If the median of the following distribution is 32.5, then find the values of x and y. Class: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70 Frequency: x, 5, 9, 12, y, 3, 2 [Total = 40]
Show Hint
For any median equation with decimals like \(2.5 = (6 - x) \times \frac{10}{12}\):
Simplify the fraction to \(\frac{5}{6}\) first:
\[ 2.5 = (6-x) \times \frac{5}{6} \implies 2.5 \times \frac{6}{5} = 6 - x \implies 3 = 6 - x \implies x = 3 \]
This algebraic cleaning reduces calculation time!
Step 1: Frame the median formula as a simple proportion instead of solving the algebraic equation from scratch.
First, from the total frequency 40:
\[ x + 5 + 9 + 12 + y + 3 + 2 = 40 \quad \Rightarrow \quad x + y = 9 \]
Since the median 32.5 falls inside the class 30-40, this is the median class, with lower limit $l = 30$, frequency $f = 12$, class width $h = 10$, and $\frac{N}{2} = 20$.
Step 2: Think of how far into the median class the median value sits, as a plain fraction, instead of substituting into the formula directly.
The median 32.5 is $32.5 - 30 = 2.5$ units into a class that is 10 units wide. So the median lies $\frac{2.5}{10} = \frac{1}{4}$ of the way through the class.
Step 3: This same fraction must equal the fraction of the class frequency that has been covered by the time we reach the median.
The cumulative frequency just before the median class is $x + 5 + 9 = x + 14$. Out of the 12 values in the median class, the number needed to reach position 20 is $20 - (x + 14) = 6 - x$.
Setting this as a fraction of the class frequency 12, equal to the position fraction $\frac{1}{4}$:
\[ \frac{6 - x}{12} = \frac{1}{4} \]
Step 4: Solve for $x$, then find $y$.
\[ 4(6 - x) = 12 \]
\[ 6 - x = 3 \]
\[ x = 3 \]
Using $x + y = 9$:
\[ y = 9 - 3 = 6 \]
Final Answer:
The values are $x = 3$ and $y = 6$, which corresponds to option (A).
\[ \boxed{x = 3,\ y = 6} \]
