CG Board Class 10 Maths Chapter 16 Solutions: Data Analysis (आँकड़ों का विश्लेषण) (2026-27)

• अवर्गीकृत आंकड़े:

Arithmetic Mean (X) = (आंकड़ों का योग)/(आंकड़ों की कुल संख्या)

X = (∑ x_i)/(N)

• अवर्गीकृत बारम्बारता बंटन:

X = (∑ f_i x_i)/(∑ f_i)

Where f_i is the frequency and x_i is the value of the observation.

• वर्गीकृत आंकड़े:

X = (∑ f_i x_i)/(∑ f_i)

Where x_i is the Class Mark / वर्ग चिह्न / मध्य बिंदु of each class interval:

x_i = (निम्न सीमा + उच्च सीमा)/(2)

### 2. माध्यिका

Median is the middlemost value of a dataset when arranged in ascending or descending order.

• अवर्गीकृत आंकड़े:
• जब पदों की संख्या n विषम हो:

Median = ((n + 1)/(2))वाँ पद

• जब पदों की संख्या n सम हो:

Median = ((n/2)th term + (n/2 + 1)th term)/(2)

• वर्गीकृत आंकड़े:

First calculate cumulative frequencies (cf) and determine the Median Class where cf ≥ n/2.

Median = l + ((n/2 – cf)/(f)) × h

• l = माध्यिका वर्ग की निम्न सीमा
• n = कुल बारम्बारता, ∑ f_i
• cf = माध्यिका वर्ग से ठीक पूर्व वर्ग की संचयी बारम्बारता
• f = माध्यिका वर्ग की बारम्बारता
• h = वर्ग अंतराल की माप

### 3. बहुलक

Mode is the observation that occurs with the highest frequency in a dataset.

• वर्गीकृत आंकड़े:

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

• l = बहुलक वर्ग की निम्न सीमा
• f₁ = बहुलक वर्ग की बारम्बारता
• f₀ = बहुलक वर्ग से ठीक पूर्व वर्ग की बारम्बारता
• f₂ = बहुलक वर्ग के ठीक बाद वाले वर्ग की बारम्बारता
• h = वर्ग अंतराल की माप

### 4. Range, Interpolation, परिसर, अंतर्वेशन एवं बहिर्वेशन

• परिसर / विस्तार: The difference between the maximum and minimum values in a dataset.

Range = Maximum Value – Minimum Value

• अंतर्वेशन: Finding or estimating data points within the known range of given observations using graphical representation.
• बहिर्वेशन: Estimating values beyond the known range of data points by extending the observed trend on a graph.

(i) To determine the most popular size of shoes sold by a shopkeeper.**

(ii) To find the average performance of a student in terminal examinations.

(iii) To select candidates for 10 vacancies out of 21 candidates based on interview marks.

(iv) To determine the average monthly expenditure of a family.

(v) To determine the most common height of students in a class.

(vi) To calculate the average rainfall in a city over a decade.

(vii) Determining the height of doors in a house based on the height of family members.

Number of months (N) = 10

Average Rainfall = (Sum of rainfall in all months)/(Total number of months)

Sum = 243.50 + 266.00 + 347.70 + 240.00 + 325.20 + 264.80 + 356.30 + 211.60 + 246.90 + 282.70

Sum = 2784.70 mm

Average Rainfall = (2784.70)/(10) = 278.47 mm

Average rainfall is 278.47 mm.

First 21 natural numbers: 1, 2, 3, 4, …, 21 (N = 21)

Sum of first n natural numbers = (n(n + 1))/(2)

Sum = (21 × 22)/(2) = 21 × 11 = 231

Average = Sum/N = 231/21 = 11

Average of first 21 natural numbers is 11.

Petrol prices = 28.5, 29.2, 28.8, 29.5, 30.0 (Total observations N = 5)

Sum = 28.5 + 29.2 + 28.8 + 29.5 + 30.0 = 146.0 Rupees

Mean Price = (146.0)/(5) = Rupees 29.2

Mean price of petrol is Rupees 29.2.

1. समांतर माध्य:

Sum = 1.80 + 1.99 + 1.95 + 2.02 + 2.02 = 9.78 meters

Mean = (9.78)/(5) = 1.956 ≈ 1.965 meters

1. माध्यिका:

Arrange in ascending order: 1.80, 1.95, 1.99, 2.02, 2.02 (n = 5, odd)

Median = ((5 + 1)/(2))th term = 3rd term = 1.99 meters

1. बहुलक:

Observation 2.02 occurs maximum number of times (2 times).

