resource Management 4 Marks

Rural Piped Water Supply

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Smart flow meters track daily household water supply (Litres) across 150 rural villages:

Water Supply (L/household)	Villages (\(f_i\))
100 − 120	15
120 − 140	25
140 − 160	32
160 − 180	48
180 − 200	20
200 − 220	10

Q1. Median Class Lower Limit 1 Mark

Find the lower limit of the median class for the piped water distribution.

(A) 140 Litres

(B) 160 Litres (class 160 − 180)

(C) 180 Litres

(D) 120 Litres

Q2. Preceding Cumulative Frequency 1 Mark

Determine cumulative frequency (cf) of the class interval preceding median class.

(A) 72

(B) 40

(C) 120

(D) 48

Q3. Median Water Supply & Mode Comparison 2 Marks

Calculate the exact median daily water supply per household across the surveyed villages.

(A) 160.50 Litres

(B) 162.00 Litres

(C) 161.25 Litres

(D) 163.75 Litres

OR (Alternative Q3)

Calculate modal daily water supply and determine whether Mode is greater or less than Median.

(A) Mode = 158.50 L; Mode < Median

(B) Mode = 167.27 L; Mode > Median (167.27 > 161.25)

(C) Mode = 161.25 L; Mode = Median

(D) Mode = 170.00 L; Mode > Median

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	cf: 15, 40, 72, 120, 140, 150. N/2 = 75 ⇒ Median class = 160 − 180, Lower limit = 160.	[0.5 Mark] for cf progression. [0.5 Mark] for lower limit = 160.
Q2	Preceding class is 140 − 160 with cumulative frequency cf = 72.	[1 Mark] for cf = 72.
Q3	Median = 160 + [(75 − 72)/48] × 20 = 160 + (3/48) × 20 = 160 + 1.25 = 161.25 Litres.	[1 Mark] for formula substitution. [1 Mark] for 161.25 Litres.
Q3 (OR)	Modal class = 160 − 180 (f₁ = 48, f₀ = 32, f₂ = 20). Mode = 160 + [(48 − 32)/(96 − 32 − 20)] × 20 = 160 + (16/44) × 20 = 160 + 7.27 = 167.27 Litres. Modal supply (167.27 L) > Median supply (161.25 L).	[1 Mark] for Mode = 167.27 L. [1 Mark] for comparison (Mode > Median).

Case Study 8 Vocational Training 4 Marks

Vocational Training Centre

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A programming certification assessment for 100 students is tabulated in a 'less than' cumulative table:

Marks	Cumulative Students (\(cf\))
Less than 10	5
Less than 20	12
Less than 30	22
Less than 40	40
Less than 50	65
Less than 60	80
Less than 70	95
Less than 80	100

Q1. Interval Conversion & Frequency 1 Mark

Convert to standard intervals and state the frequency of interval 40 − 50.

(A) 18

(B) 25 (65 − 40)

(C) 15

(D) 20

Q2. Modal Class & Midpoint 1 Mark

Identify the modal class of this distribution and compute its class mark.

(A) 40 − 50, Class mark = 45

(B) 30 − 40, Class mark = 35

(C) 50 − 60, Class mark = 55

(D) 40 − 50, Class mark = 40

Q3. Median & Mean Marks 2 Marks

Calculate the median marks scored by students in the programming assessment.

(A) 42 marks

(B) 45 marks

(C) 44 marks

(D) 46 marks

OR (Alternative Q3)

Using the standard grouped table, compute the mean marks scored by the students.

(A) 43.1 marks (Σfᵢxᵢ = 4310)

(B) 44.5 marks

(C) 41.8 marks

(D) 42.6 marks

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Frequencies: 5, 7, 10, 18, 25, 15, 15, 5. For 40 − 50: frequency = 65 − 40 = 25.	[1 Mark] for frequency = 25.
Q2	Highest frequency is 25 in class 40 − 50 ⇒ Modal class = 40 − 50, Class mark = (40 + 50)/2 = 45.	[0.5 Mark] for modal class. [0.5 Mark] for class mark = 45.
Q3	N/2 = 50. Median class is 40 − 50 (l = 40, cf = 40, f = 25, h = 10). Median = 40 + [(50 − 40)/25] × 10 = 40 + 4 = 44 marks.	[1 Mark] for formula setup. [1 Mark] for Median = 44 marks.
Q3 (OR)	Σfᵢxᵢ = 5(5) + 7(15) + 10(25) + 18(35) + 25(45) + 15(55) + 15(65) + 5(75) = 25 + 105 + 250 + 630 + 1125 + 825 + 975 + 375 = 4310. Mean = 4310 / 100 = 43.1 marks.	[1 Mark] for Σfᵢxᵢ = 4310. [1 Mark] for Mean = 43.1 marks.

Live Practice: Chapter 13 Statistics

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Case Study 1

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