1. Given frequency table

[ \begin{array}{c|ccccc} \text{Class} & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 \ \hline \text{Frequency} & \alpha & 110 & 54 & 30 & \beta \end{array} ]

Total frequency is given as [ \alpha+110+54+30+\beta=584 ] so [ \alpha+\beta+194=584 ] [ \alpha+\beta=390 \qquad ...(1) ]

2. Use the median formula

Total number of observations: [ N=584 ] Hence, [ \frac{N}{2}=292 ]

Given median is $45$.

The median class is clearly $40-50$, because median value $45$ lies in this class.

For grouped data, [ \text{Median}=l+\left(\frac{\frac{N}{2}-c_f}{f}\right)h ] where:

• $l=40$ (lower limit of median class),
• $h=10$,
• $f=30$ (frequency of median class),
• $c_f=\alpha+110+54=\alpha+164$.

Substitute in the formula: [ 45=40+\left(\frac{292-(\alpha+164)}{30}\right)10 ]

Simplify: [ 5=\left(\frac{128-\alpha}{30}\right)10 ] [ 5=\frac{128-\alpha}{3} ] [ 15=128-\alpha ] [ \alpha=113 ]

3. Find $\beta$ using (1)

[ \alpha+\beta=390 ] [ 113+\beta=390 ] [ \beta=277 ]

4. Compute $|\alpha-\beta|$

[ |\alpha-\beta|=|113-277|=164 ]

5. Final Answer

[ \boxed{164} ]

The derived answer matches the stored correct answer.