Finding Missing Frequencies Calculating Median Marks For 230 Students
Hey there, math enthusiasts! Today, we're diving into a fascinating statistical puzzle involving the median marks of 230 students. We've got a dataset with some missing frequencies, and our mission, should we choose to accept it, is to find those elusive numbers. So, buckle up and let's embark on this mathematical adventure!
Understanding the Problem: Median Marks and Missing Frequencies
The heart of our problem lies in a frequency distribution table that shows the marks obtained by 230 students. Marks are grouped into class intervals, such as 10-20, 20-30, and so on, while the frequency indicates how many students fall within each interval. However, a few frequencies are missing, denoted by dashes. Our main goal is to determine these missing frequencies, given that the median mark for the entire group is 46. Let's break down why this is important and how we're going to tackle it.
The median, as you might recall, is the middle value in a dataset when it's arranged in ascending order. In a grouped frequency distribution, like ours, the median is the value that splits the dataset into two equal halves. Knowing the median mark (46) gives us a crucial piece of information. It tells us that the middle student (or students) in our group of 230 scored around 46. This knowledge is our anchor as we navigate through the frequencies. Think of the missing frequencies as placeholders in an equation. We have some known quantities (total students, median mark, some frequencies), and we need to solve for the unknowns.
To crack this, we'll be using the median formula for grouped data. This formula involves the lower limit of the median class (the class interval containing the median), the cumulative frequency of the class preceding the median class, the frequency of the median class, the class width, and the total number of observations. It might sound like a mouthful now, but we'll break it down step by step. Remember, understanding each component of the formula is key to applying it correctly. We're not just plugging numbers into a formula; we're understanding the logic behind it. Each part of the formula represents a specific aspect of the data distribution, and when we put them together, they help us pinpoint the median. So, let's dive deeper into the data and start uncovering those missing frequencies. Are you excited? I know I am!
Setting Up the Table and Identifying Key Values
Before we jump into calculations, let's organize our data into a clear table. This will help us visualize the information and identify the values we need for our formula. Here's the table we have:
| Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
|---|---|---|---|---|---|---|---|
| Frequency | 12 | 30 | f1 | 65 | f2 | 25 | 18 |
In this table, we see the marks divided into class intervals and the corresponding frequencies. Notice the missing frequencies, which we've labeled as f1 (for the 30-40 interval) and f2 (for the 50-60 interval). These are the values we're after! We also know that the total number of students is 230. This is a crucial piece of information because it allows us to set up our first equation. The sum of all frequencies must equal the total number of students. So, we have:
12 + 30 + f1 + 65 + f2 + 25 + 18 = 230
This simplifies to:
f1 + f2 = 80
This is our first equation, a crucial step in solving the mystery. It tells us that the sum of the two missing frequencies is 80. But we need another equation to solve for f1 and f2 individually. This is where the median information comes into play. Remember, the median mark is 46. This tells us something important: the median class, the class interval that contains the median, must be the 40-50 interval. Why? Because 46 falls within this range.
Identifying the median class is a key step because it allows us to use the median formula effectively. Now, let's calculate the cumulative frequencies. Cumulative frequency is the running total of frequencies. It helps us locate the median class and apply the median formula. To find the cumulative frequencies, we add up the frequencies as we go along. For the first class (10-20), the cumulative frequency is simply 12. For the second class (20-30), it's 12 + 30 = 42. We continue this process, including the missing frequencies, to build our cumulative frequency column. This cumulative frequency will be essential when we apply the median formula, helping us pinpoint the exact location of the median within the data set. So, let's get those cumulative frequencies calculated and move closer to solving for f1 and f2!
Calculating Cumulative Frequencies and Applying the Median Formula
Now that we have our frequency table, let's add a column for cumulative frequencies. This will help us pinpoint the median class and apply the median formula effectively. Here's the updated table:
| Marks | Frequency | Cumulative Frequency |
|---|---|---|
| 10-20 | 12 | 12 |
| 20-30 | 30 | 42 |
| 30-40 | f1 |
42 + f1 |
| 40-50 | 65 | 107 + f1 |
| 50-60 | f2 |
107 + f1 + f2 |
| 60-70 | 25 | 132 + f1 + f2 |
| 70-80 | 18 | 150 + f1 + f2 |
Remember from the previous step, we already know that f1 + f2 = 80. So, we can simplify the last two cumulative frequencies: 132 + f1 + f2 = 132 + 80 = 212, and 150 + f1 + f2 = 150 + 80 = 230. The cumulative frequency represents the number of students who scored marks up to the upper limit of that class interval. For instance, 42 students scored less than 30 marks, and 107 + f1 students scored less than 50 marks.
As we discussed, the median is 46, which falls within the 40-50 class interval. This confirms that the 40-50 interval is our median class. Now, let's unleash the median formula for grouped data:
Median = L + [ (N/2 - cf) / f ] * h
Where:
- L is the lower limit of the median class (40 in our case)
- N is the total number of observations (230)
- cf is the cumulative frequency of the class preceding the median class (42 +
f1) - f is the frequency of the median class (65)
- h is the class width (10, since the class intervals are 10-20, 20-30, etc.)
