Statistics

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CLASS IX Mathematics Ch 12 of 12
Statistics

Class 9 · Mathematics · NCERT chapter notes · Akanksha Classes

💡 Big idea

Every day the world throws data at us — cricket scores, exam marks, rainfall, prices. Statistics is the friendly maths that collects, organises, shows and summarises that data so a messy pile of numbers tells a clear story!

Data

Facts or figures collected for a purpose. Raw data is unsorted.

Frequency

How many times a value (or class) appears in the data.

Graphs

Bar graphs, histograms & frequency polygons show data as pictures.

Central value

Mean, median & mode — one number that represents the whole set.

📚 Explained

1. What is data?

Data means the facts or figures we collect for a definite purpose. When we just write numbers in the order we get them, it is called raw or ungrouped data. For example, the marks of 10 students — 23, 41, 56, 38, 41, 23, 56, 41, 38, 56 — are raw data. Data collected by the investigator herself (by surveying, measuring or counting) is called primary data, while data taken from a source someone else gathered (a book, website or report) is called secondary data. The range of data is the difference between the highest and lowest values, and it tells us how spread out the values are.

2. Presentation of data

Raw data is hard to read, so we organise it. The first step is often an array — arranging values in ascending or descending order. Next we build a frequency distribution table. The frequency of an observation is the number of times it occurs. Drawing small tally marks (groups of five, with the fifth crossing the bundle) makes counting fast and error-free. A table that lists each separate value with its frequency is an ungrouped frequency distribution.

3. Grouped frequency distribution

When data has many different values (say marks from 0 to 100), listing every value is clumsy. Instead we make class intervals (groups) like 0–10, 10–20, 20–30, and count how many observations fall in each. The size of a class is its class width (here 10). Each class has a lower class limit and an upper class limit. The midpoint of a class is the class mark, found by class mark = (lower limit + upper limit) ÷ 2. There are two styles: the exclusive (continuous) form, where 10–20 means 10 included but 20 excluded (20 goes into the next class), and the inclusive form like 1–10, 11–20, where both ends are included. To draw graphs we must first change inclusive form to exclusive form by adjusting the limits with half the gap between classes.

4. Graphical representation

Pictures make data instantly understandable. Bar graphs use bars of equal width with equal gaps; the height (length) of each bar shows the frequency. They are used for discrete, separate categories like favourite sports or modes of transport. A histogram looks like a bar graph but the bars touch each other with no gaps, because it is used for continuous class intervals; the area of each bar is proportional to its frequency. When the class widths are unequal we adjust the heights so the area stays proportional. A frequency polygon is made by joining the midpoints (class marks) at the tops of the histogram bars with straight line segments; we also add a class with zero frequency at each end so the polygon closes neatly on the horizontal axis. A frequency polygon can be drawn with or without first drawing the histogram.

5. Measures of central tendency

Often we want a single number that represents the whole data set — a typical or central value. These are the measures of central tendency: mean, median and mode.

Mean (average): add up all the observations and divide by how many there are. For values x₁, x₂, …, xₙ, the mean is the sum divided by n, written as the sum of all observations ÷ total number of observations.

Median: the middle value when the data is arranged in order. First arrange the data in ascending order. If the number of observations n is odd, the median is the value at position (n+1)÷2. If n is even, the median is the average of the two middle values, at positions n÷2 and (n÷2)+1. The median splits the data into two equal halves and is not affected by very large or very small extreme values.

Mode: the observation that occurs most often — the value with the highest frequency. A set can have one mode, more than one mode, or no mode at all.

6. Choosing the right measure

The mean uses every value, so it is the most common, but a single very large or very small value (an outlier) can pull it away from the centre. The median is safer when extreme values are present, because it only cares about position, not size. The mode is the best choice when we want the most popular or most frequent item, such as the shoe size a shop sells the most. Understanding which measure fits the situation is an important skill, not just calculating them.

⚡ Key formulae & facts
  • Mean = (Sum of all observations) ÷ (Number of observations)
  • Class mark = (Lower limit + Upper limit) ÷ 2
  • Class width = Upper limit − Lower limit
  • Median (n odd) = value at the (n+1)÷2 th position
  • Median (n even) = average of the (n÷2)th and (n÷2 + 1)th values
  • Mode = observation with the highest frequency
  • Range = Highest value − Lowest value
  • Histogram bars touch (continuous data); bar-graph bars have gaps (discrete data).
📝 Worked example 1

The marks obtained by 9 students are: 52, 60, 48, 55, 60, 70, 48, 60, 65. Find the mean, median and mode.

