Shortcuts & Speed Methods

Mean From a Working Mean

A statistics question gives a list or a frequency table and asks for one number: a mean, a median, a mode or a standard deviation. The time goes in long additions, products and squares. Each method below cuts one of these, says when it works, and in some cases where it stops working.

When to use it: the values are close to one another, such as daily sales, weights or marks.

Fast way: choose a round number $A$ near the middle. Write how far each value is above $A$ (plus) or below it (minus), add these deviations and divide by the count: $$\bar{x} = A + \frac{\text{sum of deviations}}{n}$$

For example: a dairy sold 142, 138, 151, 147, 135, 149, 144 and 146 litres of milk on eight days. From $A = 145$ the deviations are $-3, -7, +6, +2, -10, +4, -1, +1$, which add up to $-8$. The mean is $145 - \frac{8}{8} = 144$ litres.

Means of Number Patterns

When to use it: the values go up in equal steps, or the question names the first $n$ natural, even or odd numbers, or their squares or cubes.

• Values in equal steps: mean $= \frac{\text{first} + \text{last}}{2}$, which is also the middle value.
• First $n$ natural numbers: $\frac{n + 1}{2}$. First $n$ even numbers: $n + 1$. First $n$ odd numbers: $n$.
• Squares of the first $n$ natural numbers: $\frac{(n + 1)(2n + 1)}{6}$. Cubes: $\frac{n(n + 1)^2}{4}$.

For example: the mean of the multiples of 9 from 18 to 180 is $\frac{18 + 180}{2} = 99$, and there is no need to count them. The mean of the first 25 odd numbers is 25, and the mean of $1^2, 2^2, \ldots, 11^2$ is $\frac{12 \times 23}{6} = 46$.

Where it fails: the steps must be equal. For the primes 2, 3, 5, 7 and 11, the first and last give 6.5, but the mean is $\frac{28}{5} = 5.6$.

One Value Wrong, Added or Removed

Work with the change, not with the whole total. Word problems of this kind, such as a teacher's age joining a class, are in Chapter 5; here are the three forms that statistics questions use.

• A value copied wrongly: new mean $=$ old mean $+ \frac{\text{correct value} - \text{wrong value}}{n}$.
• A value added, with the new mean of all $n + 1$ values given: the added value is the new mean plus $n$ times the rise in the mean.
• A value removed, with the mean of the other $n - 1$ given: the removed value is the old mean plus $(n - 1)$ times the fall in the mean.

For example: the mean of 40 marks was 62, but a mark of 86 had been entered as 41. The total was short by 45, so the mean was short by $\frac{45}{40} = 1.125$, and the correct mean is 63.125. In another case, 20 workers earn a mean of ₹ 600 a day, and a supervisor joins them. If the mean of all 21 becomes ₹ 615, the supervisor earns $615 + 20 \times 15 = 915$ rupees a day.

Combined Mean of Two or Three Groups

Fast way: the combined mean $\frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}$ lies between the two means and closer to the mean of the bigger group. Start from one mean and move $\frac{n_2}{n_1 + n_2}$ of the gap towards the other.

For example: 45 boys have a mean of 56 marks and 15 girls a mean of 64. The gap is 8 and the girls are $\frac{15}{60} = \frac{1}{4}$ of the class, so the combined mean is $56 + \frac{1}{4} \times 8 = 58$.

Turned round: when the combined mean $M$ is given, the group sizes are in the ratio $(\bar{x}_2 - M) : (M - \bar{x}_1)$. With means 56 and 64 and $M = 58$, boys to girls is $(64 - 58) : (58 - 56) = 3 : 1$, the same as 45 : 15.

For three groups, take deviations from one of the means. Sections of 40, 35 and 25 students with means 62, 70 and 58 give $62 + \frac{35 \times 8 - 25 \times 4}{100} = 63.8$.

Grouped Mean by Step Deviation

When to use it: classes of equal width with large class marks, such as wages or weights.

