Measures of central tendency — when to use mean, median, or mode

Question

The marks of 10 students are: 12, 15, 15, 18, 20, 22, 25, 28, 30, 95. Find the mean, median, and mode. Which measure best represents this data? Why?

(CBSE 10 & 11 — statistics chapter)

Solution — Step by Step

\text{Mean} = \frac{\text{Sum of all values}}{n} = \frac{12 + 15 + 15 + 18 + 20 + 22 + 25 + 28 + 30 + 95}{10} = \frac{280}{10} = \mathbf{28}

Data is already in ascending order. For $n = 10$ (even):

\text{Median} = \frac{\text{5th value} + \text{6th value}}{2} = \frac{20 + 22}{2} = \mathbf{21}

The most frequently occurring value is 15 (appears twice). All others appear once.

\text{Mode} = \mathbf{15}

The mean ( $28$ ) is pulled up by the outlier ( $95$ ). Most students scored between $12$ - $30$ , so $28$ overestimates the “typical” score.

The median ( $21$ ) is the best representative here — it’s unaffected by the extreme value and sits right in the middle of the data.

The mode ( $15$ ) is too low — it just happens to repeat but doesn’t represent the centre.

Why This Works

Each measure captures a different aspect of “centre”:

Mean uses every value — good for symmetric data, sensitive to outliers
Median is the middle value — robust to outliers, good for skewed data
Mode is the most frequent — useful for categorical data (favourite colour, shoe size)

graph TD
    A["Which central tendency?"] --> B{"Data characteristics?"}
    B -->|"Symmetric, no outliers"| C["Mean<br/>(most informative)"]
    B -->|"Skewed or has outliers"| D["Median<br/>(robust to extremes)"]
    B -->|"Categorical data"| E["Mode<br/>(most frequent category)"]
    B -->|"Open-ended classes"| F["Median<br/>(mean can't be calculated)"]
    C --> G["Use: average income<br/>of a uniform group"]
    D --> H["Use: typical house price<br/>in a city"]
    E --> I["Use: most popular<br/>shirt size"]

Alternative Method — For Grouped Data

When data is in class intervals (like CBSE problems):

Mean: Use $\bar{x} = a + h\left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right)$ (step deviation method)
Median: Use $M = l + \frac{(N/2 - cf)}{f} \times h$
Mode: Use $Mo = l + \frac{(f_1 - f_0)}{2f_1 - f_0 - f_2} \times h$

For CBSE 10 boards: the step deviation method for mean is fastest and least error-prone. Pick the assumed mean $a$ as the class mark of the class with highest frequency. This minimises the $u_i$ values and keeps arithmetic simple.

Common Mistake

In the median formula for grouped data, students confuse $cf$ (cumulative frequency of the class before the median class) with the cumulative frequency of the median class. The formula uses the cumulative frequency up to but NOT including the median class. Getting this wrong shifts your answer significantly.

Question

Solution — Step by Step

Why This Works

Alternative Method — For Grouped Data

Common Mistake

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