Q.Find the median of the following marks obtained by 5 students: 12, 15, 11, 18, 14.

Median $=Q_2=D_5=P_{50}$; the sixth decile $D_6$ is the odd one out.

The median divides the data at the 50% point. The partition values that also fall at 50% are:

• Second quartile $Q_2$ (2 of 4 parts $=50\%$),
• Fifth decile $D_5$ (5 of 10 parts $=50\%$),
• Fiftieth percentile $P_{50}$ ($50\%$).

The sixth decile $D_6$ lies at $60\%$ of the data, so $D_6\neq$ median.

Median = the central value splitting ordered data into two equal halves.

When the observations are arranged in ascending (or descending) order, the median is the value of the middle item; exactly 50% of the observations fall below it and 50% above it. It is a positional average unaffected by extreme values.

Median of the 7 ordered values is the 4th one $=32$.

Arrange in ascending order: $25,27,30,32,35,38,40$. With $n=7$ (odd), the median is the $\left(\dfrac{n+1}{2}\right)^{\text{th}}=4^{\text{th}}$ value $=32$.

Order the data and average the two central values because $n=6$ is even; the median is $13$.

This is a very-short-answer item from the Measures of Central Tendency chapter of the CHSE Odisha +2 Class-12 Business Mathematics & Statistics syllabus (aligned with the NCERT/CBSE commerce curriculum).

Step 1 — arrange the values in ascending order:

$5,\ 7,\ 12,\ 14,\ 15,\ 18$

Step 2 — count the observations: $n=6$ (even).

Step 3 — for an even number of observations the median is the average of the $\left(\dfrac{n}{2}\right)^{\text{th}}$ and $\left(\dfrac{n}{2}+1\right)^{\text{th}}$ items, i.e. the 3rd and 4th items:

$\text{Median}=\dfrac{12+14}{2}=\dfrac{26}{2}=13$

The correct word is positional.

The median is the value of the middle item when the data are arranged in order. Its value is fixed by the position of the central observation, not by the numerical magnitude of every value — so it is called a positional average. (By contrast, the arithmetic mean is a calculated/mathematical average because it uses all the magnitudes.)

Replace mean with median: $Q_2$ equals the median.

The quartiles divide an ordered series into four equal parts. The second quartile $Q_2$ lies exactly at the half-way point of the distribution — which is precisely the definition of the median. Hence the second quartile and the median of a series are equal ($Q_2=$ Median), not the mean.

Median position $=\frac{n+1}{2}=\frac{12}{2}=6$th value.

The eleven values $2,4,6,\dots,22$ are already arranged in ascending order. For an odd number of observations $n=11$, the median is the middle item at position

$\frac{n+1}{2}=\frac{11+1}{2}=6.$

So the median is the 6th value (which is $12$).

Ordered data: $98.01, 98.08, 98.1, 98.13, 98.42$; median is the middle value $98.1$.

Arrange in ascending order:

$98.01,\;98.08,\;98.1,\;98.13,\;98.42.$

For $n=5$ (odd), the median is at position $\dfrac{n+1}{2}=3$, which is $98.1$.

The correct word is positional — the median is a positional average, not a mathematical one.

Averages are of two broad kinds:

• Mathematical averages — computed using all the values through arithmetic operations: Arithmetic Mean, Geometric Mean, Harmonic Mean.
• Positional averages — located by their position in an arranged (ascending/descending) series: Median and Mode.

The median is the value of the middle item of an ordered data set, so it depends on the position of items rather than on a calculation involving every observation. Hence calling it a mathematical average is incorrect.

Median $= 190$.

Data: $198, 205, 179, 146, 210, 186, 190$. Arrange in ascending order:

$146,\; 179,\; 186,\; 190,\; 198,\; 205,\; 210$

Here $N = 7$ (odd), so the median is the value of the $\left(\dfrac{N+1}{2}\right)^{th} = 4^{th}$ item.

The $4^{th}$ value is $190$.

Correct option: (d) 83.

Data: $72, 78, 81, 88, 92, 83, 100$. Arrange in ascending order:

$72,\; 78,\; 81,\; 83,\; 88,\; 92,\; 100.$

With $N = 7$ (odd), the median is the $\left(\dfrac{N+1}{2}\right)^{th} = 4^{th}$ item, which is $83$.