Mode = 2.02 meters

Arithmetic mean = 1.965 meters, Median = 1.99 meters, Mode = 2.02 meters.

Weights = 30, 32, 33, 38, 37, 41, 35, 40 (N = 8)

Sum = 30 + 32 + 33 + 38 + 37 + 41 + 35 + 40 = 286 kg

Average Weight = 286/8 = 35.75 kg

Average weight is 35.75 kilogram.

Enrollment data = 1150, 1250, 1360, 1275, 1310 (N = 5)

Sum = 1150 + 1250 + 1360 + 1275 + 1310 = 6345 students

Average Enrollment = (6345)/(5) = 1269 students

Average enrollment over 5 years is 1269 students.

• बारम्बारता सारणी गणना:
• माध्य गणना:

X = (∑ f_i x_i)/(∑ f_i) = (1390)/(35) = 39.71428… ≈ 39.71\%

• 5 5 निष्कर्ष:

1. The mean percentage of female teachers across states is 39.71%.
2. Maximum number of states (11 states) have 25-35% female teachers.
3. Minimum number of states (1 state) falls in the highest range of 75-85% female teachers.
4. In 24 states (out of 35), the percentage of female teachers is less than 45%.
5. Only 3 states have more than 65% female teachers.

Mean = 39.71%.

• बहुलक वर्ग की पहचान:

Maximum frequency is 16, corresponding to class interval 100-150.

Therefore, Modal Class = 100-150.

• मान:
• l = 100 (Lower limit)
• f₁ = 16 (Frequency of modal class)
• f₀ = 5 (Preceding frequency)
• f₂ = 12 (Succeeding frequency)
• h = 50 (Class width)
• बहुलक सूत्र से हल:

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

Mode = 100 + [(16 – 5)/(2(16) – 5 – 12)] × 50

Mode = 100 + [(11)/(32 – 17)] × 50 = 100 + [11/15] × 50

Mode = 100 + 550/15 = 100 + 36.666… = 136.66

• 5 5 निष्कर्ष:

1. The modal number of wickets taken by bowlers is 136.66 wickets.
2. Maximum number of bowlers (16) have taken wickets between 100 and 150.
3. Total number of bowlers in the study is 4 + 5 + 16 + 12 + 3 + 2 = 42.
4. Most bowlers (28 out of 42) have taken between 50 and 200 wickets.
5. Only 2 bowlers have taken more than 250 wickets.

Mode = 136.66.

• बारम्बारता सारणी गणना:
• माध्य गणना:

X = (∑ f_i x_i)/(∑ f_i) = (2430)/(35) = 69.4285… ≈ 69.42\%

• 3 3 निष्कर्ष:

1. The average literacy rate across 35 cities is 69.42%.
2. Maximum number of cities (11 cities) fall in the literacy rate bracket of 65-75%.
3. 22 cities out of 35 have a literacy rate of 65% or higher.

Mean literacy rate is 69.42%.

• बारम्बारता सारणी गणना:
• माध्य गणना:

X = (∑ f_i x_i)/(∑ f_i) = (2830)/(80) = 35.375 ≈ 35.37 years

• 3 3 निष्कर्ष:

1. The average age of patients admitted to the hospital is 35.37 years.
2. Maximum number of patients (23) admitted fall in the age group of 35-45 years.
3. Total number of patients admitted during the year is 80.

Mean age is 35.37 years.

• संचयी बारम्बारता सारणी:
• माध्यिका वर्ग का निर्धारण:

Total frequency N = 84 ⇒ N/2 = 84/2 = 42.

Cumulative frequency just greater than 42 is 69, corresponding to class interval 30-40.

Therefore, Median Class = 30-40.

• मान:
• l = 30 (Lower limit)
• cf = 37 (Preceding cumulative frequency)
• f = 32 (Frequency of median class)
• h = 10 (Class size)
• माध्यिका सूत्र से हल:

Median = l + ((N/2 – cf)/(f)) × h

Median = 30 + ((42 – 37)/(32)) × 10 = 30 + (5/32) × 10

Median = 30 + 50/32 = 30 + 1.5625 ≈ 31.56

• 3 3 निष्कर्ष:

1. The median mark of the students is 31.56.
2. Exactly 50% of the students scored less than 31.56 marks, and 50% scored more.
3. The maximum concentration of students (32 students) scored between 30 and 40 marks.

Median = 31.56.

💡 Teacher Nudge / अध्ययन की अगली सलाह: