Example 1: $\sum (2, 3, 7) = 2 + 3 + 7 = 12$

Example 2: If $x = (2, 1, 4, 7)$, we can say $x_1 = 2, x_2 = 1, x_3 = 4, x_4 = 7$

$\sum_{i=1}^{4} x_i$ means sum all of the $x$ values starting with $x_1$ and finishing at $x_4$

So,$\sum_{i=1}^{4} x_i = 2 + 1 + 4 + 7 = 14$

Example 3: If $y = (2, 7, 3, 4, 9)$

$\sum_{i=3}^{5} y_i = 3 + 4 + 9 = 16$

Example 4: For the above $y$, we say $\sum y$ is the sum of all y values

i.e., $\sum y = 2 + 7 + 3 + 4 + 9 = 25$

Example: Calculate the median of ($2, 3, 7, 12, 16, 109$).

Given $n$ = $6$:

This means the median is at position $3.5$ in the list, i.e., halfway between the 3rd and 4th item.

Example: Find the quartiles of $(1, 7, 9, 12, 12, 15, 16, 19)$

• $n = 8 \rightarrow \frac{8+1}{2} = 4.5$
• Median has position $4.5.$

We include this, and the median lies between positions $4$ and $5$:

$(1, 7, 9, \underline{12}) , \underline{12}, 15, 16, 19$

• $LQ = \frac{7+9}{2} = 8$
• $UQ = \frac{15+16}{2} = 15.5$

Note: In the case that the original list is of odd length, the two sublists are formed with the median in the middle:

Example list: $(2, 3, 5, 9, 12)$

• LQ = 3
• UQ = 9

IQR Formula:

$\text{IQR} = UQ - LQ$