Now, let's plug in the values:
46 = 40 + [ (230/2 - (42 + f1)) / 65 ] * 10
Simplifying this equation will give us our second equation involving f1. Hold on tight, guys, we're getting closer to cracking this code! By substituting the known values into the median formula, we've transformed our statistical problem into an algebraic one. Solving this equation will give us the value of f1, and once we have that, finding f2 will be a breeze. Remember, each step we take brings us closer to the solution, so let's keep pushing forward with confidence.
Solving for the Missing Frequencies: f1 and f2
Let's simplify the median formula equation we arrived at in the previous step:
46 = 40 + [ (115 - (42 + f1)) / 65 ] * 10
First, subtract 40 from both sides:
6 = [ (115 - 42 - f1) / 65 ] * 10
Simplify the numerator inside the brackets:
6 = [ (73 - f1) / 65 ] * 10
Now, divide both sides by 10:
- 6 = (73 -
f1) / 65
Multiply both sides by 65:
- 6 * 65 = 73 -
f1
39 = 73 - f1
Now, isolate f1 by adding it to both sides and subtracting 39 from both sides:
f1 = 73 - 39
f1 = 34
Woohoo! We've found the value of f1! It's 34. Now, remember our first equation from earlier:
f1 + f2 = 80
We can now substitute the value of f1 into this equation to find f2:
34 + f2 = 80
Subtract 34 from both sides:
f2 = 80 - 34
f2 = 46
And there we have it! We've found both missing frequencies. f1 is 34, and f2 is 46. We've successfully solved the puzzle! This journey through the median formula has not only given us the answers but also deepened our understanding of how the median works in grouped data. We've seen how each component of the formula plays its part in pinpointing the central value of the distribution. Remember, math isn't just about formulas; it's about the logic and reasoning that underpins them. So, let's take a moment to celebrate our success and appreciate the power of statistical analysis!
Verification and Interpretation of Results
Now that we've found the missing frequencies, it's always a good idea to verify our results. Let's plug the values of f1 and f2 back into our frequency table and check if everything adds up correctly. Here's the completed table:
| Marks | Frequency | Cumulative Frequency |
|---|---|---|
| 10-20 | 12 | 12 |
| 20-30 | 30 | 42 |
| 30-40 | 34 | 76 |
| 40-50 | 65 | 141 |
| 50-60 | 46 | 187 |
| 60-70 | 25 | 212 |
| 70-80 | 18 | 230 |
First, let's check if the sum of the frequencies equals the total number of students: 12 + 30 + 34 + 65 + 46 + 25 + 18 = 230. Excellent! It matches our total number of students, so that's a good sign. Now, let's revisit our median calculation. With f1 = 34 and f2 = 46, the cumulative frequency just before the median class (40-50) is 76. The cumulative frequency up to the median class is 141.
Since half of the total number of students (N/2) is 115, the median lies within the 40-50 class, as 115 falls between 76 and 141. This confirms our earlier assumption about the median class. Now, let's think about what these frequencies tell us in the context of the student marks. We found that 34 students scored between 30 and 40 marks, and 46 students scored between 50 and 60 marks. This gives us a clearer picture of the distribution of scores. Most students seem to be clustered around the 40-50 range, which is reflected in the high frequency (65) for that class. The median of 46 further supports this observation, indicating that the middle score falls within this range.
Interpreting these results is as important as calculating them. It allows us to draw meaningful conclusions from the data. In this case, we can say that the marks are somewhat concentrated around the median, with a good number of students scoring in the 40-60 range. However, there's also a spread of scores across the other intervals, indicating some variability in student performance. So, we've not only solved the mathematical puzzle but also gained valuable insights into the distribution of student marks. That's the power of statistics in action!
Conclusion: Mastering Medians and Missing Frequencies
We've reached the end of our mathematical journey, and what a journey it has been! We started with a frequency distribution table with missing frequencies and a median mark of 46. Through careful application of the median formula and some algebraic manipulation, we successfully uncovered the missing frequencies: f1 = 34 and f2 = 46. But more than just finding the numbers, we've deepened our understanding of the median and how it works in grouped data. We've seen how cumulative frequencies help us locate the median class and how the median formula allows us to pinpoint the central value of a distribution.
This problem highlights the importance of not just memorizing formulas but also understanding the concepts behind them. By breaking down the problem into smaller steps, identifying key values, and applying the formula logically, we were able to solve it effectively. Remember, math isn't just about crunching numbers; it's about critical thinking, problem-solving, and logical reasoning. And today, we've exercised all those skills! So, next time you encounter a similar problem, remember the steps we took, the formulas we used, and the logical reasoning we applied. You've got this! Keep practicing, keep exploring, and keep challenging yourself. The world of mathematics is full of exciting puzzles waiting to be solved, and you now have another tool in your arsenal to tackle them. Well done, everyone! You've truly mastered the art of medians and missing frequencies.
