The effect of adding a constant to every data value on the mean, median, mode, range and standard deviation; why the two spread measures are unchanged by a shift; the effect of multiplying every value by a constant; why every statistic scales; and the general transformation that scales and then shifts.
Subject: Algebra 2 · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Algebra 2 · Chapter 11 — Data Analysis and Statistics
Apply Transformations to Data
Objectives
Five outcomes. Change every value the same way, and predict what happens to all five statistics.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.2 Apply Transformations to Data §11.2, pp. 751-753 — the lesson these objectives are drawn from
Warm-up
Lesson 11.1 gave you five statistics for any data set.
Discussion prompt
The data 7, 12, 16, 20, 20 has mean 15 and range 13. Add 10 to every value. Without recomputing from scratch, what would you expect the new mean and the new range to be?
Hint: How far apart are the values afterwards?
Answer:
\[ 17, 22, 26, 30, 30: \; \bar{x} = 25, \; \text{range} = 13 \]
The mean rose by 10, as expected. But the range did not change: the values are exactly as far apart as before, because every one of them moved the same distance.
That split — centres move, spreads do not — is the whole of this lesson's first half.
Concept
Transforming every value the same way transforms the statistics predictably. Measures of centre follow the transformation exactly. Measures of spread ignore a shift entirely but scale with a multiplier.
transformation of data — Changing every value of a data set in the same way, by adding a constant, multiplying by a constant, or both.
\[ x \to ax+b \]
The reason is that spread measures are built from differences, and a shift cancels out of every difference while a multiplier does not.
Figure (svg): Two columns comparing adding a constant with multiplying by one
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.2 Apply Transformations to Data §11.2, pp. 751-752
Section
Section 1
Concept
When the same constant is added to every value, the mean, median and mode all increase by that constant. The range and standard deviation are unchanged.
\[ \bar{x} \to \bar{x}+b; \qquad s \to s \]
No recomputation is needed. The five new statistics follow from the five old ones by one addition each, and two of them need not even be touched.
Figure (svg): A data set shifted by adding ten to every value, with the five statistics before and after
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.2 Apply Transformations to Data §11.2, pp. 751-751 — Adding a Constant to Data Values
Picture it
Five values shifted right by ten.
Figure (svg): A data set shifted by adding ten to every value, with the five statistics before and after
Every dot moved the same distance, so the pattern is identical — just relocated. The three centres moved with it and the two spreads did not.
Worked example
Example 1.
\[ \text{Eight astronauts weigh } 142,150,155,156,160,160,166,175 \text{ lb. A suit adds } 250 \text{ lb. Find all five statistics both ways.} \]
Compute the original statistics
Why: Sum 1264 over 8; middle pair 156 and 160; 160 twice.
\[ 158, 158, 160 \]
Compute the original spreads
Why: One seventy-five minus 142; squares total 694 over 8.
\[ 33\text{ and about } 9.3 \]
Add 250 to the three centres
Why: One fifty-eight plus 250, twice, and 160 plus 250.
\[ 408, 408, 410 \]
Leave the two spreads alone
Why: Every weight rose equally, so nothing spread out.
\[ 33\text{ and } 9.3 \]
Figure (svg): A data set shifted by adding ten to every value, with the five statistics before and after
\[ 408, \; 408, \; 410, \; 33, \; 9.3 \]
Verify: check the range directly
Why: The suited weights run from 392 to 425, and 425 minus 392 is 33 — exactly the original range. Adding the same amount to both extremes leaves their difference untouched, which is the range's whole definition.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.2 Apply Transformations to Data §11.2, pp. 751-751
Sorting
Centres or spreads.
Sort into buckets
Sort each statistic for a data set with 6 added to every value.
Three move and two do not, and the split is exactly the split between measures of centre and measures of dispersion.
Worked example
Guided Practice 1.
\[ \text{The MMU adds another } 300 \text{ lb. Find all five statistics for suit plus MMU.} \]
Find the total constant
Why: Two hundred fifty plus 300.
\[ 550 \]
Add it to the three centres
Why: One fifty-eight plus 550, twice, and 160 plus 550.
\[ 708, 708, 710 \]
Leave the spreads
Why: Still the same equipment for every astronaut.
\[ 33\text{ and } 9.3 \]
Note the shortcut
Why: Two shifts in a row are one shift by their total.
