Consider the following frequency distribution:

If the sum of all frequencies is 584 and median is 45, then $$|\alpha - \beta|$$ is equal to ___.

We have been given a grouped frequency distribution with class-interval width $$h = 10$$. The classes and their (as yet unknown) frequencies are

10 - 20 : $$\alpha$$,    20 - 30 : 110,    30 - 40 : 54,    40 - 50 : 30,    50 - 60 : $$\beta$$.

The total number of observations is stated to be $$N = 584$$ and the median is $$45$$.

For a continuous (grouped) distribution, the median is found by the formula

$$\text{Median} = L + \frac{\dfrac{N}{2} - \text{c.f.\,(prev)}}{f_m}\;h,$$

where $$L$$ is the lower class boundary of the median class, $$\text{c.f.\,(prev)}$$ is the cumulative frequency up to (but not including) the median class, $$f_m$$ is the frequency of the median class, and $$h$$ is the class width.

Since the median value $$45$$ lies in the interval $$40\!-\!50$$, that class is the median class. Hence $$L = 40$$, $$f_m = 30$$ and $$h = 10$$.

The cumulative frequency just before the median class is the sum of the first three class frequencies:

$$\text{c.f.\,(prev)} = \alpha + 110 + 54 = \alpha + 164.$$

Substituting all these quantities into the median formula gives

$$45 = 40 + \frac{\dfrac{584}{2} - (\alpha + 164)}{30}\;10.$$

Simplifying step by step:

$$45 - 40 = \frac{292 - (\alpha + 164)}{30}\;10$$ $$5 = \frac{292 - \alpha - 164}{30}\;10$$ $$5 = \frac{128 - \alpha}{3}$$ $$5 \times 3 = 128 - \alpha$$ $$15 = 128 - \alpha$$ $$\alpha = 128 - 15$$ $$\alpha = 113.$$

Now we use the fact that the sum of all frequencies equals $$584$$:

$$\alpha + 110 + 54 + 30 + \beta = 584.$$

Substituting $$\alpha = 113$$ we get

$$113 + 110 + 54 + 30 + \beta = 584.$$

Adding the known numbers:

$$307 + \beta = 584$$ $$\beta = 584 - 307$$ $$\beta = 277.$$

Finally, we require the absolute difference between $$\alpha$$ and $$\beta$$:

$$|\alpha - \beta| = |113 - 277| = |-164| = 164.$$

Hence, the correct answer is Option 164.