Find the median of the following series: Items: 20-30, 30-40, 40-50, 50-60, 60-70, 70-80, 80-90 Frequency: 5, 10, 16, 18, 12, 10, 8
Understand the Problem
The question asks to find the median of the provided series. The series lists items in ranges (e.g., 20-30, 30-40) along with their corresponding frequencies. The goal is to calculate the median value from this frequency distribution.
Answer
$Median = 54.72$
Answer for screen readers
$Median = 54.72$
Steps to Solve
- Calculate the Cumulative Frequencies
First, we need to determine the cumulative frequencies for each class. This is done by adding the frequency of each class to the cumulative frequency of the preceding class.
| Items | Frequency | Cumulative Frequency |
|---|---|---|
| 20-30 | 5 | 5 |
| 30-40 | 10 | 15 |
| 40-50 | 16 | 31 |
| 50-60 | 18 | 49 |
| 60-70 | 12 | 61 |
| 70-80 | 10 | 71 |
| 80-90 | 8 | 79 |
- Calculate N/2
Determine the total number of observations, $N$, which is the sum of the frequencies. Then, find $N/2$.
$N = 5 + 10 + 16 + 18 + 12 + 10 + 8 = 79$
$N/2 = 79/2 = 39.5$
- Identify the Median Class
The median class is the class interval where the cumulative frequency is greater than or equal to $N/2$. In this case, $N/2 = 39.5$, so the median class is 50-60 because its cumulative frequency (49) is the first one greater than 39.5.
- Apply the Median Formula
The formula for the median of grouped data is:
$Median = L + \frac{(N/2 - CF)}{f} \times h$
Where:
- $L$ is the lower boundary of the median class (50)
- $N$ is the total frequency (79)
- $CF$ is the cumulative frequency of the class preceding the median class (31)
- $f$ is the frequency of the median class (18)
- $h$ is the class width (10)
- Substitute the values into the formula
$Median = 50 + \frac{(39.5 - 31)}{18} \times 10$
$Median = 50 + \frac{8.5}{18} \times 10$
$Median = 50 + \frac{85}{18}$
$Median = 50 + 4.72$
$Median = 54.72$ (approximately)
$Median = 54.72$
More Information
The median represents the middle value in a dataset. In a grouped frequency distribution, it's the point that divides the distribution into two equal halves.
Tips
A common mistake is to incorrectly identify the median class. It's crucial to choose the class where the cumulative frequency first exceeds $N/2$. Another mistake is using the cumulative frequency of the median class instead of the class preceding it in the formula.
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