Figure (svg): The solution to Worked example adding the manoeuvring unit shown as a ladder of expressions, one row per algebraic move
\[ 708, \; 708, \; 710, \; 33, \; 9.3 \]
Verify: check that the shifts combine
Why: Adding 250 and then 300 gives the same result as adding 550 once, since addition is associative. The spreads survived both shifts, so they would survive any number of them — a shift can never change how spread out data is.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.2 Apply Transformations to Data §11.2, pp. 752-752
Trap
\[ s = 10, \text{ then } 3 \text{ is added to every value} \]
Add 3 to the standard deviation as well
Why: Every statistic is treated the same way.
\[ s = 13 \quad \text{(wrong)} \]
Adding 3 to every value moves them all together, so no value ended up further from the mean than before.
\[ s \text{ stays } 10 \]
Leave both spread measures alone
Why: A shift changes no distance, and both spreads are built from distances.
\[ (x+3)-(\bar{x}+3) = x-\bar{x} \]
This is lesson exercise 9's printed error. The mean also rose by 3, so every deviation is exactly what it was.
Fill the middle
Example 1.
Fill in the blanks
158+250 = 408
Why: The mean rises by exactly the constant added, so 158 becomes 408. The same addition applies to the median and the mode.
Matching
Adding 250 to every weight.
Match the pairs
Why: Two of the four changed and two did not, which is the whole rule. Nothing here required recomputing a single statistic from the raw data.
Prediction
Commit before reasoning.
Predict first
Every value in a data set has 5 subtracted from it. What happens to the statistics?
Correct: The three centres fall by 5 and the two spreads are unchanged.
\[ b = -5: \; \bar{x} \to \bar{x}-5, \; s \to s \]
Why: Subtracting 5 is adding negative 5, so the same rule applies with a negative constant. The values all move left together, so distances between them are untouched. The rule never depends on the sign of the constant, only on the fact that the same amount is added to everything.
Section
Section 2
Concept
The range is a difference of two values and the standard deviation is built from differences from the mean. Adding the same constant to both parts of any difference leaves it unchanged.
\[ (x+b)-(\bar{x}+b) = x-\bar{x} \]
The mean shifts along with the data, which is the key step: if the mean stayed put, the deviations would change and so would the standard deviation.
Figure (svg): Why a shift leaves the spread measures alone, shown through the deviations
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.2 Apply Transformations to Data §11.2, pp. 751-751 — Adding a Constant to Data Values
Picture it
The same five values before and after a shift of ten.
Figure (svg): Why a shift leaves the spread measures alone, shown through the deviations
The deviation row is identical top and bottom, so every squared deviation is identical and so is their average.
Worked example
Proving the rule for the standard deviation.
\[ \text{Show that adding } b \text{ to every value leaves } s \text{ unchanged.} \]
Find the new mean
Why: The sum rises by n times b, and dividing by n gives b.
Find a new deviation
Why: The new value minus the new mean.
Simplify
Why: The two b's cancel.
Conclude
Why: Every deviation is unchanged, so every square is too.
Figure (svg): Why a shift leaves the spread measures alone, shown through the deviations
\[ (x+b)-(\bar{x}+b) = x-\bar{x} \]
Verify: check the range the same way
Why: The range is the largest value minus the smallest, and both rise by b, so their difference is unchanged. Both spread measures are differences at heart, which is exactly why both are immune to a shift.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.2 Apply Transformations to Data §11.2, pp. 751-751
Fill the middle
The deviation after adding b.
Fill in the blanks
(x+b)-(\barxbar+b) = x-___
Why: The two copies of b cancel, leaving the original deviation. That single line explains why both spread measures are immune to a shift.
Worked example
Lesson exercise 3.
\[ \text{For } 14,15,17,17,19,21,23 \text{ and the same data plus } 6, \text{ find all five statistics.} \]
Original centres
Why: Sum 126 over 7; middle value; most frequent.
\[ 18, 17, 17 \]
Original spreads
Why: Twenty-three minus 14; squares total 62 over 7.
\[ 9\text{ and about } 3.0 \]
Shifted centres
Why: Add 6 to each.
\[ 24, 23, 23 \]
Shifted spreads
Why: Unchanged by the rule.
\[ 9\text{ and about } 3.0 \]
Figure (svg): The solution to Worked example check the rule numerically shown as a ladder of expressions, one row per algebraic move
\[ 24, \; 23, \; 23, \; 9, \; 3.0 \]
Verify: recompute one shifted spread from scratch
Why: The shifted data is 20, 21, 23, 23, 25, 27, 29 with mean 24, so the deviations are negative 4, negative 3, negative 1, negative 1, 1, 3 and 5 — identical to before. Doing it the long way once is worth it to see the rule is not a shortcut but a fact.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.2 Apply Transformations to Data §11.2, pp. 753-753
Error analysis
A student argues that the standard deviation must change when a constant is added.
