For a normal distribution, standard empirical relationships link Quartile Deviation (QD), Mean Deviation (MD), and Standard Deviation (SD / $\sigma$):

• $\text{Mean Deviation} = \sqrt{\frac{2}{\pi}}\sigma \approx 0.7979\sigma \approx \frac{4}{5}\sigma \implies 5 \times \text{MD} = 4 \times \text{SD}$ (Option 1 is correct).
• $\text{Quartile Deviation} = 0.6745\sigma \approx \frac{2}{3}\sigma \implies 3 \times \text{QD} = 2 \times \text{SD}$ (Option 2 is correct).
  The overall ratio of dispersion measures is: $$\text{QD} : \text{MD} : \text{SD} = \frac{2}{3}\sigma : \frac{4}{5}\sigma : \sigma = 10 : 12 : 15$$
  Because both Option 1 and Option 2 are mathematically and practically valid for a normal distribution, NTA's official final answer key correctly recognized both '1' and '2' as valid answers.

When equal distances are traversed at different speeds ($s_1$ and $s_2$), the average speed is given by the Harmonic Mean (HM) of the speeds:

Given:
$s_1 = 45\text{ km/hr}$
$s_2 = 55\text{ km/hr}$

Calculation:

Note: Using the arithmetic mean $\frac{45 + 55}{2} = 50\text{ km/hr}$ is a classic conceptual mistake because the rider spends more time travelling at the slower speed.

The median (as well as quartiles, deciles, and percentiles) of a grouped frequency distribution is determined graphically using Ogives (Cumulative Frequency Curves):
By drawing a horizontal line from $N/2$ on the cumulative frequency axis to intersect the ogive and dropping a perpendicular to the horizontal axis, or
By the point of intersection of the 'less than' ogive and the 'more than' ogive.
In contrast, the mode is determined graphically from a Histogram.