Q.For the data $10, 20, 30, 40, 50, 60, 70, 80, 90$, compute $Q_2$, $D_5$ and $P_{50}$ and hence verify that $Q_2 = D_5 = P_{50} = \text{Median}$.

All three middle partition values sit at the same central position; compute each and compare.

The data are ascending, $n = 9$, $n + 1 = 10$.

$Q_2$: position $= \dfrac{2(10)}{4} = 5$ → 5th value $= 50$.

$D_5$: position $= \dfrac{5(10)}{10} = 5$ → 5th value $= 50$.

$P_{50}$: position $= \dfrac{50(10)}{100} = 5$ → 5th value $= 50$.

Median: with $n = 9$ (odd), the median is the $\tfrac{9+1}{2} = 5$th value $= 50$.

All four give the $5$th observation, $50$, so $Q_2 = D_5 = P_{50} = \text{Median} = 50$. …

The equality can be seen without any data: each of $\frac{2}{4}$, $\frac{5}{10}$, $\frac{50}{100}$ equals $\frac{1}{2}$, so all three locate the same central position. Th …

Treating $Q_2$, $D_5$ and $P_{50}$ as three separate calculations that might differ — they are the same value by construction. The exercise is really abou …