Annotate
On: \( \text{new deviation} = (x+3)-\bar{x} = (x-\bar{x})+3 \)
Both the value and the mean move together, which is the whole reason the deviations survive. Shifting one without the other breaks the argument.
Prediction
Commit before reasoning.
Predict first
Why does adding b to every value raise the mean by exactly b?
Correct: Because the sum rises by n times b, and dividing by n gives b.
\[ \frac{\sum x + nb}{n} = \bar{x}+b \]
Why: Each of the n values gains b, so the total gains n times b, and the mean is the total over n. That is why the increase in the mean is exactly b regardless of how many values there are or how large b is. The median and mode shift for a different reason: their positions in the ordering do not change, so the value at that position is simply the old one plus b.
Comparison
Fill the blanks. All three shift together.
Comparison matrix
| Quantity | Before | After adding b |
|---|---|---|
| A data value | x | x + b |
| The mean | the mean | the mean + b |
| The deviation | x - mean | x - mean, unchanged |
| The standard deviation | s | s, unchanged |
The third row is where the b disappears, and every consequence for the spread measures follows from that one cancellation.
Sorting
Which statistics are distances?
Sort into buckets
Sort each statistic.
The two lists are exactly the spreads and the centres, so knowing which kind a statistic is tells you at once how a shift affects it.
Section
Section 3
Concept
When every value is multiplied by a constant, all five statistics are multiplied by that same constant — the two spread measures included.
\[ \bar{x} \to a\bar{x}; \qquad s \to as \]
This is what makes unit conversion painless: convert the statistics rather than the data, since both give the same answer.
Figure (svg): A data set converted from metres to feet, with every statistic scaled by the same factor
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.2 Apply Transformations to Data §11.2, pp. 752-752 — Multiplying Data Values by a Constant
Picture it
Example 2: Olympic triple jump distances converted at 3.28 feet per metre.
Figure (svg): A data set converted from metres to feet, with every statistic scaled by the same factor
All five statistics multiply by 3.28. The spread measures scale because the distances between values scale with everything else.
Worked example
Example 2.
\[ \text{Eleven jump distances have mean } 17.53, \text{ median } 17.39, \text{ mode } 17.35, \text{ range } 1.32, \; s = 0.37 \text{ m. Convert to feet.} \]
Identify the factor
Why: One metre is about 3.28 feet.
\[ a = 3.28 \]
Scale the three centres
Why: Multiply each by 3.28.
\[ 57.50, 57.04, 56.91 \]
Scale the range
Why: One point three two times 3.28.
\[ \text{about } 4.33 \]
Scale the standard deviation
Why: Point three seven times 3.28.
\[ \text{about } 1.21 \]
Figure (svg): A data set converted from metres to feet, with every statistic scaled by the same factor
\[ 57.50, \; 57.04, \; 56.91, \; 4.33, \; 1.21 \]
Verify: check one value directly
Why: The longest jump was 18.17 metres, which is 59.6 feet, and the shortest 16.85 metres, or 55.3 feet. Their difference is 4.3 feet, matching the scaled range. Converting the raw data and then computing gives the same answers as converting the statistics.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.2 Apply Transformations to Data §11.2, pp. 752-752
Sorting
Every value tripled.
Sort into buckets
Sort each statistic.
Every statistic in the same units as the data scales with it. Only counts, which have no units, are untouched.
Worked example
Guided Practice 2.
\[ \text{Convert the same statistics to yards, at } 1.09 \text{ yards per metre.} \]
Scale the mean and median
Why: Seventeen point five three and 17.39 times 1.09.
\[ \text{about } 19.11\text{ and } 18.95 \]
Scale the mode
Why: Seventeen point three five times 1.09.
\[ \text{about } 18.91 \]
Scale the range
Why: One point three two times 1.09.
\[ \text{about } 1.44 \]
Scale the standard deviation
Why: Point three seven times 1.09.