Fast way: take $A$ as a class mark near the middle and $h$ as the class width. Each class gets $u = \frac{x - A}{h}$, a small whole number, and $$\bar{x} = A + h \times \frac{\sum fu}{\sum f}$$

For example: the weights of 50 mangoes.

The class marks are 320, 360, 400, 440 and 480, so with $A = 400$ and $h = 40$ the values of $u$ are as shown. $\sum fu = -12 - 11 + 0 + 10 + 10 = -3$, and $\bar{x} = 400 + 40 \times \frac{-3}{50} = 397.6$ g.

Where it fails: when the classes have different widths, $u$ is no longer a whole number. Use $d = x - A$ without dividing by $h$.

A Missing Frequency: Deviations From the Given Mean

The deviations from the mean, each multiplied by its frequency, add up to zero. So when the mean is given and one frequency is missing, take deviations from the given mean itself. $\sum f(x - \bar{x}) = 0$ is one equation in the unknown, with no division at the end.

For example: the mean of this table is 16. Find $p$.

The deviations from 16 are $-11, -6, -1, 4$ and $9$. So $4(-11) - 6p + 10(-1) + 9(4) + 6(9) = 0$, which is $36 - 6p = 0$, and $p = 6$.

Where it fails: if the missing frequency belongs to a value equal to the mean, its deviation is 0 and the mean cannot fix it. Such a question must also give the total.

[SSC Tip] Put the answer back before marking it: with $p = 6$ the total is 35 and $\sum fx = 560$, which is $16 \times 35$.

Median of a List: Count Before You Sort

Fast way: count the values first. With $n$ odd, the median is the $\frac{n + 1}{2}$-th value in order. With $n$ even, it is the mean of the $\frac{n}{2}$-th and the $\left(\frac{n}{2} + 1\right)$-th values. Once you know which positions you need, sort only until you reach them.

For example: a batsman's scores in ten innings are 38, 7, 66, 52, 0, 61, 25, 52, 84 and 15. With $n = 10$ the median is the mean of the 5th and 6th values. In order the list is 0, 7, 15, 25, 38, 52, 52, 61, 66, 84, so the median is $\frac{38 + 52}{2} = 45$ runs.

Taking the 5th and 6th values as written, 0 and 61, gives 30.5, so the list must be put in order first. A value that repeats, like 52 here, is counted each time it appears.

Median of a Grouped Table in Four Lines

1. Find $\frac{N}{2}$ (for grouped data it is $\frac{N}{2}$, not $\frac{N + 1}{2}$).
2. Write the cumulative frequencies. The first class whose cf reaches $\frac{N}{2}$ is the median class.
3. Note $L$, its lower boundary, $f$, its frequency, $h$, its width, and $cf$, the cumulative frequency of the class before it.
4. Median $= L + \frac{\frac{N}{2} - cf}{f} \times h$. The fraction says how far into the class the middle value lies.

For example: the monthly savings of 60 families.

$\frac{N}{2} = 30$, and the cf first reaches 30 in 40-50. There are 20 families before this class, so the 30th family is 10 places into its 20: median $= 40 + \frac{10}{20} \times 10 = 45$, that is ₹ 4,500.

Inclusive classes such as 10-19, 20-29 must first become 9.5-19.5, 19.5-29.5, so $L$ drops by 0.5.

Two Missing Frequencies From the Median

When to use it: the total $N$ and the median are given, and two frequencies are missing.

Fast way: the value of the median tells you the median class straight away. Put the known numbers into the formula; the only unknown left is inside $cf$, so the formula gives one frequency, and the total gives the other.

For example: $N = 70$ and the median is 152.

152 lies in 140-160, so $L = 140$, $f = 25$, $h = 20$ and $cf = 7 + x$. Then $152 = 140 + \frac{35 - 7 - x}{25} \times 20$, so $12 = \frac{4(28 - x)}{5}$, which gives $28 - x = 15$ and $x = 13$. The total gives $y = 70 - 7 - 13 - 25 - 16 = 9$.

Mode of Grouped Data: Which Way Does It Lean?

