Coded Data

AS · S1 · 14 min

Coding means replacing each value $x$ by a simpler one, usually $x - a$ for some convenient number $a$ , before doing the arithmetic. The syllabus asks you to find the mean and standard deviation "from coded totals $\sum (x - a)$ and $\sum (x - a)^2$ ", and to use them in problems with up to two data sets. This appears on almost every Paper 5: a short question gives coded totals and asks for $\bar{x}$ and the standard deviation, or asks you to recover $\sum x$ and $\sum x^2$ , or to combine two sets that were coded differently.

Why coding works

Suppose the masses of six bags of flour, in grams, are

1012, \ 1005, \ 998, \ 1010, \ 1003, \ 1008.

Squaring numbers like $1012$ is slow and error-prone. Subtract $1000$ from each value instead:

12, \ 5, \ -2, \ 10, \ 3, \ 8.

These coded values are exactly the original values slid $1000$ units to the left on the number line. Sliding every value by the same amount moves the mean by that amount but does not change how far apart the values are. So:

the mean of the coded values is $1000$ less than the mean of the original values;
the standard deviation of the coded values is the same as the standard deviation of the original values.

The coded values have $\sum (x - 1000) = 36$ and $\sum (x - 1000)^2 = 144 + 25 + 4 + 100 + 9 + 64 = 346$ . Their mean is $6$ , so the mean mass is $1000 + 6 = 1006\ \text{g}$ . Their variance is $\tfrac{346}{6} - 6^2 = 21.667$ , so the standard deviation of the masses is $\sqrt{21.667} = 4.65\ \text{g}$ , with no large numbers anywhere.

The coded formulas

Key result

For $n$ values coded as $x - a$ :

\bar{x} = a + \frac{\sum (x - a)}{n}

\text{variance of } x = \frac{\sum (x - a)^2}{n} - \left(\frac{\sum (x - a)}{n}\right)^2, \qquad \text{sd of } x = \sqrt{\frac{\sum (x - a)^2}{n} - \left(\frac{\sum (x - a)}{n}\right)^2}.

Subtracting $a$ changes the mean by $a$ and leaves the variance and standard deviation unchanged.

The variance formula is just the usual "mean of the squares minus the square of the mean", applied to the coded values. The $a$ appears only when you convert the mean back.

Watch out

Adding $a$ to the standard deviation. The most common error in coding questions is writing $\text{sd of } x = a + \text{sd of } (x - a)$ . Shifting values does not change their spread. Only the mean gets $a$ added back.

Watch out

Subtracting $a^2$ in the variance. The variance is $\dfrac{\sum (x - a)^2}{n} - \left(\dfrac{\sum (x - a)}{n}\right)^2$ . The second term is the square of the coded mean, not $\bar{x}^2$ and not $a^2$ .

Converting coded totals to $\sum x$ and $\sum x^2$

Some questions, especially ones that combine data sets, need the uncoded totals. Expand the brackets:

Key result

\sum (x - a) = \sum x - na \qquad\Longleftrightarrow\qquad \sum x = \sum (x - a) + na

\sum (x - a)^2 = \sum x^2 - 2a\sum x + na^2 \qquad\Longleftrightarrow\qquad \sum x^2 = \sum (x - a)^2 + 2a \sum x - na^2

In the second line the " $-2a\sum x$ " comes from the cross term: $(x - a)^2 = x^2 - 2ax + a^2$ , and summing $n$ copies of $a^2$ gives $na^2$ . There is no need to memorise these; write out $(x - a)^2 = x^2 - 2ax + a^2$ and sum each term.

An equivalent route, often quicker, is to find $\bar{x}$ and the standard deviation from the coded totals first, then use $\sum x = n\bar{x}$ and $\sum x^2 = n(\sigma^2 + \bar{x}^2)$ from the standard deviation note.

Watch out

$\sum (x - a) \ne \sum x - a$ . Subtracting $a$ from every one of $n$ values subtracts $na$ in total. Likewise $\sum (x - a)^2 \ne \sum x^2 - a^2$ .

Combining two coded data sets

Totals can only be added when they are on the same code. Two cases:

Same $a$ . Add the $n$ s, add the $\sum (x - a)$ s and add the $\sum (x - a)^2$ s, then use the coded formulas.
Different $a$ s. Convert one set to the other's code, or convert both to $\sum x$ and $\sum x^2$ , then add.

Method

Check whether both sets use the same coding constant.
If not, convert: either find $\sum x$ and $\sum x^2$ for each set, or re-code one set. To change $\sum (x - b)$ to $\sum (x - a)$ , write $x - a = (x - b) + (b - a)$ , so $\sum (x - a) = \sum (x - b) + n(b - a)$ and $\sum (x - a)^2 = \sum (x - b)^2 + 2(b - a)\sum (x - b) + n(b - a)^2$ .
Add the totals for the two sets.
Find the combined mean and standard deviation from the combined totals.