\[ \text{about } 0.40 \]
Figure (svg): The solution to Worked example convert to yards shown as a ladder of expressions, one row per algebraic move
\[ 19.11, \; 18.95, \; 18.91, \; 1.44, \; 0.40 \]
Verify: sanity-check the factor
Why: A yard is slightly shorter than a metre, so the numbers should be slightly larger — and every one of them is, by about 9 percent. Feet are much shorter still, which is why the same distances came out around three times larger in feet than in metres.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.2 Apply Transformations to Data §11.2, pp. 752-752
Trap
\[ \text{convert } 17.53 \text{ m to feet, but leave } s = 0.37 \]
Scale only the centres
Why: The rule from adding a constant is carried across.
\[ \bar{x} = 57.50 \text{ ft}, \; s = 0.37 \quad \text{(wrong)} \]
A standard deviation of 0.37 feet against a mean of 57.50 feet would mean the jumps were within about four inches of each other, which is not what the data shows.
\[ s = 3.28 \times 0.37 \approx 1.21 \text{ ft} \]
Scale every statistic, spreads included
Why: Multiplying stretches the distances between values as well as their positions.
\[ \text{units must match: feet throughout} \]
A units check catches this immediately: a spread measured in metres cannot sit beside a mean measured in feet.
Fill the middle
Lesson exercise 16.
Fill in the blanks
\text63 21, \text___ \;\Longrightarrow\; \text___ = ___
Why: Twenty-one times 3 is 63. Both extremes triple, so their difference triples with them.
Matching
Metres to feet at 3.28.
Match the pairs
Why: All four used the same multiplication, unlike the addition rule where two statistics were left alone. Scaling treats every statistic identically.
Prediction
Commit before reasoning.
Predict first
Every value in a data set is halved. What happens to the standard deviation?
Correct: It halves as well.
\[ \sqrt{\left(\tfrac{1}{2}\right)^2 s^2} = \tfrac{1}{2}s \]
Why: Halving is multiplying by one half, so every statistic including the standard deviation is halved. The squared deviations each fall by a factor of four, but the square root at the end turns that back into a factor of two — which is exactly why the rule comes out as a single factor rather than its square.
Section
Section 4
Concept
Multiplying by a scales the mean, so every deviation scales by a as well. Each squared deviation therefore scales by a squared, and the square root at the end brings that back to a single factor of a.
\[ \sqrt{\frac{\sum (ax-a\bar{x})^2}{n}} = \sqrt{a^2}\,s = as \]
The rule needs a to be positive for the square root to give a rather than its absolute value, which is the usual case for unit conversions and scale factors.
Figure (svg): Why multiplying scales every statistic, shown through the deviations
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.2 Apply Transformations to Data §11.2, pp. 752-752 — Multiplying Data Values by a Constant
Picture it
Why each statistic multiplies by a.
Figure (svg): Why multiplying scales every statistic, shown through the deviations
The first four are direct; the standard deviation needs the extra observation that a squared inside a square root comes back out as a.
Worked example
Lesson exercise 17.
\[ \text{Show that the mean of } ax_1, ax_2, \dots, ax_n \text{ is } a\bar{x}. \]
Write the new mean
Why: The sum of the scaled values over n.
\[ \frac{a x _{1} +... + a x _{n}}{n} \]
Factor out a
Why: Every term has a factor of a.
\[ a(x _{1} +... + x _{n}) / n \]
Recognise the old mean
Why: What remains is the original mean.
\[ a \times x - b a r \]
State the conclusion
Why: The new mean is a times the old.
\[ a x - b a r \]
Figure (svg): The solution to Worked example prove the rule for the mean shown as a ladder of expressions, one row per algebraic move
\[ \frac{a\sum x}{n} = a\bar{x} \]
Verify: check with the jump data
Why: The metre mean was 17.53 and the foot mean 57.50, and 3.28 times 17.53 is 57.50. Factoring the constant out of a sum is the same distributive step that made the proof work, so the numerical check and the algebra are the same argument.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.2 Apply Transformations to Data §11.2, pp. 753-753
Fill the middle
The scaled deviation.
Fill in the blanks
ax-a\bara = ___(x-\bar___)
Why: Factoring out a shows that every deviation scales by exactly a. That single step drives the rule for both the range and the standard deviation.
