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Statistics - Median of Grouped Data

Grade 10CBSE

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

🔑Concepts

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The Median is the measure of central tendency which gives the value of the middle-most observation in the data. For grouped data, it divides the frequency distribution into two equal parts.

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Cumulative Frequency (cf): To calculate the median, we first create a cumulative frequency table (of 'less than' type). Visually, this is a third column where the value for each row is the sum of its frequency and the frequencies of all preceding rows, representing a running total.

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The Median Class: This is the class interval whose cumulative frequency is just greater than (and nearest to) n2\frac{n}{2}, where nn is the total number of observations (total frequency). Identifying this class is the first step in applying the formula.

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Formula Variables: Once the median class is identified, we define ll as the lower limit of the median class, ff as the frequency of the median class, and cfcf as the cumulative frequency of the class preceding the median class.

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Class Size (hh): This represents the width of the class intervals. It is crucial that all class intervals are of equal width and continuous. If the data has gaps (e.g., 10−19,20−2910-19, 20-29), they must be converted to continuous classes (e.g., 9.5−19.5,19.5−29.59.5-19.5, 19.5-29.5) by adjusting boundaries by 0.50.5.

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Empirical Relationship: There is a mathematical relationship between the three measures of central tendency: 3×Median=Mode+2×Mean3 \times \text{Median} = \text{Mode} + 2 \times \text{Mean}. This is useful for checking the consistency of your results.

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Graphical Representation: The median can be estimated visually using Ogives (Cumulative Frequency Curves). The x-coordinate of the point of intersection of the 'less than type' ogive and 'more than type' ogive gives the Median.

📐Formulae

Median=l+(n2−cff)×h\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h

n=∑fin = \sum f_i

3×Median=Mode+2×Mean3 \times \text{Median} = \text{Mode} + 2 \times \text{Mean}

💡Examples

Problem 1:

Calculate the median for the following distribution: Class: 0−10,10−20,20−30,30−40,40−500-10, 10-20, 20-30, 30-40, 40-50; Frequency (ff): 5,8,20,15,75, 8, 20, 15, 7.

Solution:

  1. Find Cumulative Frequency (cfcf):
  • 0−10:50-10: 5
  • 10−20:5+8=1310-20: 5+8=13
  • 20−30:13+20=3320-30: 13+20=33
  • 30−40:33+15=4830-40: 33+15=48
  • 40−50:48+7=5540-50: 48+7=55

  1. Here, n=55n = 55. Therefore, n2=552=27.5\frac{n}{2} = \frac{55}{2} = 27.5.

  2. Identify Median Class: The cfcf just greater than 27.527.5 is 3333. So, the median class is 20−3020-30.

  3. Identify parameters:

  • l=20l = 20
  • cf=13cf = 13 (cumulative frequency of the preceding class)
  • f=20f = 20
  • h=10h = 10
  1. Apply Formula: Median=20+(27.5−1320)×10\text{Median} = 20 + \left( \frac{27.5 - 13}{20} \right) \times 10 Median=20+(14.520)×10\text{Median} = 20 + \left( \frac{14.5}{20} \right) \times 10 Median=20+14.52=20+7.25=27.25\text{Median} = 20 + \frac{14.5}{2} = 20 + 7.25 = 27.25

Explanation:

We first calculate the cumulative frequencies to find the middle position. Since the 27.5th27.5^{th} value falls within the 20−3020-30 range, we use the specific properties of that class (l,f,hl, f, h) and the previous class (cfcf) to interpolate the exact median value.

Problem 2:

If the median of a distribution is 28.528.5 and the total frequency is 6060, find the missing frequency xx for the class 10−2010-20 if the frequencies are: 0−10:5,10−20:x,20−30:20,30−40:15,40−50:y,50−60:50-10: 5, 10-20: x, 20-30: 20, 30-40: 15, 40-50: y, 50-60: 5.

Solution:

  1. Form cfcf table:
  • 0−10:50-10: 5
  • 10−20:5+x10-20: 5+x
  • 20−30:25+x20-30: 25+x
  • 30−40:40+x30-40: 40+x
  • 40−50:40+x+y40-50: 40+x+y
  • 50−60:45+x+y50-60: 45+x+y

  1. Given n=60n=60, so 45+x+y=60⇒x+y=1545+x+y=60 \Rightarrow x+y=15.

  2. Median is 28.528.5, which lies in class 20−3020-30. So, Median Class =20−30= 20-30.

  3. Parameters:

  • l=20,f=20,cf=5+x,h=10,n2=30l=20, f=20, cf=5+x, h=10, \frac{n}{2}=30

  1. Substitute in formula: 28.5=20+(30−(5+x)20)×1028.5 = 20 + \left( \frac{30 - (5+x)}{20} \right) \times 10 8.5=25−x28.5 = \frac{25-x}{2} 17=25−x17 = 25 - x x=8x = 8

  2. Since x+y=15x+y=15, 8+y=15⇒y=78+y=15 \Rightarrow y=7.

Explanation:

When the median is given, we determine the median class by checking which interval contains the median value (28.528.5). We then set up an algebraic equation using the median formula to solve for the unknown frequencies.