Presentation of Data — Class 11 CBSE Statistics (Chapter 4)

Why We Classify and Present Data

Raw data (also called ungrouped data) is the information collected directly from the field — unorganised and unprocessed. For example, marks of 30 students written in the order they were checked. Such data is difficult to analyse directly. Classification is the process of arranging data into groups or classes according to similarities, so that all observations sharing a common characteristic fall into one class. Once classified, data is displayed as a frequency distribution — a table showing how many observations (frequency) fall into each class. This organisation reduces bulk, brings out hidden patterns, allows comparison, and becomes the base for calculating averages, dispersion and correlation later. Two key methods of presenting classified data are: (1) Tabular presentation — arranging into rows and columns, and (2) Diagrammatic presentation — bar diagrams, pie charts, histograms, frequency polygons and ogives, which appeal to the eye and communicate trends instantly. A good presentation is condensed, clear, and free from ambiguity.

Key Terms You Must Know

Raw Data / Array

Data as originally collected, unorganised, is raw data. When arranged in ascending or descending order, it is called an array.

Class Limits

The two extreme values of a class — lower class limit (smallest value) and upper class limit (largest value), e.g., in 20–30, 20 is lower limit and 30 is upper limit.

Class Interval / Class Width

The difference between the upper and lower limits of a class, i.e., width = upper limit − lower limit.

Class Mark (Mid-value)

The middle value of a class, calculated as (lower limit + upper limit) ÷ 2. Used to represent the whole class in calculations.

Frequency

The number of observations falling within a particular class interval, denoted by f.

Exclusive Series

A classification where the upper limit of one class is the same as the lower limit of the next (e.g., 10–20, 20–30). A value equal to the upper limit is included in the NEXT class.

Inclusive Series

A classification where both limits belong to the same class (e.g., 10–19, 20–29), leaving a gap between classes that must be adjusted before further calculation.

Cumulative Frequency

The running total of frequencies up to a given class — 'less than' type accumulates from the start, 'more than' type accumulates from the end.

Steps to Construct a Frequency Distribution Table

1. Step 1: Collect and note the raw data — Write down all observations exactly as recorded, e.g., marks of 20 students: 45,32,78,56,23,67,89,12,34,56,78,90,23,45,67,89,12,34,56,78.
2. Step 2: Find the Range — Range = Maximum value − Minimum value. Here, Max = 90, Min = 12, so Range = 90 − 12 = 78.
3. Step 3: Decide the number of classes — Use Sturges' Rule: k = 1 + 3.322 log₁₀N, where N is the number of observations. For a rough estimate, 5 to 15 classes are commonly chosen depending on data size.
4. Step 4: Determine class width (i) — Class width i = Range ÷ Number of classes, rounded to a convenient number (like 10, 20). Here we choose width = 10, giving 8–9 classes to cover 12 to 90.
5. Step 5: Form mutually exclusive class intervals — Start from a value slightly below the minimum: 0–10, 10–20, 20–30, ..., 90–100, ensuring every data point fits exactly one class with no overlap.
6. Step 6: Tally the observations — Go through each raw value once and place a tally mark (|) in its correct class interval; group every 5th tally as ||||̸ for easy counting.
7. Step 7: Count frequencies and verify total — Convert tally marks into frequency numbers for each class. The sum of all frequencies MUST equal the total number of observations, N.

Fully Worked Examples

• Example 1 — Building the Frequency Distribution: Using the 20 marks (45,32,78,56,23,67,89,12,34,56,78,90,23,45,67,89,12,34,56,78) with class width 10: Step 1: Range = 90 − 12 = 78. Step 2: Choose classes of width 10 starting at 0: 0–10,10–20,...,90–100. Step 3: Tally each value into its class. 0–10: 0 | 10–20: 12,12 → 2 | 20–30: 23,23 → 2 | 30–40: 32,34,34 → 3 | 40–50: 45,45 → 2 | 50–60: 56,56,56 → 3 | 60–70: 67,67 → 2 | 70–80: 78,78,78 → 3 | 80–90: 89,89 → 2 | 90–100: 90 → 1. Step 4: Check total frequency = 2+2+3+2+3+2+3+2+1 = 20 = N. ✓ Final table is the required exclusive-type frequency distribution.
• Example 2 — Converting Inclusive to Exclusive Series: Given inclusive classes 10–19, 20–29, 30–39, 40–49 with frequencies 4, 8, 10, 6, convert to continuous (exclusive) form. Step 1: Find the gap between upper limit of one class and lower limit of next: 20 − 19 = 1. Step 2: Correction factor = gap ÷ 2 = 0.5. Step 3: Subtract 0.5 from every lower limit and add 0.5 to every upper limit: 9.5–19.5, 19.5–29.5, 29.5–39.5, 39.5–49.5. Step 4: Frequencies remain unchanged: 4, 8, 10, 6. Final answer: The continuous series is 9.5–19.5 (f=4), 19.5–29.5 (f=8), 29.5–39.5 (f=10), 39.5–49.5 (f=6), suitable for drawing a histogram.
• Example 3 — Constructing 'Less Than' Cumulative Frequency: Using the frequency table from Example 1 (classes of width 10, frequencies 0,2,2,3,2,3,2,3,2,1), build the less than cumulative frequency (c.f.) table. Step 1: 'Less than 10' c.f. = 0. Step 2: 'Less than 20' c.f. = 0+2 = 2. Step 3: 'Less than 30' c.f. = 2+2 = 4. Step 4: 'Less than 40' c.f. = 4+3 = 7. Step 5: Continue adding: Less than 50 = 9, Less than 60 = 12, Less than 70 = 14, Less than 80 = 17, Less than 90 = 19, Less than 100 = 20. Final answer: The last cumulative frequency (20) equals total N, confirming correctness — this table is plotted to draw a 'less than' Ogive curve.