Mode $= L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$, where $f_1$ is the frequency of the modal class, $f_0$ that of the class before it and $f_2$ that of the class after it.

Fast check: the mode lies inside the modal class and leans towards the neighbour with the bigger frequency. If $f_0 = f_2$, it is exactly the middle of the class. This can rule out options before any working.

For example: the marks of 70 students.

The modal class is 40-60. The class after it (18) beats the class before it (15), so the mode is above the middle, 50. Mode $= 40 + \frac{24 - 15}{48 - 15 - 18} \times 20 = 40 + \frac{9}{15} \times 20 = 52$.

If $f_0$ and $f_2$ are swapped, the formula gives 48, below the middle, and the check shows at once that something is wrong.

The Empirical Relation in Three Forms

For a moderately skewed distribution, $\text{Mode} = 3\,\text{Median} - 2\,\text{Mean}$. Rearranged, the same relation gives:

• $\text{Mean} - \text{Mode} = 3(\text{Mean} - \text{Median})$;
• $\text{Median} = \frac{2\,\text{Mean} + \text{Mode}}{3}$, so the median lies a third of the way from the mean to the mode;
• $\text{Mean} = \frac{3\,\text{Median} - \text{Mode}}{2}$.

For example: a mean of 45.6 and a median of 44.2 give $\text{Mode} = 132.6 - 91.2 = 41.4$. A mode of 30 and a mean of 36 give $\text{Median} = \frac{72 + 30}{3} = 34$.

The 3 goes with the median, the measure that lies between the other two. Writing $3\,\text{Mean} - 2\,\text{Median}$ gives a wrong value that often appears as an option. The relation is also only approximate: if a question gives raw data and asks for the actual mode, count the values.

What a Change to Every Value Does

So for $y = ax + b$: $\bar{y} = a\bar{x} + b$, $\sigma_y = \lvert a \rvert \sigma_x$ and $\sigma_y^2 = a^2 \sigma_x^2$.

For example: a list has mean 40 and SD 6. For $y = 15 - 2x$, the mean is $15 - 80 = -65$, the SD is $2 \times 6 = 12$ (an SD is never negative) and the variance is 144. The 15 changes nothing in the spread.

[SSC Tip] The coefficient of variation divides the SD by the mean, so adding a constant does change it, even though the SD stays the same.

Variance Without Large Squares: Shift First

Subtracting the same number from every value does not change the variance. So subtract a round number $A$ first, work with the small deviations $d = x - A$, and use $$\sigma^2 = \frac{\sum d^2}{n} - \left(\frac{\sum d}{n}\right)^2$$ The second term takes care of $A$ not being the mean.

For example: five packets weigh 495, 499, 501, 503 and 507 g. From $A = 500$ the deviations are $-5, -1, 1, 3, 7$, with $\sum d = 5$ and $\sum d^2 = 25 + 1 + 1 + 9 + 49 = 85$. So $\sigma^2 = \frac{85}{5} - 1^2 = 16$ and $\sigma = 4$ g. Squaring 495 and 507 was never needed.

Where it fails: dropping the $\left(\frac{\sum d}{n}\right)^2$ term gives 17, which is the variance only when $A$ is the mean itself.

Ready Results for Spread

• Two values $a$ and $b$: SD $= \frac{\lvert a - b \rvert}{2}$.
• $n$ consecutive whole numbers, such as the first $n$ natural numbers: variance $= \frac{n^2 - 1}{12}$.
• $n$ values in equal steps of $d$: variance $= \frac{d^2(n^2 - 1)}{12}$, so the SD is $d$ times the SD of $1, 2, \ldots, n$.
• All values equal: SD $= 0$.

For example: the SD of 18 and 30 is $\frac{12}{2} = 6$. The variance of 31, 32, …, 37 (seven values) is $\frac{49 - 1}{12} = 4$, so their SD is 2. The values 5, 10, 15, …, 35 are also seven in number, in steps of 5, so their SD is $5 \times 2 = 10$.

