NCERT Class 10 Maths Chapter 14: Statistics Exercise 14.3

Understanding the Median of Grouped Data

The Median is a measure of central tendency that gives the value of the middlemost observation in the data. For ungrouped data, we simply arrange the numbers in ascending order and find the middle term. However, for grouped frequency distributions, we cannot identify the exact observations directly. To solve this, we construct a Cumulative Frequency ($cf$) table.

There are two types of cumulative frequencies:

1. 'Less than' type: Obtained by adding the frequencies of all previous classes up to the current upper limit.
2. 'More than' type: Obtained by subtracting preceding class frequencies from the total frequency starting from the lower limit.

To find the median, we identify the Median Class—the class interval whose cumulative frequency is greater than and closest to $n/2$ (where $n$ is the total frequency). Once the median class is identified, we use the following standard interpolation formula:

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

Where:

• $l$ = lower limit of the median class
• $n$ = total number of observations (sum of frequencies)
• $cf$ = cumulative frequency of the class preceding the median class
• $f$ = frequency of the median class
• $h$ = class size (assuming class widths are equal)

How to Calculate the Median of Grouped Data

1. Construct the Cumulative Frequency Column — Create a new column in your table named Cumulative Frequency ($cf$). Calculate each row's cumulative frequency by adding its frequency to the sum of all preceding frequencies.
2. Calculate the Half-Frequency Value — Sum all the frequencies to find $n$. Then, compute the value of $n/2$.
3. Locate the Median Class — Look down the cumulative frequency column to find the first value that is greater than or equal to $n/2$. The corresponding class interval is your Median Class.
4. Extract Values and Apply Formula — Identify $l$ (lower limit of the median class), $cf$ (cumulative frequency of the preceding class), $f$ (frequency of the median class), and $h$ (class width). Substitute these values into the median formula and calculate the final result.

Step-by-Step Solved Core Examples

• Example 1: Basic Median Calculation Calculate the median for the following distribution: - Class Interval: 0-10, 10-20, 20-30, 30-40, 40-50 - Frequency ($f$): 5, 8, 12, 7, 8 Solution: 1. Prepare the Cumulative Frequency table: - 0-10: $f = 5 \rightarrow cf = 5$ - 10-20: $f = 8 \rightarrow cf = 5 + 8 = 13$ - 20-30: $f = 12 \rightarrow cf = 13 + 12 = 25$ - 30-40: $f = 7 \rightarrow cf = 25 + 7 = 32$ - 40-50: $f = 8 \rightarrow cf = 32 + 8 = 40$ 2. Find $n$: Here, $n = 40$. Therefore, $n/2 = 20$. 3. Locate Median Class: The $cf$ just greater than $20$ is $25$. The corresponding class is 20-30. Thus, the Median Class is 20-30. 4. Extract parameters: - $l = 20$ - $cf = 13$ (from the preceding class 10-20) - $f = 12$ - $h = 10$ 5. Apply Formula: $\text{Median} = 20 + \left(\frac{20 - 13}{12}\right) \times 10 = 20 + \left(\frac{7}{12}\right) \times 10 = 20 + 5.83 = 25.83$ Final Answer: The median is 25.83.

Pro Board Exam Tips & Common Pitfalls

• The '$cf
  Value Trap: A very common mistake is using the cumulative frequency of the median class itself rather than the preceding class. Remember: $cf$ is always taken from the row above your median class!
• Ensure Continuous Classes: If your class intervals are given in a discontinuous format (e.g., 11-15, 16-20), you must convert them to continuous intervals (e.g., 10.5-15.5, 15.5-20.5) by subtracting 0.5 from the lower limits and adding 0.5 to the upper limits before solving.
• The Empirical Formula: Keep this formula memorized to check your calculations or quickly solve short-answer questions:

$3 \times \text{Median} = \text{Mode} + 2 \times \text{Mean}$
If you have calculated any two, you can easily verify or calculate the third using this relationship.

