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Statistics - Quartiles and Interquartile Range

Grade 10ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Quartiles are measures of central tendency that divide a sorted data set into four equal parts. Imagine a horizontal number line where all data points are plotted from smallest to largest; the three points that divide this line into four segments of equal frequency are the quartiles: Q1Q_1 (Lower), Q2Q_2 (Median), and Q3Q_3 (Upper).

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The Lower Quartile (Q1Q_1) marks the 25th25^{th} percentile of the data. Visually, if you have a cumulative frequency curve (Ogive), Q1Q_1 is found by locating the value N4\frac{N}{4} on the vertical y-axis and identifying the corresponding value on the horizontal x-axis.

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The Median (Q2Q_2) is the middle value of the data set, representing the 50th50^{th} percentile. On a graph, this is the point where the horizontal line from the frequency N2\frac{N}{2} intersects the Ogive and drops down to the x-axis.

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The Upper Quartile (Q3Q_3) marks the 75th75^{th} percentile of the data. It is the value below which 75% of the observations lie. On an Ogive, it is determined by finding the x-value corresponding to the cumulative frequency 3N4\frac{3N}{4} on the y-axis.

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The Interquartile Range (IQR) represents the spread of the middle 50% of the data. It is the numerical distance between the upper and lower quartiles. Visually, in a box-and-whisker plot, the IQR is represented by the length of the central box.

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The Semi-Interquartile Range, also known as the Quartile Deviation, is exactly half of the Interquartile Range. It provides a measure of how much the data deviates from the median, focusing on the central portion of the distribution.

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The Ogive (Cumulative Frequency Curve) is the primary visual tool for finding quartiles in grouped data for the ICSE curriculum. It is a smooth curve drawn by plotting upper class boundaries on the x-axis and cumulative frequencies on the y-axis. The curve always slopes upwards from left to right, resembling an elongated 'S' shape.

📐Formulae

Interquartile Range (IQR)=Q3−Q1\text{Interquartile Range (IQR)} = Q_3 - Q_1

Semi-Interquartile Range (Quartile Deviation)=Q3−Q12\text{Semi-Interquartile Range (Quartile Deviation)} = \frac{Q_3 - Q_1}{2}

For Raw Data (if n is odd): Q1=(n+14)th term\text{For Raw Data (if } n \text{ is odd): } Q_1 = (\frac{n+1}{4})^{th} \text{ term}

For Raw Data (if n is odd): Q3=(3(n+1)4)th term\text{For Raw Data (if } n \text{ is odd): } Q_3 = (\frac{3(n+1)}{4})^{th} \text{ term}

For Raw Data (if n is even): Q1=(n4)th term\text{For Raw Data (if } n \text{ is even): } Q_1 = (\frac{n}{4})^{th} \text{ term}

For Raw Data (if n is even): Q3=(3n4)th term\text{For Raw Data (if } n \text{ is even): } Q_3 = (\frac{3n}{4})^{th} \text{ term}

For Grouped Data (using Ogive): Q1=value at frequency N4\text{For Grouped Data (using Ogive): } Q_1 = \text{value at frequency } \frac{N}{4}

For Grouped Data (using Ogive): Q3=value at frequency 3N4\text{For Grouped Data (using Ogive): } Q_3 = \text{value at frequency } \frac{3N}{4}

💡Examples

Problem 1:

Find the Interquartile Range for the following set of marks: 18,12,25,14,21,30,22,15,2818, 12, 25, 14, 21, 30, 22, 15, 28.

Solution:

  1. Arrange the data in ascending order: 12,14,15,18,21,22,25,28,3012, 14, 15, 18, 21, 22, 25, 28, 30.
  2. Count the number of observations: n=9n = 9 (which is odd).
  3. Calculate Q1Q_1: Q1=(n+14)th term=(9+14)th=2.5th termQ_1 = (\frac{n+1}{4})^{th} \text{ term} = (\frac{9+1}{4})^{th} = 2.5^{th} \text{ term}. Taking the average of the 2nd2^{nd} and 3rd3^{rd} terms: Q1=14+152=14.5Q_1 = \frac{14 + 15}{2} = 14.5.
  4. Calculate Q3Q_3: Q3=(3(n+1)4)th term=(3(10)4)th=7.5th termQ_3 = (\frac{3(n+1)}{4})^{th} \text{ term} = (\frac{3(10)}{4})^{th} = 7.5^{th} \text{ term}. Taking the average of the 7th7^{th} and 8th8^{th} terms: Q3=25+282=26.5Q_3 = \frac{25 + 28}{2} = 26.5.
  5. Calculate IQR: IQR=Q3−Q1=26.5−14.5=12IQR = Q_3 - Q_1 = 26.5 - 14.5 = 12.

Explanation:

To find quartiles in raw data, the data must first be sorted. Since the positions resulted in decimals (2.5 and 7.5), we took the arithmetic mean of the adjacent values to find the precise quartile points.

Problem 2:

In a frequency distribution of 40 students, the cumulative frequency (cfcf) table shows that the 10th10^{th} value is 35 marks and the 30th30^{th} value is 72 marks. Find the Semi-Interquartile Range.

Solution:

  1. Identify NN: N=40N = 40 (even).
  2. Find Q1Q_1 position: N4=404=10th term\frac{N}{4} = \frac{40}{4} = 10^{th} \text{ term}. Given: Q1=35Q_1 = 35.
  3. Find Q3Q_3 position: 3N4=3×404=30th term\frac{3N}{4} = \frac{3 \times 40}{4} = 30^{th} \text{ term}. Given: Q3=72Q_3 = 72.
  4. Calculate Semi-Interquartile Range: Semi−IQR=Q3−Q12Semi-IQR = \frac{Q_3 - Q_1}{2} Semi−IQR=72−352=372=18.5Semi-IQR = \frac{72 - 35}{2} = \frac{37}{2} = 18.5.

Explanation:

For even NN in frequency distributions, Q1Q_1 and Q3Q_3 correspond to the values at the N/4N/4 and 3N/43N/4 positions. The Semi-Interquartile Range is then calculated by halving the difference between these two values.