Notes for SSC CGL examinations — read every chart correctly, locate the centre and spread of data, then turn uncertainty into exact probability.
Data are recorded observations. Before calculation, ask:
Then identify the requested comparison: total, difference, ratio, percentage, average or percentage change.
For combined categories, add the raw values before forming a ratio or percentage. If values are indices or percentages with different bases, convert to actual quantities first.
Bar height or length represents magnitude. Bars have equal width and separate categories.
Always read axis scale and truncated origins.
A sector represents a fraction of the full \(360^\circ\) circle.
Since \(1\%=3.6^\circ\), divide angle by \(3.6\) to get percentage.
If a rate rises from \(20\%\) to \(25\%\), the rise is \(5\) percentage points but \(25\%\) relative change.
Chart questions often provide both as distractors.
A histogram uses touching rectangles because adjacent continuous class intervals share boundaries. For equal widths, height is frequency. For unequal widths, compare frequency density.
Area of each bar, not height alone, represents frequency when widths differ.
Plot each class midpoint against its frequency, then join successive points with straight lines. To close the polygon, add a zero-frequency class midpoint before the first and after the last.
A frequency polygon can be drawn over a histogram or independently.
The mean uses every value, so it is sensitive to extreme observations. If every value rises by \(a\), mean rises by \(a\); if every value is multiplied by \(b\), mean is multiplied by \(b\).
Convert means back to totals before combining.
If one observation \(x\) is missing from \(n\) values with known mean \(\bar x\):
If a value is corrected from wrong \(w\) to right \(r\), new mean \(=\) old mean \(+\frac{r-w}{n}\).
First arrange values. For \(n\) odd, median is the \((n+1)/2\)-th value. For \(n\) even, average the \(n/2\)-th and \((n/2+1)\)-th values.
Median resists extreme values better than mean.
Find the class whose cumulative frequency first reaches or exceeds \(N/2\). That is the median class.
\(L\): lower class boundary; \(c_f\): cumulative frequency before median class; \(f\): median-class frequency; \(h\): class width.
Ungrouped mode is the value occurring most often. A distribution can be unimodal, bimodal, multimodal or have no mode.
For grouped data:
\(f_1\) is modal-class frequency; \(f_0,f_2\) are preceding/following frequencies.
| Measure | Best use | Main strength | Main weakness |
|---|---|---|---|
| Mean | balanced numerical data | uses every value; algebra-friendly | pulled by extremes |
| Median | skewed data, open-ended classes | position-based; robust | ignores much magnitude information |
| Mode | most common category/size | works with qualitative labels | may be absent or non-unique |
For a moderately skewed unimodal distribution, the empirical relation is an approximation:
Deviation is \(x_i-\bar x\). Deviations sum to zero, so square them before averaging.
For a frequency table:
Lower standard deviation means greater consistency around the mean.
If every value increases by constant \(a\), SD is unchanged. If every value is multiplied by \(b\), SD is multiplied by \(|b|\).
When two series have different means or units of scale, compare relative spread:
Smaller CV means greater relative consistency, provided means are positive and comparison is meaningful.
For class intervals, use midpoint \(x_i\) and prepare columns:
Then \(\bar x=\sum f_ix_i/\sum f_i\), followed by the shortcut variance. Midpoint treatment assumes values concentrate at class marks.
For finite equally likely outcomes:
| Experiment | Total equally likely outcomes | Typical favourable count |
|---|---|---|
| one fair coin | \(2:\{H,T\}\) | one head: \(1\) |
| two fair coins | \(4:\{HH,HT,TH,TT\}\) | exactly one head: \(2\) |
| one fair die | \(6:\{1,2,3,4,5,6\}\) | even: \(3\) |
| two fair dice | \(36\) ordered pairs | sum \(7\): \(6\) |
| one card from standard deck | \(52\) | one suit: \(13\); one rank: \(4\) |
Ordered outcomes matter: \((2,5)\) and \((5,2)\) are different two-dice outcomes.
The complement \(A^c\) contains all outcomes not in \(A\).
“At least one” is often fastest by subtracting “none” from \(1\).
If events are mutually exclusive, \(A\cap B=\varnothing\), so simply add.
If \(A,B\) are independent, \(P(B\mid A)=P(B)\):
With replacement usually preserves independence; without replacement changes the second denominator.
| Idea | Meaning | Test |
|---|---|---|
| Mutually exclusive | cannot occur together | \(P(A\cap B)=0\) |
| Independent | one does not change the other’s probability | \(P(A\cap B)=P(A)P(B)\) |
Non-zero mutually exclusive events cannot be independent: knowing one occurred makes the other impossible.
Use systematic listing, product rule, or simple combinations when outcomes remain equally likely.
Example: choose \(2\) objects from \(n\) without order:
Probability still equals favourable count divided by total count only after both counts use the same ordering convention.
Estimate the ratio from the chart before exact arithmetic. It rejects impossible options and catches unit errors.
If options are widely separated, estimation may finish the question.
Recognise sectors instead of dividing by \(3.6\) each time.
When two group means and the combined mean are known:
The group sizes are inverse to their distances from the combined mean.
Replace wrong \(w\) by correct \(r\): net correction \(=r-w\). Multiple corrections add algebraically.
For small integer data, compute \(\sum x\) and \(\sum x^2\) directly. This is usually faster than squaring every deviation.
For repeated independent trials with failure probability \(q\):
Example: at least one head in \(3\) fair tosses \(=1-(1/2)^3=7/8\).
For two dice, sum counts rise to \(7\) then fall symmetrically:
These correspond to sums \(2\) through \(12\), over \(36\) ordered outcomes.
Use as a trigger, then check overlap/independence:
Simple add/multiply works only for mutually exclusive/independent cases respectively.
Percentage change divides by the old value, not new value or simple difference.
A \(72^\circ\) sector is \(20\%\), not \(72\%\).
Axis marks may represent \(5,10,1000\) or percentages. Reading bar height without multiplying by the scale gives a polished wrong answer.
When class widths differ, taller bar does not necessarily mean greater frequency. Compare bar area or read frequency density.
The middle written observation is not necessarily the median. Arrange values in order first.
Simple average of two means works only when group sizes are equal.
Weight each mean by its group size.
Variance is the average squared deviation; SD is its square root. Their units differ.
If variance is \(25\), SD is \(5\), not \(25\).
Favourable/total counting works only when elementary outcomes are equally likely. A biased coin cannot be solved by counting heads and tails as \(1/2\).
After one object is removed, both favourable and total counts may change. The second draw is conditional.
Mutually exclusive means “cannot coexist”; independent means “no influence.” They are not synonyms.
Categories are separate, so bar gaps remain. Continuous intervals touch, so histogram bars touch.
Every slice is its share of the full circle. Keep the fraction triangle in mind: part/total = angle/360 = percentage/100.
Mean redistributes total equally; median stands in the ordered middle; mode repeats most often.
Think of points tied to the mean with elastic bands. Short bands mean low SD; long bands mean high SD.
Addition moves every point together, leaving SD unchanged. Multiplication stretches every gap.
The denominator must list all equally likely elementary outcomes; the numerator keeps only those satisfying the event.
It is easier to count the single all-failure path and subtract it from certainty.
OR collects branches, so add and correct overlap. AND follows one path through stages, so multiply conditional branch probabilities.