Scaling as well as shifting

Sometimes every value is multiplied by a constant, for instance converting units or giving everyone a percentage pay rise. Multiplying every value by $b$ multiplies the distances between values by $b$ too.

Key result

If every value is transformed to $y = bx + c$ (with $b > 0$ ):

\bar{y} = b\bar{x} + c, \qquad \text{sd of } y = b \times \text{sd of } x, \qquad \text{variance of } y = b^2 \times \text{variance of } x.

Adding $c$ affects the mean only; multiplying by $b$ affects both.

The same idea run backwards handles codes such as $y = \dfrac{x - a}{b}$ : then $x = by + a$ , so $\bar{x} = b\bar{y} + a$ and $\text{sd of } x = b \times \text{sd of } y$ . Paper 5 coding questions mostly use the shift $x - a$ alone, but scaled versions do appear in context (temperature conversions, pay rises, unit changes).

Choosing a coding constant

If a question asks you to choose the code yourself, pick $a$ close to the middle of the data (or a round number near the mean). The coded values are then small and both positive and negative, so the squares stay small. Any value of $a$ gives the same final answers; a good choice just reduces the arithmetic.

Worked examples

Mean and standard deviation from coded totals (routine)

For $20$ values of $x$ , $\sum (x - 50) = 64$ and $\sum (x - 50)^2 = 1052$ . Find the mean and standard deviation of $x$ .

Solution

\bar{x} = 50 + \frac{64}{20} = 50 + 3.2 = 53.2

\text{sd} = \sqrt{\frac{1052}{20} - 3.2^2} = \sqrt{52.6 - 10.24} = \sqrt{42.36} = 6.51 \text{ (3 s.f.)}

Note that the $3.2$ being squared is the coded mean, not $53.2$ .

Recovering uncoded totals

For $30$ values of $x$ , $\sum (x - 10) = 45$ and $\sum (x - 10)^2 = 320$ .

(a) Find $\sum x$ and $\sum x^2$ .

(b) Find the variance of $x$ in two ways, and confirm that they agree.

Solution

(a) $\sum x = \sum (x - 10) + 30 \times 10 = 45 + 300 = 345$ .

Expand: $\sum (x - 10)^2 = \sum x^2 - 20\sum x + 30 \times 100$ , so

320 = \sum x^2 - 20(345) + 3000 \ \Rightarrow \ \sum x^2 = 320 + 6900 - 3000 = 4220.

(b) From coded totals: $\dfrac{320}{30} - \left(\dfrac{45}{30}\right)^2 = 10.667 - 2.25 = 8.42$ (3 s.f.).

From uncoded totals: $\bar{x} = \tfrac{345}{30} = 11.5$ , and $\dfrac{4220}{30} - 11.5^2 = 140.667 - 132.25 = 8.42$ (3 s.f.).

Both give $8.416\ldots$ , as they must: coding by subtraction does not change the variance.

Coded totals from the mean and standard deviation

The marks of $40$ students in an examination have mean $62.5$ and standard deviation $4.2$ . Find $\sum (x - 60)$ and $\sum (x - 60)^2$ .

Solution

The coded values $x - 60$ have mean $62.5 - 60 = 2.5$ and the same standard deviation, $4.2$ .

\sum (x - 60) = 40 \times 2.5 = 100

For the coded values, $4.2^2 = \dfrac{\sum (x - 60)^2}{40} - 2.5^2$ , so

\sum (x - 60)^2 = 40(4.2^2 + 2.5^2) = 40(17.64 + 6.25) = 40 \times 23.89 = 955.6.

Finding the coding constant

For $20$ values of $x$ , $\sum (x - a) = 70$ and $\sum (x - a)^2 = 830$ . The mean of $x$ is $13.5$ . Find $a$ and the standard deviation of $x$ .

Solution

13.5 = a + \frac{70}{20} = a + 3.5 \ \Rightarrow \ a = 10.

\text{sd} = \sqrt{\frac{830}{20} - 3.5^2} = \sqrt{41.5 - 12.25} = \sqrt{29.25} = 5.41 \text{ (3 s.f.)}

Two data sets with different codes (exam-hard)

The lengths, $x$ cm, of $12$ fish of species A satisfy $\sum (x - 20) = 36$ and $\sum (x - 20)^2 = 250$ . The lengths of $18$ fish of species B satisfy $\sum (x - 25) = -27$ and $\sum (x - 25)^2 = 300$ . Find the mean and standard deviation of the lengths of all $30$ fish.