Common Mistakes and Board Exam Tips

Students frequently lose marks by: (1) forgetting to convert an inclusive series into an exclusive one before drawing a histogram — histograms need continuous class boundaries with no gaps; (2) miscounting tally marks by missing one observation, so ALWAYS verify Σf = N; (3) confusing class mark (midpoint) with class limit in calculations; (4) mixing up 'less than' and 'more than' ogive construction — less than ogive uses upper limits and rises, more than ogive uses lower limits and falls; (5) drawing a bar diagram instead of a histogram for continuous data (bars must touch for histograms, since class boundaries are continuous, but stand separated for bar diagrams of discrete/categorical data). Always label axes, mention the scale used, and write the correction factor calculation explicitly if converting inclusive to exclusive limits — examiners award step marks for this.

Practice Questions with Solutions

• Q: The weights (in kg) of 50 students range from a minimum of 32 kg to a maximum of 74 kg. Using Sturges' Rule, find the approximate number of classes and the class width (round width to nearest 5). A: Step 1: Range = Maximum − Minimum = 74 − 32 = 42. Step 2: Sturges' Rule: k = 1 + 3.322 log₁₀N = 1 + 3.322 log₁₀50. Step 3: log₁₀50 ≈ 1.699, so k = 1 + 3.322 × 1.699 ≈ 1 + 5.645 = 6.645 ≈ 7 classes. Step 4: Class width i = Range ÷ k = 42 ÷ 7 = 6, rounded to a convenient value of 5 (so we may use i = 5 giving about 9 classes, or keep i = 6 for 7 classes as computed). Final answer: Approximately 7 classes with class width ≈ 6 (or adjusted to 5 for convenience).
• Q: Convert the inclusive class interval 30–39 (frequency 12) into its exclusive (continuous) form. A: Step 1: Identify the gap between classes: next class starts at 40, so gap = 40 − 39 = 1. Step 2: Correction factor = gap ÷ 2 = 0.5. Step 3: New lower limit = 30 − 0.5 = 29.5; new upper limit = 39 + 0.5 = 39.5. Step 4: Frequency remains unchanged at 12. Final answer: The exclusive class interval is 29.5–39.5 with frequency 12.
• Q: Find the class mark (mid-value) of the class interval 45–65. A: Step 1: Class mark formula = (Lower limit + Upper limit) ÷ 2. Step 2: Substitute values: (45 + 65) ÷ 2 = 110 ÷ 2. Step 3: Simplify: 110 ÷ 2 = 55. Final answer: The class mark is 55.
• Q: Given the frequency distribution — Classes: 0–10, 10–20, 20–30, 30–40 with frequencies 5, 8, 12, 5 — construct the 'more than' type cumulative frequency table. A: Step 1: Total frequency N = 5+8+12+5 = 30. Step 2: 'More than 0' c.f. = total N = 30 (all values are ≥ 0). Step 3: 'More than 10' c.f. = 30 − 5 = 25 (subtract frequency of first class). Step 4: 'More than 20' c.f. = 25 − 8 = 17. Step 5: 'More than 30' c.f. = 17 − 12 = 5, and this equals the last class frequency (5), confirming accuracy. Final answer: More than 0 = 30, More than 10 = 25, More than 20 = 17, More than 30 = 5.
• Q: Distinguish between a histogram and a simple bar diagram, and state one situation where each is used. A: Step 1: A histogram is drawn for continuous, grouped (frequency distribution) data, where bars are erected on class intervals (which touch each other with no gaps) and the area of each bar is proportional to frequency. Step 2: A simple bar diagram is drawn for discrete or categorical data (like sales by year or population by state), where bars are of equal width but are kept separated by gaps to show they represent distinct, non-continuous categories. Step 3: Example usage — a histogram suits presenting marks of students grouped in class intervals (0–10, 10–20...); a bar diagram suits comparing the population of 5 different states. Final answer: Histogram = continuous grouped data with touching bars; Bar diagram = discrete/categorical data with separated bars.

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