More Consistent? Compare the SD With the Mean

The coefficient of variation is $\text{CV} = \frac{\sigma}{\bar{x}} \times 100$. The smaller CV belongs to the more consistent (less variable) set. To choose between two sets you need not find either percentage: compare the fractions $\frac{\sigma_1}{\bar{x}_1}$ and $\frac{\sigma_2}{\bar{x}_2}$ by cross-multiplying.

For example: over a term, student P's test marks have a mean of 48 with SD 12, and student Q's a mean of 36 with SD 8. $12 \times 36 = 432$ and $8 \times 48 = 384$, so Q's fraction is smaller and Q is more consistent. The CVs are 25% and about 22.2%.

The SDs alone decide only when the means are equal. P does have the bigger SD here, but that is not why Q is chosen.

Rule Out Options Before Working

• The mean, the median and the mode all lie between the smallest and the largest value.
• Variance and SD are never negative, and they are 0 only when every value is the same.
• The SD is at most half the range.
• For symmetric data, such as values in equal steps, the mean equals the median.
• When the variance is asked, an option equal to the SD is a distractor, and the other way round.

For example: the SD of 12, 15, 18, 20 and 25 cannot be 8, because the range is 13 and the SD is at most 6.5. Worked out, the mean is 18, the squared deviations add up to 98, the variance is 19.6 and the SD is about 4.43.

Solved Examples

Each of these fifteen questions is solved by one of the methods above. Give yourself a minute for each before you read the answer.

Question 1

The heights (in cm) of nine students are 152, 147, 158, 149, 161, 155, 146, 153 and 156. Find their mean height.

Answer. I took 150 as the working mean. The deviations are $+2, -3, +8, -1, +11, +5, -4, +3, +6$. Adding them: $2 - 3 + 8 - 1 + 11 + 5 - 4 + 3 + 6 = 27$. So the mean is $150 + \frac{27}{9} = 150 + 3 =$ 153 cm.

Question 2

The mean of 21 consecutive even numbers is 64. What is the smallest of them?

Answer. The numbers go up in equal steps, so the mean is the middle number, the 11th. There are 10 numbers before it, each 2 less than the next, so the smallest is $64 - 10 \times 2 =$ 44.

Question 3

The mean of 30 observations was found to be 52. Later it was seen that two of them, 46 and 73, had been copied as 64 and 37. Find the correct mean.

Answer. I don't need the total. The correct values add up to $46 + 73 = 119$ and the copied ones to $64 + 37 = 101$, so the total was 18 too small. Spread over 30 values that is $\frac{18}{30} = 0.6$. The correct mean is $52 + 0.6 =$ 52.6.

Time check: about 20 seconds.

Question 4

The mean weight of 24 students of section A is 52 kg, and the mean weight of 36 students of section B is 47 kg. Find the mean weight of all 60 students.

Answer. The answer lies between 47 and 52, nearer 47 because B is the bigger group. The gap is 5 kg and A has $\frac{24}{60} = \frac{2}{5}$ of the students, so the mean moves $\frac{2}{5} \times 5 = 2$ kg up from 47. The mean weight is 49 kg.

Question 5

The table shows the time taken by 40 students to finish a test.

Find the mean time taken.

Answer. The class marks are 20, 30, 40, 50 and 60. I took $A = 40$ and $h = 10$, so $u$ is $-2, -1, 0, 1, 2$.

$\sum fu = 2$, so the mean is $40 + 10 \times \frac{2}{40} = 40 + 0.5 =$ 40.5 minutes.

Question 6

The table shows how many cups of tea each person in a group drinks in a day. The mean is 2.5 cups. Find $p$.

Answer. I took deviations from the mean, 2.5. They are halves, so I doubled them to get whole numbers: $-5, -3, -1, 1, 3, 5$. The doubled deviations times the frequencies must still add up to 0: $$3(-5) + 7(-3) + 12(-1) + p(1) + 6(3) + 2(5) = 0$$ That is $-48 + p + 28 = 0$, so $p - 20 = 0$ and $p = 20$.