Solution

The codes differ, so convert species B to the code $x - 20$ . Write $x - 20 = (x - 25) + 5$ .

\sum (x - 20) = \sum (x - 25) + 18 \times 5 = -27 + 90 = 63

\sum (x - 20)^2 = \sum (x - 25)^2 + 2 \times 5 \sum (x - 25) + 18 \times 5^2 = 300 - 270 + 450 = 480

Combine with species A on the same code: $n = 30$ , $\sum (x - 20) = 36 + 63 = 99$ , $\sum (x - 20)^2 = 250 + 480 = 730$ .

\bar{x} = 20 + \frac{99}{30} = 20 + 3.3 = 23.3\ \text{cm}

\text{sd} = \sqrt{\frac{730}{30} - 3.3^2} = \sqrt{24.333 - 10.89} = \sqrt{13.443} = 3.67\ \text{cm (3 s.f.)}

Check by the other route. Species A: $\sum x = 36 + 12 \times 20 = 276$ and $\sum x^2 = 250 + 40 \times 276 - 12 \times 400 = 6490$ . Species B: $\sum x = -27 + 18 \times 25 = 423$ and $\sum x^2 = 300 + 50 \times 423 - 18 \times 625 = 10200$ . Together $\sum x = 699$ and $\sum x^2 = 16690$ , so $\bar{x} = 23.3$ and the variance is $\tfrac{16690}{30} - 23.3^2 = 556.333 - 542.89 = 13.443$ , as before. Re-coding one set involved smaller numbers, which is why it is usually the faster method.

Scaling and shifting in context

The daily midday temperatures in a town during one month had mean $15\ ^\circ\text{C}$ and standard deviation $4\ ^\circ\text{C}$ . The temperature in degrees Fahrenheit is $F = 1.8C + 32$ . Find the mean and standard deviation of the temperatures in $^\circ\text{F}$ .

Solution

Mean: $1.8 \times 15 + 32 = 59\ ^\circ\text{F}$ .

Standard deviation: $1.8 \times 4 = 7.2\ ^\circ\text{F}$ . Adding $32$ shifts every temperature equally and so does not change the spread.

Exam tip

State the uncoded mean clearly: examiners see many answers that stop at the coded mean ( $3.2$ instead of $53.2$ ).
Show the coded formula with numbers in it, for example $\sqrt{\tfrac{1052}{20} - 3.2^2}$ . This earns the method mark even if the arithmetic goes wrong.
When asked for $\sum x^2$ , write out $\sum (x - a)^2 = \sum x^2 - 2a\sum x + na^2$ before substituting; the $na^2$ term is where marks are usually lost.
When combining sets, say which code you are converting to. Combining totals with different codes is a zero-mark error.
Give answers to 3 significant figures, but keep exact values ( $3.3$ , $24.333\ldots$ ) in the working.

Summary

Coding $x \to x - a$ makes arithmetic easier and does not change the answers.
$\bar{x} = a + \dfrac{\sum (x - a)}{n}$ : add $a$ back to the coded mean.
Variance and standard deviation are unchanged by subtracting $a$ : use $\dfrac{\sum (x - a)^2}{n} - \left(\dfrac{\sum (x - a)}{n}\right)^2$ .
$\sum x = \sum (x - a) + na$ and $\sum (x - a)^2 = \sum x^2 - 2a\sum x + na^2$ .
Totals on different codes cannot be added until one is converted.
For $y = bx + c$ : mean $\times b$ then $+ c$ ; standard deviation $\times b$ only; variance $\times b^2$ .

Practice questions

Question

For $15$ values of $x$ , $\sum (x - 100) = -45$ and $\sum (x - 100)^2 = 1200$ . Find the mean and standard deviation of $x$ .
The masses, $x$ kg, of six parcels are $2.43, 2.47, 2.51, 2.38, 2.46, 2.45$ . Using the coding $y = 100(x - 2.4)$ , find the mean and standard deviation of the masses.
The ages, $x$ years, of $25$ people have mean $23.1$ and standard deviation $2.2$ . Find $\sum (x - 20)$ and $\sum (x - 20)^2$ .
For $50$ values, $\sum (x - 30) = -120$ and $\sum (x - 30)^2 = 2088$ . Find the mean and standard deviation of $x$ .
Ten values of $x$ have mean $7.6$ and standard deviation $2.3$ . Find $\sum (x - 5)$ and $\sum (x - 5)^2$ .
The mean of $20$ values of $x$ is $8$ , and $\sum (x - 3)^2 = 680$ . Find the standard deviation of $x$ .
Each of the $12$ employees in a small firm earns a weekly wage. The wages have mean $1200 and standard deviation $150. Every wage is increased by $4\%$ and then by a further $50. Find the new mean and standard deviation.
For $16$ values of $x$ , $\sum (x - a) = -12$ , $\sum (x - a)^2 = 148$ and $\sum x = 388$ . Find $a$ , and the mean and standard deviation of $x$ .
Set A has $10$ values with $\sum (x - 50) = 24$ and $\sum (x - 50)^2 = 420$ . Set B has $15$ values with $\sum (x - 40) = 180$ and $\sum (x - 40)^2 = 2600$ . Find the mean and standard deviation of all $25$ values.
For $12$ values of $x$ , $\sum (x - c) = 30$ and $\sum (x - c)^2 = 375$ . Also $\sum (x - 2c) = -42$ . Find $c$ , the mean of $x$ and the standard deviation of $x$ .

