Chapter 3, Data Handling (आँकड़ों का प्रबंधन), teaches students how to organize, represent, and interpret data using measures of central tendency (Mean, Median, Mode) and probability.
Table of Contents
Exercise 3.1: Mean (Average)
This exercise introduces the Arithmetic Mean.
| Q. No. | Type | Example (Step-by-Step Solution) | Key Concept |
| 1 | Finding Mean | Find the mean of the first five whole numbers ($0, 1, 2, 3, 4$). Sum $= 0 + 1 + 2 + 3 + 4 = 10$ No. of observations $= 5$ Mean $= \frac{\text{Sum}}{\text{No. of observations}} = \frac{10}{5} = \mathbf{2}$ | Mean: Sum of all observations divided by the number of observations. |
| 2 | Finding Range | A cricket player scored 6, 15, 120, 50, 100, 80 in 6 innings. Find the range. Highest Score $= 120$ Lowest Score $= 6$ Range $= 120 – 6 = \mathbf{114}$ | Range: The difference between the highest and the lowest observation. |
| 3 | Application | A student’s height (in cm) is 150, 145, 155. What is the mean height? Sum $= 150 + 145 + 155 = 450$ Mean $= \frac{450}{3} = \mathbf{150 \text{ cm}}$ | Real-world application of average. |
Exercise 3.2: Mode and Median
This focuses on two other measures of central tendency: Mode and Median.
| Q. No. | Type | Example (Step-by-Step Solution) | Key Concept |
| 1 | Finding Mode | Find the mode of the data: $2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2$. Count occurrences: 2 appears 3 times, 3 appears 3 times, 4 appears 2 times, etc. Mode: 2 and 3 (Bimodal) | Mode: The observation that occurs most frequently. |
| 2 | Finding Median | Find the median of the data: $13, 16, 12, 14, 19, 12, 14, 13, 14$. 1. Arrange in ascending order: $12, 12, 13, 13, \mathbf{14}, 14, 14, 16, 19$. 2. Number of observations ($N$) = 9 (Odd). Median is the $\frac{N+1}{2}$-th term = 5th term. Median: $\mathbf{14}$ | Median: The middle value when the data is arranged in order. |
| 3 | Median (Even N) | Data: $1, 2, 4, 5, 6, 8$. Find the median. $N=6$ (Even). Median is the average of the $\frac{N}{2}$-th (3rd) and $(\frac{N}{2}+1)$-th (4th) terms. Median $= \frac{4 + 5}{2} = \frac{9}{2} = \mathbf{4.5}$ | For an even number of observations, the median is the average of the two middle terms. |
Exercise 3.3: Bar Graphs
This focuses on reading and interpreting Bar Graphs.
| Q. No. | Type | Example (Step-by-Step Solution) | Key Concept |
| 1 | Reading Bar Graphs | Based on a bar graph showing students’ favorite subjects, which subject is most popular? Solution: Look for the tallest bar. (e.g., Mathematics, if the bar is highest) | Interpretation of the height of bars to compare data values. |
| 2 | Double Bar Graphs | A double bar graph compares the performance of a class in two consecutive tests. If the Science bar increased from 60 to 80, what is the improvement? Solution: Improvement $= 80 – 60 = \mathbf{20}$ marks. | Double Bar Graph: Used to compare two sets of data simultaneously. |
Exercise 3.4: Probability
This introduces the basic concepts of Chance and Probability.
| Q. No. | Type | Example (Step-by-Step Solution) | Key Concept |
| 1 | Certainty | Tell whether the event is certain to happen, impossible, or can happen but not certain: You are older today than yesterday. Answer: Certain to happen. | Events can be Certain, Impossible, or Possible. |
| 2 | Calculating Probability | There are 6 marbles in a box with numbers 1 to 6 marked on each of them. What is the probability of drawing a marble with number 2? Total outcomes = 6 Favourable outcome (getting a 2) = 1 Probability $= \frac{\text{Favourable outcomes}}{\text{Total outcomes}} = \mathbf{\frac{1}{6}}$ | Probability: A measure of the likelihood of an event occurring. |
| 3 | Coin Toss | If you toss a coin, what is the probability of getting a head? Total outcomes (Head or Tail) = 2 Favourable outcome (Head) = 1 Probability: $\mathbf{\frac{1}{2}}$ | Probability ranges from 0 (impossible) to 1 (certain). |