A student‘s marks were wrongly entered as 65 instead of 45. Due to this, the average marks for the class got increased by 1/3. The number of students in the class is:
60
Understanding the Impact of Wrong Data Entry on Average
This problem involves understanding how changing one value in a dataset affects the overall average (mean). We are given a scenario where a student's mark was incorrectly recorded, leading to an increase in the class average. We need to find the total number of students in the class.
Analyzing the Given Information
- Correct mark of the student: 45
- Wrong mark entered: 65
- Difference in mark entry: Wrong mark - Correct mark = 65 - 45 = 20
- Increase in the class average due to this error: \( \frac{1}{3} \)
The error caused an increase in the total sum of marks by 20 (since 65 was entered instead of 45). This increase in the total sum, when divided by the number of students, resulted in an average increase of \( \frac{1}{3} \).
Setting up the Problem
Let:
- \( n \) be the number of students in the class.
- \( T \) be the original total marks of all students (if the mark was correctly entered as 45).
- Original Average = \( \frac{T}{n} \)
When the mark was wrongly entered as 65, the total marks became \( T - 45 + 65 = T + 20 \).
- New Average = \( \frac{T + 20}{n} \)
Formulating the Equation
According to the problem, the new average is greater than the original average by \( \frac{1}{3} \). We can write this as an equation:
\[ \text{New Average} = \text{Original Average} + \frac{1}{3} \] \[ \frac{T + 20}{n} = \frac{T}{n} + \frac{1}{3} \]Solving for the Number of Students
Now, we solve the equation for \( n \), the number of students.
We can rewrite the left side of the equation:
\[ \frac{T}{n} + \frac{20}{n} = \frac{T}{n} + \frac{1}{3} \]Subtract \( \frac{T}{n} \) from both sides of the equation:
\[ \frac{20}{n} = \frac{1}{3} \]To find \( n \), we can cross-multiply:
\[ 20 \times 3 = n \times 1 \] \[ 60 = n \]So, the number of students in the class is 60.
Detailed Calculation Steps
| Step | Description | Calculation/Equation |
|---|---|---|
| 1 | Identify the difference in the entered mark. | Difference = 65 - 45 = 20 |
| 2 | Let \( n \) be the number of students. | \( n \) |
| 3 | Let \( T \) be the original total marks. | \( T \) |
| 4 | Write the original average. | \( \frac{T}{n} \) |
| 5 | Write the new total marks. | \( T + 20 \) |
| 6 | Write the new average. | \( \frac{T + 20}{n} \) |
| 7 | Set up the equation based on the average increase. | \( \frac{T + 20}{n} = \frac{T}{n} + \frac{1}{3} \) |
| 8 | Simplify the equation. | \( \frac{20}{n} = \frac{1}{3} \) |
| 9 | Solve for \( n \). | \( n = 20 \times 3 = 60 \) |
The number of students in the class is 60.
Revision Table: Key Concepts
| Concept | Explanation | Formula Example |
|---|---|---|
| Average (Mean) | Sum of all values divided by the number of values. | \( \text{Average} = \frac{\text{Sum}}{\text{Count}} \) |
| Effect of Adding/Removing Value | Adding a value changes the sum, thus changing the average. The change in average is the change in sum divided by the count. | If sum changes by \( \Delta S \) and count is \( n \), average changes by \( \frac{\Delta S}{n} \). |
Additional Information: Average and Data Errors
Understanding the concept of average is fundamental in statistics. The average is sensitive to changes in the data. If even one data point is incorrect, it can skew the average. In this problem, a positive error (entering a higher mark) increased the total sum and therefore increased the average. A negative error (entering a lower mark) would decrease the total sum and the average.
The key insight used here is that the difference in the total sum caused by the error is distributed among all students when calculating the average. The increase in total marks is 20. This total increase divided by the number of students (\( n \)) equals the increase in the average (\( \frac{1}{3} \)).
This principle is useful in solving problems where you need to find the number of items (like students, observations) when the average changes due to a known change in the sum of values.
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