Answers

$\bar{x} = 100 + \tfrac{-45}{15} = 100 - 3 = 97$ . Standard deviation $= \sqrt{\tfrac{1200}{15} - (-3)^2} = \sqrt{80 - 9} = \sqrt{71} = 8.43$ (3 s.f.).
The coded values are $3, 7, 11, -2, 6, 5$ : $\sum y = 30$ , $\sum y^2 = 9 + 49 + 121 + 4 + 36 + 25 = 244$ . $\bar{y} = 5$ , and the standard deviation of $y$ is $\sqrt{\tfrac{244}{6} - 25} = \sqrt{15.667} = 3.958$ . Since $x = \tfrac{y}{100} + 2.4$ : mean mass $= \tfrac{5}{100} + 2.4 = 2.45\ \text{kg}$ ; standard deviation $= \tfrac{3.958}{100} = 0.0396\ \text{kg}$ (3 s.f.).
Coded mean $= 23.1 - 20 = 3.1$ , so $\sum (x - 20) = 25 \times 3.1 = 77.5$ . $\sum (x - 20)^2 = 25(2.2^2 + 3.1^2) = 25(4.84 + 9.61) = 361.25$ .
Coded mean $= \tfrac{-120}{50} = -2.4$ , so $\bar{x} = 30 - 2.4 = 27.6$ . Standard deviation $= \sqrt{\tfrac{2088}{50} - (-2.4)^2} = \sqrt{41.76 - 5.76} = \sqrt{36} = 6$ .
Coded mean $= 7.6 - 5 = 2.6$ , so $\sum (x - 5) = 26$ . $\sum (x - 5)^2 = 10(2.3^2 + 2.6^2) = 10(5.29 + 6.76) = 120.5$ .
Coded mean $= 8 - 3 = 5$ . Variance $= \tfrac{680}{20} - 5^2 = 34 - 25 = 9$ , so the standard deviation is $3$ .
New wage $= 1.04x + 50$ . New mean $= 1.04 \times 1200 + 50 = 1298$ , i.e. $1298. New standard deviation $= 1.04 \times 150 = 156$ , i.e. $156 (adding $50 does not change the spread).
$\sum (x - a) = \sum x - 16a$ , so $-12 = 388 - 16a$ and $a = 25$ . Mean $= 25 + \tfrac{-12}{16} = 25 - 0.75 = 24.25$ . Standard deviation $= \sqrt{\tfrac{148}{16} - 0.75^2} = \sqrt{9.25 - 0.5625} = \sqrt{8.6875} = 2.95$ (3 s.f.).
Convert set B to the code $x - 50$ , using $x - 50 = (x - 40) - 10$ : $\sum (x - 50) = 180 - 15 \times 10 = 30$ ; $\sum (x - 50)^2 = 2600 - 2 \times 10 \times 180 + 15 \times 100 = 2600 - 3600 + 1500 = 500$ . Combined: $n = 25$ , $\sum (x - 50) = 24 + 30 = 54$ , $\sum (x - 50)^2 = 420 + 500 = 920$ . Mean $= 50 + \tfrac{54}{25} = 52.16$ . Standard deviation $= \sqrt{\tfrac{920}{25} - 2.16^2} = \sqrt{36.8 - 4.6656} = \sqrt{32.1344} = 5.67$ (3 s.f.).
$\sum (x - 2c) = \sum (x - c) - 12c$ , so $-42 = 30 - 12c$ and $c = 6$ . Mean $= 6 + \tfrac{30}{12} = 8.5$ . Standard deviation $= \sqrt{\tfrac{375}{12} - 2.5^2} = \sqrt{31.25 - 6.25} = \sqrt{25} = 5$ .

Why coding works

The coded formulas

Converting coded totals to ∑x\sum x∑x and ∑x2\sum x^2∑x2

Combining two coded data sets

Scaling as well as shifting

Choosing a coding constant

Worked examples

Practice questions

Console

Converting coded totals to $\sum x$ and $\sum x^2$