1. Class marks (mid-points)

For the classes $10$–$20$, $20$–$30$, $30$–$40$, the class marks are:

$15,\; 25,\; 35$

The corresponding frequencies are:

$2,\; x,\; 2$

So total frequency is

$N = 2 + x + 2 = x+4$

2. Mean of the distribution

By symmetry, since frequencies at $15$ and $35$ are equal, the mean is

$\bar{x} = 25$

(We can also verify using formula:)

$\bar{x} = \frac{2(15) + x(25) + 2(35)}{x+4} = \frac{30 + 25x + 70}{x+4} = \frac{25x+100}{x+4} = 25$

3. Variance formula

Variance is

$\sigma^2 = \frac{\sum f_i (m_i-\bar{x})^2}{\sum f_i}$

Here,

• for $15$: $(15-25)^2 = 100$
• for $25$: $(25-25)^2 = 0$
• for $35$: $(35-25)^2 = 100$

Therefore,

$\sigma^2 = \frac{2(100) + x(0) + 2(100)}{x+4}$

$\sigma^2 = \frac{200+200}{x+4} = \frac{400}{x+4}$

Given variance is $50$, so

$\frac{400}{x+4} = 50$

4. Solve for $x$

$400 = 50(x+4)$

$400 = 50x + 200$

$200 = 50x$

$x = 4$

5. Final answer

$\boxed{4}$

The derived answer matches the stored correct answer.