Practice Questions with Solutions

• Q: Calculate the median lifetime of 400 neon lamps from the following data: - Lifetime (in hours): 1500-2000, 2000-2500, 2500-3000, 3000-3500, 3500-4000, 4000-4500, 4500-5000 - Number of lamps ($f$): 14, 56, 60, 86, 74, 62, 48 A: Step 1: Prepare the cumulative frequency column. - 1500-2000: $f = 14 \rightarrow cf = 14$ - 2000-2500: $f = 56 \rightarrow cf = 70$ - 2500-3000: $f = 60 \rightarrow cf = 130$ - 3000-3500: $f = 86 \rightarrow cf = 216$ - 3500-4000: $f = 74 \rightarrow cf = 290$ - 4000-4500: $f = 62 \rightarrow cf = 352$ - 4500-5000: $f = 48 \rightarrow cf = 400$ Step 2: Calculate $n/2$. $n = 400$, so $n/2 = 200$. Step 3: Identify the median class. The $cf$ just greater than 200 is 216, which corresponds to the class 3000-3500. Step 4: Identify values and solve. - $l = 3000$ - $cf = 130$ - $f = 86$ - $h = 500$ $\text{Median} = 3000 + \left(\frac{200 - 130}{86}\right) \times 500$ $\text{Median} = 3000 + \left(\frac{70}{86}\right) \times 500 = 3000 + 406.98 = 3406.98$ Final answer: The median lifetime of the neon lamps is 3406.98 hours.
• Q: If the median of the distribution given below is 28.5, find the values of $x$ and $y$. Total frequency is 60. - Class: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60 - Frequency ($f$): 5, $x$, 20, 15, $y$, 5 A: Step 1: Create cumulative frequency ($cf$) expressions. - 0-10: $cf = 5$ - 10-20: $cf = 5 + x$ - 20-30: $cf = 25 + x$ - 30-40: $cf = 40 + x$ - 40-50: $cf = 40 + x + y$ - 50-60: $cf = 45 + x + y$ Step 2: Set up equations. We know total frequency $n = 60$. Thus: $45 + x + y = 60 \Rightarrow x + y = 15$ --- (Equation 1) Step 3: Identify the median class. Since the given median is 28.5, which lies in the interval 20-30, the median class is 20-30. - $l = 20$ - $cf = 5 + x$ - $f = 20$ - $h = 10$ - $n/2 = 30$ Step 4: Solve for $x$ and $y$. $28.5 = 20 + \left(\frac{30 - (5 + x)}{20}\right) \times 10$ $8.5 = \frac{25 - x}{2}$ $17 = 25 - x \Rightarrow x = 8$ Substitute $x = 8$ in Equation 1: $8 + y = 15 \Rightarrow y = 7$ Final answer: $x = 8$ and $y = 7$.
• Q: A life insurance agent found the following data for distribution of ages of 100 policyholders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years: - Age (in years): Below 20, Below 25, Below 30, Below 35, Below 40, Below 45, Below 50, Below 55, Below 60 - Number of policyholders ($cf$): 2, 6, 24, 45, 78, 89, 92, 98, 100 A: Step 1: Convert the 'Below' distribution to continuous class intervals and find actual frequencies ($f$). - 15-20: $f = 2$ - 20-25: $f = 6 - 2 = 4$ - 25-30: $f = 24 - 6 = 18$ - 30-35: $f = 45 - 24 = 21$ - 35-40: $f = 78 - 45 = 33$ - 40-45: $f = 89 - 78 = 11$ - 45-50: $f = 92 - 89 = 3$ - 50-55: $f = 98 - 92 = 6$ - 55-60: $f = 100 - 98 = 2$ Step 2: Find $n/2$ and median class. $n = 100$, so $n/2 = 50$. The $cf$ just greater than 50 is 78. This corresponds to the median class 35-40. Step 3: Extract values and apply formula. - $l = 35$ - $cf = 45$ - $f = 33$ - $h = 5$ $\text{Median} = 35 + \left(\frac{50 - 45}{33}\right) \times 5 = 35 + \left(\frac{5}{33}\right) \times 5 = 35 + 0.76 = 35.76$ Final answer: The median age of the policyholders is 35.76 years.
• Q: Using the empirical relationship between the three measures of central tendency, find the Mean of a distribution whose Median is 45 and Mode is 48. A: Step 1: State the empirical formula. The empirical relationship is: $3 \times \text{Median} = \text{Mode} + 2 \times \text{Mean}$ Step 2: Substitute the given values. Given: - Median = 45 - Mode = 48 $3 \times (45) = 48 + 2 \times \text{Mean}$ $135 = 48 + 2 \times \text{Mean}$ Step 3: Solve for Mean. $2 \times \text{Mean} = 135 - 48$ $2 \times \text{Mean} = 87$ $\text{Mean} = 87 / 2 = 43.5$ Final answer: The Mean of the distribution is 43.5.

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