  1. Mean: add all marks → 52 + 60 + 48 + 55 + 60 + 70 + 48 + 60 + 65 = 518.
  2. There are n = 9 students, so Mean = 518 ÷ 9 = 57.55 (approximately 57.6).
  3. Median: arrange in ascending order → 48, 48, 52, 55, 60, 60, 60, 65, 70.
  4. n = 9 is odd, so median is the (9+1)÷2 = 5th value. The 5th value is 60.
  5. Mode: count frequencies — 60 appears 3 times, more than any other value, so mode = 60.
Answer: Mean ≈ 57.6, Median = 60, Mode = 60.
📝 Worked example 2

The heights (in cm) of 8 plants are: 6, 12, 18, 24, 30, 36, 42, 48. Make a grouped frequency table with class size 12 (classes 0–12, 12–24, …) and find the class marks.

  1. Decide the classes using the exclusive form: 0–12, 12–24, 24–36, 36–48, 48–60. (In exclusive form the upper limit goes into the next class.)
  2. Place each value: 6 → 0–12; 12 and 18 → 12–24; 24 and 30 → 24–36; 36 and 42 → 36–48; 48 → 48–60.
  3. Count the frequencies: 0–12 has 1, 12–24 has 2, 24–36 has 2, 36–48 has 2, 48–60 has 1. Total = 8. ✓
  4. Find class marks using (lower + upper) ÷ 2: for 0–12 it is (0+12)÷2 = 6; for 12–24 it is 18; for 24–36 it is 30; for 36–48 it is 42; for 48–60 it is 54.
Answer: Frequencies 1, 2, 2, 2, 1 (total 8); class marks 6, 18, 30, 42, 54.
🧠 Memory hack

Remember the three M’s: Mean = Move it all together (add & divide), Median = Middle (line them up, pick the centre), Mode = Most (the one that shows up most often). And for graphs: Bars with Breaks = bar graph, Hugging bars = Histogram.

🔥 Rapid fire
Frequency = how many timesClass mark = midpointHistogram bars touchMedian = middle valueMode = most frequentMean = averageRange = max − min
⚠ Don’t lose marks

The most common mistake is finding the median without first arranging the data in ascending order — the middle value of an unsorted list is wrong! Always sort first. Also, when n is even, the median is the average of the two middle values, not just the first middle one. And never leave gaps between histogram bars for continuous data.

🎯 Important questions (with answers)

Q1. Find the mean of the first five even natural numbers.

Answer: The first five even natural numbers are 2, 4, 6, 8, 10. Their sum = 2 + 4 + 6 + 8 + 10 = 30. Number of values n = 5. Mean = 30 ÷ 5 = 6.

Q2. The runs scored in 10 matches are: 12, 25, 30, 12, 18, 25, 12, 40, 35, 21. Find the median.

Answer: Arrange in ascending order → 12, 12, 12, 18, 21, 25, 25, 30, 35, 40. Here n = 10 (even), so the median is the average of the (n÷2)th = 5th value and the (n÷2 + 1)th = 6th value. The 5th value is 21 and the 6th value is 25. Median = (21 + 25) ÷ 2 = 46 ÷ 2 = 23.

Q3. The mean of 6 observations is 15. If five of them are 12, 16, 18, 10 and 20, find the sixth observation.

Answer: Mean = (sum of observations) ÷ n, so sum = mean × n = 15 × 6 = 90. Sum of the five known values = 12 + 16 + 18 + 10 + 20 = 76. Sixth observation = total sum − known sum = 90 − 76 = 14.

Q4. In a class interval 25–35 of a continuous frequency distribution, find the class mark and the class width.

Answer: Class mark = (lower limit + upper limit) ÷ 2 = (25 + 35) ÷ 2 = 60 ÷ 2 = 30. Class width = upper limit − lower limit = 35 − 25 = 10. So the interval is centred at 30 and is 10 units wide.

✅ Quick recap
  • ✅ Data is organised using frequency distribution tables (ungrouped or grouped with class intervals).
  • ✅ Bar graphs (with gaps) suit discrete data; histograms (bars touching) and frequency polygons suit continuous data.
  • ✅ Class mark = (lower + upper) ÷ 2; class width = upper − lower.
  • ✅ Mean = sum ÷ number; Median = middle value (sort first!); Mode = most frequent value.
  • ✅ Use the median when extreme values exist, and the mode for the most popular item.
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