Question 7

The rainfall (in mm) at a place on 12 days was 4, 0, 12, 7, 0, 23, 9, 15, 3, 7, 18 and 11. Find the median rainfall.

Answer. There are 12 values, so I need the 6th and 7th in order. Sorting: 0, 0, 3, 4, 7, 7, 9, 11, 12, 15, 18, 23. The 6th is 7 and the 7th is 9, so the median is $\frac{7 + 9}{2} =$ 8 mm. Both zeros count as values.

Question 8

The table shows the number of goals a team scored in each of its 40 matches.

Find the median number of goals.

Answer. $N = 40$ is even, so the median is the mean of the 20th and 21st values. The cumulative frequencies are 6, 17, 26, 33, 38, 40. The 18th to 26th values are all 2, so the 20th and 21st are both 2 and the median is 2 goals.

Question 9

The table shows the marks of 50 students.

Find the median marks.

Answer. The classes are inclusive, so first I made them continuous: 19.5-29.5, 29.5-39.5, 39.5-49.5 and so on. $\frac{N}{2} = 25$, and the cumulative frequencies are 5, 18, 38, so the median class is 39.5-49.5, with 18 students before it. $$\text{Median} = 39.5 + \frac{25 - 18}{20} \times 10 = 39.5 + 3.5 = 43$$ The median is 43 marks. Taking $L = 40$ would have given 43.5.

Question 10

The table shows the monthly tuition fees paid by 60 students.

Find the modal fee.

Answer. The modal class is 12-16, with $f_1 = 20$, $f_0 = 14$ and $f_2 = 11$. Since $f_0$ is bigger than $f_2$, the mode must lie below the middle of the class, 14. $$\text{Mode} = 12 + \frac{20 - 14}{40 - 14 - 11} \times 4 = 12 + \frac{6}{15} \times 4 = 13.6$$ The modal fee is ₹ 13.6 thousand, that is ₹ 13,600, and it does lie below 14.

Question 11

In a moderately skewed distribution of daily wages, the mean is ₹ 524 and the median is ₹ 518. Find the mode.

Answer. $\text{Mode} = 3 \times 518 - 2 \times 524 = 1554 - 1048 = 506$. So the mode is ₹ 506. The median sits between the mean and the mode, as it should.

Time check: about 15 seconds.

Question 12

For a distribution, the mode is 63.5 and the median is 60.5. Using the empirical relation, find the mean.

Answer. From $\text{Mode} = 3\,\text{Median} - 2\,\text{Mean}$, I get $2\,\text{Mean} = 3 \times 60.5 - 63.5 = 181.5 - 63.5 = 118$. So the mean is 59.

Question 13

The heights of five plants are 142, 148, 150, 156 and 159 cm. Find the standard deviation of their heights.

Answer. I subtracted 150 from each: $-8, -2, 0, 6, 9$. These add up to 5, and their squares add up to $64 + 4 + 0 + 36 + 81 = 185$. $$\sigma^2 = \frac{185}{5} - \left(\frac{5}{5}\right)^2 = 37 - 1 = 36$$ So the standard deviation is 6 cm.

Question 14

A set of numbers has mean 32 and standard deviation 5. Each number is multiplied by 4 and then 9 is subtracted. Find the new mean and the new variance.

Answer. The mean goes through the same change: $4 \times 32 - 9 = 119$. The SD is multiplied by 4 and the subtraction does nothing to it, so it becomes 20 and the variance is $20^2 = 400$. The answer is mean 119, variance 400.

Question 15

Over a month, the daily output of factory A had mean 40 tonnes and SD 8 tonnes, and that of factory B had mean 20 tonnes and SD 5 tonnes. Which factory's output was more consistent?

Answer. I compared $\frac{8}{40}$ with $\frac{5}{20}$ by cross-multiplying: $8 \times 20 = 160$ and $5 \times 40 = 200$. A's fraction is smaller, so factory A was more consistent, though its SD is larger. In percent the CVs are 20% and 25%.