Case Study Questions on Statistics Class 10

Case Study Questions on Statistics Class 10

Introduction to Statistics Case Studies

In Class 10 Mathematics, Math Case Study Questions Class 10 Statistics play a vital role in testing students’ understanding. These questions present real-life data for analysis, making learning more engaging. By practicing Case Study Questions on Statistics , learners develop better logical and analytical skills. Moreover, solving such questions improves interpretation abilities, which are highly useful during board examinations.

NCERT and CBSE Statistics Case Studies

Case Study Questions are designed to strengthen concepts through practical applications. Similarly, Ncert Case Study Questions Class 10 Maths Statistics provide accurate practice aligned with the curriculum. These case studies include data-based problems on mean, mode, and median. Consequently, students get the opportunity to apply theoretical knowledge while solving questions. Therefore, practicing NCERT and CBSE problems regularly is essential for exam success.

Practice with Solved Problems

Working with Class 10 Maths Case Study Problems on Statistics boosts confidence and accuracy. In addition, Solved Case Study Questions Statistics Class 10 offer step-by-step solutions, which help students learn methods effectively. This approach not only builds clarity but also ensures faster problem-solving. As a result, learners are better prepared to attempt case studies in their board examinations.

Case Study 3: Sapling Growth Analysis

Case Study 3 Statistics

A school gardening club decided to study the growth pattern of a batch of 45 saplings over a period of three months. The club recorded the heights (in centimetres) of the saplings at the end of the period and grouped the continuous data into class intervals of width 10 cm. The purpose was to determine the central tendency and to see whether most plants reached a healthy height bracket. The club also wanted to identify the modal and median classes to decide which group of plants might need extra care and which are thriving. Additionally, students were asked to prepare cumulative frequency tables and an ogive to visually read off the median. This kind of grouped data analysis helps the club plan targeted watering and fertiliser schedules and gives practice in using mean, median, mode and cumulative frequencies to make data driven decisions in a real life context. Based on the grouped frequency table below, answer the questions that follow.

Height (cm) Number of Saplings (Frequency)
20-30 4
30-40 6
40-50 10
50-60 12
60-70 8
70-80 5

1. What is the total number of saplings recorded?

Answer: (c) 45

Solution: Sum of frequencies = 4 + 6 + 10 + 12 + 8 + 5 = 45.

2. What is the cumulative frequency up to the class interval 40-50 (inclusive)?

Answer: (c) 20

Solution: Cumulative frequency up to 40-50 = 4 + 6 + 10 = 20.

3. Which class contains the median?

Answer: (b) 50-60

Solution: Total N = 45, so N/2 = 22.5. Cumulative frequencies are 4, 10, 20, 32, 40, 45. The 22.5th observation lies in the class where cumulative frequency first equals or exceeds 22.5, which is 50-60 (cumulative 32). Hence the median class is 50-60.

4. Using the grouped median formula with class boundaries, what is the approximate median height? Use the formula:
Median ≈ l + ((N/2 – c)/f) × h
where l is the lower class boundary of the median class, c is cumulative frequency before the median class, f is frequency of the median class and h is class width.

Answer: (b) 51.58 cm

Solution: Median class is 50-60. For continuous class boundaries, lower boundary l = 49.5, class width h = 10, cumulative frequency before median class c = 20, frequency of median class f = 12, and N/2 = 22.5.
Substitute:
Median ≈ 49.5 + ((22.5 – 20)/12) × 10
= 49.5 + (2.5/12) × 10
= 49.5 + 25/12 = 49.5 + 2.0833 ≈ 51.58 cm

5. What is the approximate mean height of the saplings using class marks?

Answer: (b) 51.44 cm

Solution: Class marks: 25, 35, 45, 55, 65, 75. Multiply by frequencies:
4×25 = 100
6×35 = 210
10×45 = 450
12×55 = 660
8×65 = 520
5×75 = 375
Sum of products = 100 + 210 + 450 + 660 + 520 + 375 = 2315
Mean = 2315/45 ≈ 51.444 cm ≈ 51.44 cm

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