Mathematics 9709 · Probability & Statistics 1

Coded data

Coding means: make the numbers friendlier, calculate, then undo the code.
Subtract $b$ from every value and the mean drops by exactly $b$ .
So after working with $x-b$ , add $b$ back at the end.
Most exam questions hand you a coded total and test whether you can un-shift it.

Un-shift a coded total

$\bar x=\dfrac{\sum (x-b)}{n}+b\qquad\text{and}\qquad \sum(x-b)=\sum x-nb$

The second identity recovers $\sum x$ itself — which is what lets you combine two coded data sets (un-shift each to its total first).

Read the code (e.g. $x-b$ ).
Recover the true total: $\sum x=\sum(x-b)+nb$ .
Mean: $\bar x=\sum x/n$ , or directly $\bar x=\sum(x-b)/n+b$ .
To combine sets, un-shift each to its $\sum x$ first, then add.

Worked example

Start simple: the core un-shift

Ten boys' ages 5 years ago summed to 127: $\sum(b-5)=127$ . Today's mean age?

$\bar b=\tfrac{127}{10}+5$ , or $\sum b=127+10(5)=177\Rightarrow \bar b=\tfrac{177}{10}$ .

Answer

\bar b=17.7\text{ years}

Worked example

Grouped coded data

40 values coded as $x-3$ with coded mid-values $9,21,28$ and $f=9,13,18$ : $\sum(x-3)f=81+273+504=858$ .

Answer

\bar x=\tfrac{858}{40}+3=24.45

Worked example

Combine two coded sets

$\sum(b-4)=96$ for 10 boys and $\sum(g+5)=351$ for 15 girls. Un-shift each: $\sum b=96+10(4)=136$ ; $\sum g=351-15(5)=276$ . Combined mean $\dfrac{136+276}{25}=\dfrac{412}{25}$ .

Answer

16.48\text{ years}

Coding by ×a as well: undo −b first, then ÷a

Worked example

A multiplicative code: ax − b

For 20 values coded as $2x-3$ it is given that $\sum(2x-3)=104$ . The coded mean is $\tfrac{104}{20}=5.2$ . Undo the code in reverse order — add 3, then divide by 2:

$\bar x=\dfrac{5.2+3}{2}=4.1$ .

Cross-check via totals: $2\sum x=104+20(3)=164\Rightarrow\sum x=82\Rightarrow\bar x=\tfrac{82}{20}=4.1$ .

Answer

\bar x=4.1

Coding just slides the numbers; slide the answer back by +b — and keep the n in Σx = Σ(x−b) + nb. For ax − b, undo −b before ÷a.

The code	Undo it by…
$x-b$	add b at the end
$\sum(x-b)$	$\sum x=\sum(x-b)+nb$
grouped (x−b)	coded mid-values, then +b
$ax-b$	add b, then ÷a (in that order)

Examiner note

Show the

\sum x=\sum(x-b)+nb

line explicitly — that is the method mark; do not jump straight to a decoded mean.

Common mistake

Two slips: writing

\sum(x-b)=\sum x-b

instead of

\sum x-nb

(the

n

gets dropped, so you are out by

(n-1)b

), and forgetting to decode (leaving the answer in coded units).

Common mistake

For an

ax-b

code, dividing by

a

before adding

b

. The order is fixed: undo the −b, then the ÷a (the reverse of how the code was built).

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Coded data

Coding means: make the numbers friendlier, calculate, then undo the code.
Subtract $b$ from every value and the mean drops by exactly $b$ .
So after working with $x-b$ , add $b$ back at the end.
Most exam questions hand you a coded total and test whether you can un-shift it.

Un-shift a coded total

$\bar x=\dfrac{\sum (x-b)}{n}+b\qquad\text{and}\qquad \sum(x-b)=\sum x-nb$

The second identity recovers $\sum x$ itself — which is what lets you combine two coded data sets (un-shift each to its total first).

Read the code (e.g. $x-b$ ).
Recover the true total: $\sum x=\sum(x-b)+nb$ .
Mean: $\bar x=\sum x/n$ , or directly $\bar x=\sum(x-b)/n+b$ .
To combine sets, un-shift each to its $\sum x$ first, then add.

Worked example

Start simple: the core un-shift

Ten boys' ages 5 years ago summed to 127: $\sum(b-5)=127$ . Today's mean age?

$\bar b=\tfrac{127}{10}+5$ , or $\sum b=127+10(5)=177\Rightarrow \bar b=\tfrac{177}{10}$ .

Answer

\bar b=17.7\text{ years}

Worked example

Grouped coded data

40 values coded as $x-3$ with coded mid-values $9,21,28$ and $f=9,13,18$ : $\sum(x-3)f=81+273+504=858$ .

Answer

\bar x=\tfrac{858}{40}+3=24.45

Worked example

Combine two coded sets

$\sum(b-4)=96$ for 10 boys and $\sum(g+5)=351$ for 15 girls. Un-shift each: $\sum b=96+10(4)=136$ ; $\sum g=351-15(5)=276$ . Combined mean $\dfrac{136+276}{25}=\dfrac{412}{25}$ .

Answer

16.48\text{ years}

Coding by ×a as well: undo −b first, then ÷a

Worked example

A multiplicative code: ax − b

For 20 values coded as $2x-3$ it is given that $\sum(2x-3)=104$ . The coded mean is $\tfrac{104}{20}=5.2$ . Undo the code in reverse order — add 3, then divide by 2:

$\bar x=\dfrac{5.2+3}{2}=4.1$ .

Cross-check via totals: $2\sum x=104+20(3)=164\Rightarrow\sum x=82\Rightarrow\bar x=\tfrac{82}{20}=4.1$ .

Answer

\bar x=4.1

Coding just slides the numbers; slide the answer back by +b — and keep the n in Σx = Σ(x−b) + nb. For ax − b, undo −b before ÷a.

The code	Undo it by…
$x-b$	add b at the end
$\sum(x-b)$	$\sum x=\sum(x-b)+nb$
grouped (x−b)	coded mid-values, then +b
$ax-b$	add b, then ÷a (in that order)

Examiner note

Show the

\sum x=\sum(x-b)+nb

line explicitly — that is the method mark; do not jump straight to a decoded mean.

Common mistake

Two slips: writing

\sum(x-b)=\sum x-b

instead of

\sum x-nb

(the

n

gets dropped, so you are out by

(n-1)b

), and forgetting to decode (leaving the answer in coded units).

Common mistake

For an

ax-b

code, dividing by

a

before adding

b

. The order is fixed: undo the −b, then the ÷a (the reverse of how the code was built).