The frequency distribution of the age of students in a class of 40 students is given below: Age: 15,

Q: Wait, why do we take the average of the 20th and 21st terms for the median? Why not just the 20th?

That is a fantastic question! In statistics, when the total number of observations N is even, there isn't a single middle term. Imagine lining up 40 students by age; there is no one person standing exactly in the middle. Instead, there is a middle pair—the 20th and the 21st students. By convention, we take the arithmetic mean of these two values to represent the central tendency. If N were odd, we would simply take the (N+1)/2 term.

Q: Why does the absolute value sign appear in the Mean Deviation formula? Can't we just subtract?

If we didn't use the absolute value, the positive and negative deviations from the median would cancel each other out, often resulting in a sum of zero! The Mean Deviation is designed to measure the 'average spread' or 'dispersion' of the data points from the center, regardless of whether they are older or younger than the median. The absolute value ensures we are measuring the magnitude of the distance, which is always a positive quantity.

Q: I have two variables, x and y. How do I know I have enough information to solve for both?

In algebra, to solve for two unknowns, you need two independent pieces of information—a system of equations. We derived the first equation from the total sum of frequencies (x + y = 10) and the second from the given Mean Deviation value (x + 2y = 14). Because these two equations are independent, they provide a unique solution for x and y. Always look for that second constraint when you see two unknowns!

Question

The frequency distribution of the age of students in a class of 40 students is given below: $$\begin{array}{|c|c|c|c|c|c|c|} \hline 	ext{Age} & 15 & 16 & 17 & 18 & 19 & 20 \ \hline 	ext{No of Students} & 5 & 8 & 5 & 12 & x & y \ \hline \end{array}$$ If the mean deviation about the median is 1.25, then $4x + 5y$ is equal to :

Options:

(A) 46

(B) 43

(C) 44

(D) 47

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are not just solving a math problem; we are peeling back the layers of a frequency distribution to reveal the hidden variables within. Statistics is the language of uncertainty, and mastering it is a rite of passage for every JEE aspirant.

The universe of this problem is defined by the total number of students, $N = 40$. This is our anchor. The sum of all frequencies must be exactly $40$.

We write this as:
$$5 + 8 + 5 + 12 + x + y = 40$$

Simplifying this, we find $30 + x + y = 40$, which gives us our first elegant equation:
$$x + y = 10$$

Keep this close; it is the foundation upon which we will build our solution.

### The Hunt for the Median

Now, we must find the median. The median is the heart of the data—the point that splits our distribution into two equal halves.

Since $N = 40$ is an even number, the median is the average of the $\frac{N}{2}$th and $(\frac{N}{2} + 1)$th observations. That is, the average of the $20$th and $21$st students.

We construct our cumulative frequency (C.F.) table: $5, 13, 18, 30, 30+x, 40$. Looking at this, we see that up to age $17$, we have $18$ students.

The $19$th through the $30$th students all fall into the age category of $18$. Therefore, both the $20$th and $21$st students are $18$ years old. Thus, our median $M$ is $18$.

### The Mean Deviation

We are given that the Mean Deviation (M.D.) about the median is $1.25$. The formula is our compass:
$$M.D. = \frac{\sum f_i |x_i - M|}{N}$$

Substituting our known values, we get:
$$\frac{5|15-18| + 8|16-18| + 5|17-18| + 12|18-18| + x|19-18| + y|20-18|}{40} = 1.25$$

Let us calculate the deviations carefully. The absolute differences are: $|15-18|=3$, $|16-18|=2$, $|17-18|=1$, $|18-18|=0$, $|19-18|=1$, and $|20-18|=2$.

Plugging these back in:
$$5(3) + 8(2) + 5(1) + 12(0) + x(1) + y(2) = 1.25 	imes 40$$

This simplifies to $15 + 16 + 5 + 0 + x + 2y = 50$. Combining the constants, we get $36 + x + 2y = 50$, or:
$$x + 2y = 14$$

### The Final Resolution

We now have a system of two linear equations:
1) $x + y = 10$
2) $x + 2y = 14$

Subtracting the first from the second, we find $y = 4$. Substituting this back into the first equation, we get $x + 4 = 10$, so $x = 6$.

The final step is simply to evaluate the expression $4x + 5y$. Plugging in our values:
$$4(6) + 5(4) = 24 + 20 = 44$$

You have navigated the logic, handled the algebra, and arrived at the truth. The final answer is **44**.

Animated Solution for Mathematics - Statistics: The frequency distribution of the age of students in a class of 40 students is given below: $Age No of Students 1551681751812 19 x 20 y$ If the mean deviation about the median is 1.25, then $4 x + 5 y$ is equal to :

Select Answer:

Visualized Solution

Total Frequency Equation

Calculating Cumulative Frequency

Determining the Median

Mean Deviation Formula

Setting up the Calculation

Simplifying the Equation

Solving for $x$ and $y$

Final Calculation

The Sigma Insight: Measures of Dispersion

Analyzing the Setup

The Hunt for the Median

The Mean Deviation

The Final Resolution

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