Statistics: data & averages

Sum of the values ÷ number of values. So the total is mean × number of values.

Sort the values first, then take the middle one, at position $\frac{n + 1}{2}$. With an even number of values, go halfway between the two middle ones.

65, the value, not 12, its frequency.

$\bar x = \dfrac{\sum fx}{\sum f}$. Divide by the total frequency, not by the number of columns.

Boundaries 20.5 and 30.5; width $30.5 - 20.5 = 10$ (not 9); class mark $\frac{21 + 30}{2} = 25.5$.

Use the class marks as $x$: $\bar x = \dfrac{\sum fx}{\sum f}$.

Pick an assumed mean $A$, a class mark near the middle, and let $d = x - A$. Then $\bar x = A + \dfrac{\sum fd}{\sum f}$.

$L + \left(\dfrac{\frac N2 - F}{f_m}\right) \times c$, where $L$ is the lower boundary of the median class, $F$ the cumulative frequency before it, $f_m$ its frequency and $c$ its width.

$L + \left(\dfrac{\Delta_1}{\Delta_1 + \Delta_2}\right) \times c$, where $\Delta_1$ and $\Delta_2$ are how far the modal class's frequency is above the classes before and after it.

$\dfrac{\text{frequency}}{\text{total}} \times 360^\circ$. With percentages, $1\% = 3.6^\circ$. Draw the sectors with a protractor.

Separate values (scores 1 to 6): a bar chart, with gaps. Classes of continuous data: a histogram, with bars drawn from the class boundaries and no gaps.