Data interpretation

For parts of one whole (a pie, a stacked bar, a column of a table with its total). The shares of all the parts add to 100%, apart from rounding in the display.

The base is the old (earlier) value unless the question names another. A positive result is an increase and a negative one a decrease. For a quantity that cannot go below zero, a decrease can never exceed 100%; an increase can (a rise of 100% is a doubling).

For a change in a quantity that is itself a percentage (a share, a pass rate, an interest rate). From 20% to 25% is a rise of 5 percentage points, but a 25% increase in the rate, because $\frac{25-20}{20}=0.25$.

For positive $A$ and $B$ with $A>B$. The base is the quantity after 'than'. The two percentages differ: with $A=125$ and $B=100$, $A$ is 25% more than $B$ but $B$ is 20% less than $A$.

The first holds for any $r\ge 0$; the second for $0\le r<100$. A 25% rise is undone by a 20% fall, and a 20% fall needs a 25% rise to recover.

Each change acts on the value left by the previous one. Take a decrease as negative. The order of the changes does not affect the final value.

For changes of $a\%$ then $b\%$, decreases negative. A rise of 20% followed by a fall of 20% gives $20-20-\frac{400}{100}=-4$, a net fall of 4%. The same result gives the change in a product: price up 10% and quantity sold down 10% changes revenue by $10-10-1=-1$, a 1% fall.

When the value after a change of $r\%$ is known ($r$ negative for a decrease, with $r>-100$). A value of 132 after a 10% rise was $\frac{132}{1.1}=120$, not $132-13.2=118.8$.

The base is the expenditure (the cost) unless the question names another; a negative result is a loss. Rearranged, Income $=$ Expenditure $\times \left(1+\frac{p}{100}\right)$ for a profit of $p\%$. An income of Rs. 69 lakh on an expenditure of Rs. 60 lakh is a profit of $\frac{9}{60}\times 100=15\%$.

For nested data: if 40% of a firm's staff are engineers and 25% of the engineers are women, women engineers are $\frac{40\times 25}{100}=10\%$ of the staff.

When $A$ and $B$ are $p_A\%$ and $p_B\%$ of the SAME total, their ratio needs no total. It fails in general across two different totals (two pies, two years): convert to amounts first.

Each part is its term of the ratio over the sum of the terms. Rs. 9 lakh divided in the ratio $2:3:4$ gives Rs. 2 lakh, Rs. 3 lakh and Rs. 4 lakh.

$p$ is the part's percentage share. 1% corresponds to $3.6^{\circ }$, 10% to $36^{\circ }$, 25% to $90^{\circ }$ and 50% to $180^{\circ }$.

Needs the total of THAT pie. The angles of all the sectors add to $360^{\circ }$ and their percentages to 100.

Only within one pie. Across two pies, convert each sector to an amount using its own pie's total first. The difference between two sectors is $\frac{{\theta }_1-{\theta }_2}{360^{\circ }}\times \text{Total}$.

The bottom segment's lower edge is zero. The upper edge of any higher segment is a running total, not that segment's value. The top of the whole stack is the total of all segments.

The bar totals must come from elsewhere (a table or a label). Without them, segments of different bars can be compared only as shares.

In units per period (for example tonnes per year). On a line graph it is the slope of the segment joining the two points, computed from the axis values. How steep a segment looks depends on the scales chosen for the axes.

The percentage change over one period, on the earlier value as base. A falling growth rate does not mean a falling value: a line that keeps rising but flattens has positive, shrinking growth rates.

Growth of $r\%$ in each of $n$ periods, each period's growth applied to the value at the start of that period. For a steady decline, $r$ is negative.

For positive $V_0$ and $V_n$. $n$ is the number of periods BETWEEN the two values: from 2020 to 2025, $n=5$, not 6. CAGR is the constant rate that would carry $V_0$ to $V_n$; it ignores the path in between.

The CAGR comes from the geometric mean of the growth factors. Provided every yearly rate is above $-100\%$ (no value falls to zero), it never exceeds the arithmetic mean of the yearly rates $g_1,\dots ,g_n$, and the two are equal only when every yearly rate is the same.

Use only when a question asks for the average yearly increase as a percentage of the starting value (simple, not compound). For positive $V_0$ and $V_n$ it is never smaller than the CAGR over the same span, and equals it only when $n=1$ or $V_n=V_0$.

The base period has index 100. The difference $I_2-I_1$ is a percentage change only when the earlier period is the base ($I_1=100$). Indices of 125 and 150 differ by 25 points but mean a rise of $\frac{25}{125}\times 100=20\%$.

When a graph or table shows running totals $C_t$ ('total sales up to month $t$'), the value for period $t$ is the difference of consecutive readings. A running total of non-negative amounts never falls, so a line that falls anywhere is not a cumulative line of such amounts.

Each item counts equally. Rearranged, sum of values $=n\bar{x}$, which recovers a missing value when the mean of all the items is given.

The weights $w_i$ are the group sizes (students, units sold, workers). With positive weights, the weighted mean always lies between the smallest and largest $x_i$, pulled towards the values with larger weights. For a frequency table the weights are the frequencies; for grouped classes, taking each class mid-point as $x_i$ gives an estimate of the mean, not its exact value.

$p_i$ is group $i$'s percentage and $n_i$ its size. The plain mean of the $p_i$ equals $p_{\text{overall}}$ whenever all the $n_i$ are equal; otherwise it generally differs.

The combined mean divides the gap between the two group values in the inverse ratio of the weights. The second form finds the ratio of group sizes from a known combined mean and needs $x_1<{\bar{x}}_w<x_2$. Groups with means 60 and 80 in sizes $3:1$ have combined mean $60+\frac{1}{4}(80-60)=65$.

An overall rate (revenue per unit sold, output per worker, marks per student) is total divided by total. In general the mean of the individual ratios is a different number; the two agree when all the $b_i$ are equal.

A bar of height 6 per unit over the class 20 to 30 stands for $6\times 10=60$ items. Comparing the heights of bars of different widths compares densities, not frequencies.

Holds only when the plotted origin really is $(0,0)$. If either axis is offset (starts above zero), compute $\frac{y}{x}$ from the values read off instead.

Both in the same unit. The horizontal distance is the ground distance, so convert the map distance with the scale first. Often written as '1 in $N$', meaning a rise of 1 unit for every $N$ units horizontally.

Convert units last: 1 km $=1,000$ m $=1,00,000$ cm.

Areas scale with the SQUARE of the linear scale. On a $1:50,000$ map, 1 cm represents 0.5 km, so 1 cm² represents $(0.5{)}^2=0.25$ km².