Median of Grouped Data: Formula & Calculation
The median of grouped data is the middle value in a dataset that has been organized into class intervals or groups. Unlike ungrouped data, where you can easily arrange individual data points, grouped data provides frequencies for ranges of values. Calculating the median requires identifying the median class (the class interval containing the median) and using a formula to estimate the median’s precise value within that class.
Understanding the median in grouped data is crucial because it summarizes a dataset without being overly influenced by extreme values (outliers), making it a robust measure of central tendency, especially when the data distribution is skewed.
Formulae
The formula to calculate the median of grouped data is:
$Median = L + \left[ \frac{\frac{N}{2} – CF}{f} \right] \times w$
Where:
- $L$ = Lower limit of the median class
- $N$ = Total number of observations (sum of all frequencies)
- $CF$ = Cumulative frequency of the class preceding the median class
- $f$ = Frequency of the median class
- $w$ = Width of the median class (upper limit – lower limit)
Examples
Let’s consider two examples to illustrate how to calculate the median.
Example-1:
Suppose we have the following frequency distribution of exam scores of 40 students:
Class Interval | Frequency (f) | Cumulative Frequency (CF) |
---|---|---|
30-40 | 5 | 5 |
40-50 | 8 | 13 |
50-60 | 12 | 25 |
60-70 | 10 | 35 |
70-80 | 5 | 40 |
1. Calculate N/2: $N = 40$, so $N/2 = 20$
2. Identify the median class: The median class is the class interval where the cumulative frequency first exceeds 20. This is the 50-60 class.
3. Identify the values for the formula:
- $L = 50$ (lower limit of the median class)
- $CF = 13$ (cumulative frequency of the class preceding the median class)
- $f = 12$ (frequency of the median class)
- $w = 10$ (class width: 60-50 = 10)
4. Apply the formula:
$Median = 50 + \left[ \frac{20 – 13}{12} \right] \times 10 = 50 + \left[ \frac{7}{12} \right] \times 10 = 50 + 5.83 = 55.83$
Therefore, the median exam score is approximately 55.83.
Example-2:
Consider a dataset of heights of 50 plants (in cm), grouped as:
Height (cm) | Frequency | Cumulative Frequency |
---|---|---|
10-20 | 4 | 4 |
20-30 | 6 | 10 |
30-40 | 10 | 20 |
40-50 | 15 | 35 |
50-60 | 12 | 47 |
60-70 | 3 | 50 |
1. $N = 50$, so $N/2 = 25$
2. Median Class: 40-50 (cumulative frequency first exceeds 25)
3. Values:
- $L = 40$
- $CF = 20$
- $f = 15$
- $w = 10$
4. Calculation:
$Median = 40 + \left[ \frac{25 – 20}{15} \right] \times 10 = 40 + \left[ \frac{5}{15} \right] \times 10 = 40 + 3.33 = 43.33$
So, the median height of the plants is approximately 43.33 cm.
Common mistakes by students
Students often make the following mistakes:
- Incorrectly calculating the cumulative frequency (CF): Make sure to add the frequencies cumulatively.
- Using the wrong class interval: Carefully identify the median class where the cumulative frequency first exceeds N/2.
- Using the frequency from the wrong class: Always use the frequency of the median class ($f$).
- Forgetting to multiply by the class width (w): This is a crucial part of the formula.
- Confusing median with mean or mode: Remember the formula is specific to the median calculation.
Real Life Application
The median of grouped data is used in many real-life scenarios:
- Economics and Finance: Analyzing income distributions or house prices. The median provides a better representation of the “typical” income or house price than the mean, because it’s less affected by extremely high incomes or property values.
- Healthcare: Analyzing patient age at diagnosis, or the time taken for a patient to recover.
- Education: Calculating the median score of a class on a test, to understand the central performance level.
- Market Research: Analyzing consumer spending habits grouped into various spending brackets.
Fun Fact
The median is a “resistant” statistic, meaning it’s less sensitive to outliers (extreme values) than the mean. This makes it a valuable tool when analyzing datasets that may contain unusual or atypical values that could skew the results.
Recommended YouTube Videos for Deeper Understanding
Q.1 A frequency distribution table is given below: Class Interval | Frequency —|— 0-10 | 5 10-20 | 8 20-30 | 12 30-40 | 10 40-50 | 5 What is the median class?
Check Solution
Ans: C
First calculate the cumulative frequency. The cumulative frequencies are 5, 13, 25, 35, 40. The total frequency (N) is 40. N/2 = 20. The class corresponding to the cumulative frequency that is greater than or equal to 20 is 20-30.
Q.2 The following data represents the marks obtained by 50 students in a test: Marks | Number of Students —|— 0-10 | 4 10-20 | 6 20-30 | 10 30-40 | 15 40-50 | 8 50-60 | 7 Calculate the median mark.
Check Solution
Ans: C
Here, $N = 50$. Therefore, $N/2 = 25$. The cumulative frequencies are 4, 10, 20, 35, 43, 50. The median class is 30-40. Using the formula, $Median = l + \frac{\frac{N}{2} – cf}{f} \times h$, where $l = 30, N/2 = 25, cf = 20, f = 15, h = 10$. So, $Median = 30 + \frac{25 – 20}{15} \times 10 = 33.33$. The closest option is 33.
Q.3 The median of a grouped data is calculated using the formula $Median = l + \frac{\frac{N}{2} – cf}{f} \times h$. Which of the following represents ‘h’ in the formula?
Check Solution
Ans: C
‘h’ represents the class size of the median class, which is the difference between the upper and lower limits of the class.
Q.4 For a given grouped data, the median is 25, the lower limit of the median class is 20, the class size is 10, the frequency of the median class is 12, and the cumulative frequency of the class preceding the median class is 15. What is the value of $N/2$?
Check Solution
Ans: A
Using the formula $Median = l + \frac{\frac{N}{2} – cf}{f} \times h$, we have $25 = 20 + \frac{\frac{N}{2} – 15}{12} \times 10$. Simplifying, $5 = \frac{\frac{N}{2} – 15}{12} \times 10$. $60 = 10(\frac{N}{2} – 15)$. Therefore, $6 = \frac{N}{2} – 15$, and $\frac{N}{2} = 21$.
Q.5 The following table shows the distribution of weights of 40 students. What is the median weight? Weight (kg) | Number of Students —|— 40-45 | 4 45-50 | 8 50-55 | 12 55-60 | 9 60-65 | 7
Check Solution
Ans: B
Here, $N = 40$, so $N/2 = 20$. The median class is 50-55. The cumulative frequencies are 4, 12, 24, 33, 40. Using the formula $Median = l + \frac{\frac{N}{2} – cf}{f} \times h$, we have $Median = 50 + \frac{20 – 12}{12} \times 5 = 50 + \frac{8}{12} \times 5 = 50 + 3.33 = 53.33 \approx 53.75$.
Next Topic: Cumulative Frequency Distribution
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