1. Key Concepts and Formulas

• Total Frequency (N): For a discrete frequency distribution, the total number of observations is the sum of all frequencies, $N = \sum f_i$.
• Median (M): The middle value of the data when arranged in ascending order.
  • To find the median for a discrete frequency distribution, we first calculate the cumulative frequencies ($cf_i$).
  • If $N$ is odd, the median is the value of the $\left(\frac{N+1}{2}\right)^{th}$ observation.
  • If $N$ is even, the median is the average of the $\left(\frac{N}{2}\right)^{th}$ and $\left(\frac{N}{2}+1\right)^{th}$ observations.
• Mean Deviation about the Median (MD(M)): This measures the average absolute difference between each data point and the median. It is calculated using the formula: $MD(M) = \frac{\sum_{i=1}^{k} f_i |x_i - M|}{N}$ where $x_i$ are the distinct data points, $f_i$ are their corresponding frequencies, and $k$ is the number of distinct data points.

2. Step-by-Step Solution

Step 1: Formulate the first equation using the total number of students. We are given the frequency distribution of the age of students in a class of 40 students. Let's list the ages ($x_i$) and their corresponding frequencies ($f_i$):

The total number of students ($N$) is given as 40. The sum of all frequencies must equal $N$: $\sum f_i = 5 + 8 + 5 + 12 + x + y$ $40 = 30 + x + y$ $x + y = 10 \quad \text{(Equation 1)}$ Explanation: This initial step uses the total number of observations to establish a fundamental relationship between the two unknown frequencies, $x$ and $y$.

Step 2: Determine the Median (M). To find the median, we need to calculate the cumulative frequencies ($cf_i$). The cumulative frequency for a given age is the sum of frequencies up to that age.

Explanation: Cumulative frequencies help us locate the position of each observation when the data is arranged in ascending order. For example, the $18^{th}$ student has age 17, and all students from the $19^{th}$ to the $30^{th}$ position have age 18.

Since $N = 40$ (an even number), the median is the average of the $\left(\frac{N}{2}\right)^{th}$ and $\left(\frac{N}{2}+1\right)^{th}$ observations. Here, $N/2 = 40/2 = 20$, and $(N/2)+1 = 21$. From the cumulative frequency table:

• The $18^{th}$ observation is 17.
• The $19^{th}$ observation is 18.
• The $20^{th}$ observation is 18.
• The $21^{st}$ observation is 18. Therefore, the median $M = \frac{18 + 18}{2} = 18$. Explanation: The median is found to be 18. This value is crucial for calculating the mean deviation. In this case, the values of $x$ and $y$ do not influence the median as the middle observations fall within the age 18 group.

Step 3: Calculate the Mean Deviation about the Median and formulate the second equation. Now that we have the median $M=18$, we can calculate the sum of $f_i |x_i - M|$.

| Age ($x_i$) | No. of Students ($f_i$) | $|x_i - M| = |x_i - 18|$ | $f_i |x_i - M|$ | | :---------- | :---------------------- | :------------------------ | :------------------ | | 15 | 5 | $|15-18|=3$ | $5 \times 3 = 15$ | | 16 | 8 | $|16-18|=2$ | $8 \times 2 = 16$ | | 17 | 5 | $|17-18|=1$ | $5 \times 1 = 5$ | | 18 | 12 | $|18-18|=0$ | $12 \times 0 = 0$ | | 19 | $x$ | $|19-18|=1$ | $x \times 1 = x$ | | 20 | $y$ | $|20-18|=2$ | $y \times 2 = 2y$ |

Summing the values in the last column: $\sum f_i |x_i - M| = 15 + 16 + 5 + 0 + x + 2y = 36 + x + 2y$ We are given that the mean deviation about the median is $1.25$. Using the formula: $MD(M) = \frac{\sum f_i |x_i - M|}{N}$ $1.25 = \frac{36 + x + 2y}{40}$ Multiply both sides by 40: $1.25 \times 40 = 36 + x + 2y$ $50 = 36 + x + 2y$ $x + 2y = 50 - 36$ $x + 2y = 14 \quad \text{(Equation 2)}$ Explanation: This step applies the mean deviation formula. By substituting the calculated median, the sum of absolute deviations, and the total frequency, we derive our second linear equation involving $x$ and $y$.

Step 4: Solve the system of equations for x and y. We now have a system of two linear equations:

1. $x + y = 10$
2. $x + 2y = 14$

Subtract Equation 1 from Equation 2: $(x + 2y) - (x + y) = 14 - 10$ $y = 4$ Substitute the value of $y$ into Equation 1: $x + 4 = 10$ $x = 6$ Explanation: Solving this system of equations provides the specific numerical values for the unknown frequencies $x$ and $y$. These values are positive integers, which is consistent with them representing the number of students.

Step 5: Calculate the target expression $4x+5y$. Now that we have $x=6$ and $y=4$, we can evaluate the expression $4x+5y$: $4x + 5y = 4(6) + 5(4)$ $= 24 + 20$ $= 44$ Explanation: The final step involves substituting the determined values of $x$ and $y$ into the expression required by the problem.

3. Common Mistakes & Tips

• Median for Even N: For an even number of observations, the median is the average of the two middle observations. Ensure these observations are correctly identified using cumulative frequencies.
• Absolute Value: Remember that $|x_i - M|$ always represents a non-negative distance. Errors can occur if negative signs are retained during this calculation.
• System of Equations: Carefully solve the simultaneous equations. A small calculation error here can lead to incorrect values for $x$ and $y$.
• Understanding MD(M): Mean deviation provides a measure of data dispersion. It's the average of the absolute differences between each data point and the median, giving a robust measure of spread.

4. Summary

We began by setting up the first equation based on the total number of students. Then, we calculated the cumulative frequencies to accurately determine the median of the distribution. Using the median, we computed the sum of the absolute deviations from the median, weighted by their frequencies, and equated it to $N \times MD(M)$ to obtain the second equation. Solving the system of these two linear equations yielded $x=6$ and $y=4$. Finally, substituting these values into the expression $4x+5y$ gave the result of 44.

5. Final Answer

The final answer is $\boxed{44}$.