Statistics Class 9: Mean, Median, Mode & Frequency Distribution

Statistics is the branch of mathematics that deals with collecting, organising, analysing and interpreting data. In Class 9, NCERT Chapter 14 introduces you to the foundational tools of statistics.

Why care? Every field — from cricket rankings to election predictions to medical research — uses statistics. Understanding how to summarise data with a single number (mean, median or mode) is a life skill, not just an exam topic.

The chapter covers three main areas:
1. Collection and presentation of data (frequency distribution tables)
2. Graphical representation (bar graphs, histograms, frequency polygons)
3. Measures of central tendency (mean, median, mode)

Let's go through each one.

Raw data is unorganised information. For example, the marks of 30 students in a test:

$56, 42, 78, 65, 42, 89, 56, 73, 42, 91, 78, 56, 65, 42, 78, 89, 56, 73, 65, 91, 42, 56, 78, 65, 73, 89, 42, 56, 78, 91$

Looking at this, it is hard to draw any conclusion. That's why we organise data.

Frequency: The number of times a particular value occurs in the data.

Ungrouped Frequency Distribution Table:

Now the data makes much more sense at a glance!

When data has a large range, we group it into class intervals (also called bins).

Key Terms:
- Class interval: A range like $10$-$20$, $20$-$30$, etc.
- Class size / width: Upper limit $-$ Lower limit. For $10$-$20$, class size $= 10$.
- Class mark (mid-value): $\frac{\text{Upper limit} + \text{Lower limit}}{2}$. For $10$-$20$, class mark $= 15$.
- Frequency: Number of observations falling in each class interval.

Example: The marks of 50 students (out of 100) are grouped:

Convention: In $20$-$40$, the observation $20$ is included but $40$ is not. This ensures each observation falls in exactly one class.

Cumulative Frequency: The running total of frequencies.

Graphs make data visual and easy to compare. Class 9 covers three types.

1. Bar Graph:
- Used for ungrouped or categorical data.
- Bars of equal width with uniform gaps between them.
- Height of each bar $=$ frequency of that category.

2. Histogram:
- Used for grouped (continuous) data.
- Bars are drawn without gaps (because the class intervals are continuous).
- Width of each bar $=$ class size; height $=$ frequency.
- If class sizes are unequal, use frequency density on the y-axis: $\text{Frequency density} = \frac{\text{Frequency}}{\text{Class width}}$.

3. Frequency Polygon:
- Formed by joining the mid-points of the tops of the bars in a histogram with straight lines.
- Can also be drawn independently: plot class marks on the x-axis and frequencies on the y-axis, then connect the points.
- Add a class interval with zero frequency on each side to close the polygon.

Key Difference: Bar graphs have gaps between bars; histograms do not. Bar graphs work for discrete data; histograms work for continuous data.

The mean is the most commonly used measure of central tendency. It gives the 'average' value.

For raw data:

For ungrouped frequency distribution:

Solved Example:
The following table shows the number of goals scored by a football team in 20 matches.

The team scored an average of $2.1$ goals per match.

The median is the middle value when the data is arranged in ascending (or descending) order. It divides the data into two equal halves.

How to find the median:
1. Arrange all observations in ascending order.
2. If $n$ (number of observations) is odd: Median $= \left(\frac{n+1}{2}\right)^{\text{th}}$ observation.
3. If $n$ is even: Median $= \frac{1}{2}\left[\left(\frac{n}{2}\right)^{\text{th}} + \left(\frac{n}{2} + 1\right)^{\text{th}}\right]$ observations.

**Solved Example 1 (Odd $n$):**
Find the median of: $3, 7, 2, 9, 5, 11, 4$.

Arranged: $2, 3, 4, 5, 7, 9, 11$. Here $n = 7$ (odd).

Median $= \left(\frac{7+1}{2}\right)^{\text{th}} = 4^{\text{th}}$ observation $= 5$.

**Solved Example 2 (Even $n$):**
Find the median of: $12, 8, 15, 20, 6, 10$.

Arranged: $6, 8, 10, 12, 15, 20$. Here $n = 6$ (even).

Median $= \frac{1}{2}\left[3^{\text{rd}} + 4^{\text{th}}\right] = \frac{10 + 12}{2} = 11$.

When is median better than mean? When data has extreme values (outliers). For example, if incomes are $\{20000, 25000, 22000, 30000, 500000\}$, the mean ($119400$) is skewed by the outlier, but the median ($25000$) better represents the typical income.

The mode is the value that occurs most frequently in the data. It is the simplest measure of central tendency.

How to find the mode: Count the frequency of each value. The value with the highest frequency is the mode.

Solved Example:
Find the mode of: $2, 3, 5, 3, 7, 3, 8, 5, 3, 2, 5$.

The value $3$ has the highest frequency ($4$).

$\therefore$ Mode $= 3$.

Special Cases:
- No mode: If all values occur with equal frequency (e.g., $1, 2, 3, 4, 5$).
- Bimodal: If two values have the same highest frequency (e.g., in $\{1, 2, 2, 3, 3, 4\}$, both $2$ and $3$ are modes).

When is mode useful? When you want the most 'popular' or 'common' value — e.g., the most common shoe size in a class or the most frequently ordered dish in a restaurant.

Example 1: Mean from a frequency table

The runs scored by 11 players in a cricket match are: $6, 15, 120, 50, 100, 80, 10, 15, 8, 7, 1$. Find the mean and median.

Solution:

Arranged: $1, 6, 7, 8, 10, 15, 15, 50, 80, 100, 120$. $n = 11$ (odd).

Median $= 6^{\text{th}}$ value $= 15$.

Notice the big difference between the mean ($37.45$) and median ($15$). The mean is pulled up by the century-scorers ($100, 120$). In this case, the median better represents the typical score.

Here is a compact recap of everything from NCERT Chapter 14.

Data Presentation: Raw data $\rightarrow$ Frequency distribution (ungrouped / grouped) $\rightarrow$ Graphical representation (bar graph / histogram / frequency polygon).

Measures of Central Tendency:
- Mean $= \frac{\sum f_i x_i}{\sum f_i}$
- Median $=$ Middle value (after sorting)
- Mode $=$ Most frequent value

Key Insight: Mean is affected by extreme values; median and mode are not. Choose the right measure based on the data.

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