Worked example
Extending the argument to the spread.
\[ \text{Show that the standard deviation of the scaled data is } as, \text{ for } a > 0. \]
Write a new deviation
Why: The scaled value minus the scaled mean.
\[ a x - a(x - b a r) \]
Factor out a
Why: The deviation scales by a.
\[ a(x - x - b a r) \]
Square it
Why: The square picks up a squared.
\[ a ^{2}(x - x - b a r) ^{2} \]
Average and take the root
Why: The a squared comes out of the root as a.
Figure (svg): Why multiplying scales every statistic, shown through the deviations
\[ \sqrt{a^2 \cdot \frac{\sum(x-\bar{x})^2}{n}} = as \]
Verify: check the jump data again
Why: The metre standard deviation was 0.37 and the foot value 1.21, and 3.28 times 0.37 is 1.21. Note the factor is a, not a squared — the squaring inside and the root outside cancel each other, which is why the answer stays in the original units.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.2 Apply Transformations to Data §11.2, pp. 752-752
Trap
\[ \text{every deviation scales by } a, \text{ so every square by } a^2 \]
Conclude the standard deviation scales by a squared
Why: The factor from the squares is carried to the final answer.
\[ s \to a^2 s \quad \text{(wrong)} \]
The square root at the end has not been applied. Squaring and then rooting returns a factor of a, not a squared.
\[ \sqrt{a^2 s^2} = as, \text{ for } a > 0 \]
Follow the a squared through the square root
Why: The root is the last step of the standard deviation formula.
\[ 3.28^2(0.37)^2 \text{ under a root gives } 3.28(0.37) \]
A units check settles it: a standard deviation is measured in the same units as the data, so a conversion factor can appear only once.
Ranking
Proving the standard deviation scales by a.
Put in order
Why: Step one has to come first: without knowing the mean scales, the deviations could not be shown to scale. The last step is where the squaring is undone, and skipping it produces the a squared error.
Comparison
Fill the blanks. What happens to a deviation.
Comparison matrix
| Transformation | New deviation | Effect on s |
|---|---|---|
| Add b | (x + b) - (mean + b) = x - mean | unchanged |
| Multiply by a | ax - a(mean) = a(x - mean) | multiplied by a |
| Why they differ | b cancels; a factors out | cancelling changes nothing |
| The range | same reasoning, on max minus min | unchanged, then multiplied by a |
One transformation cancels out of the differences and the other survives as a common factor, which is the entire explanation for both rules.
Prediction
Commit before reasoning.
Predict first
Every value is multiplied by negative 2. What happens to the standard deviation?
Correct: It is multiplied by 2, since a standard deviation is never negative.
\[ \sqrt{(-2)^2 s^2} = |-2|\,s = 2s \]
Why: The square root of a squared is the absolute value of a, so the factor is 2 rather than negative 2. The mean does become negative 2 times its old value, since it reports a position and positions can be negative — but a spread is a distance and cannot be. This is why the rule is usually stated for positive multipliers, which covers every unit conversion.
Section
Section 5
Concept
Transforming every value by a times x plus b combines the two rules: the three centres are scaled and then shifted, while the two spread measures are only scaled.
\[ \bar{x} \to a\bar{x}+b; \qquad s \to as \]
The b never reaches the spread measures, because it cancels out of every difference exactly as it did on its own.
Figure (svg): The general transformation multiply then add, with its effect on each statistic
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.2 Apply Transformations to Data §11.2, pp. 751-752 — Transformations of data
Picture it
The effect of scaling and then shifting.
Figure (svg): The general transformation multiply then add, with its effect on each statistic
Centres take the whole transformation and spreads take only the multiplier, which is the shortest statement of the whole lesson.
Worked example
Lesson exercise 18.
\[ \text{Nine salaries are } 39,29,42.5,28.5,48,45,38,36.5,28.5 \text{ thousand. Each gets a } 1.2 \text{ thousand bonus.} \]
Compute the original centres
Why: Sum 335 over 9; fifth of nine sorted; 28.5 twice.
\[ \text{about } 37.2, 38, 28.5 \]
Compute the original spreads
Why: Forty-eight minus 28.5; squares total about 428.6 over 9.
\[ 19.5\text{ and about } 6.9 \]
Add the bonus to the centres
Why: One point two added to each.
\[ 38.4, 39.2, 29.7 \]
Leave the spreads
Why: Everyone gained the same amount.
\[ 19.5\text{ and } 6.9 \]
Figure (svg): The solution to Worked example salaries with a bonus shown as a ladder of expressions, one row per algebraic move
\[ 38.4, \; 39.2, \; 29.7, \; 19.5, \; 6.9 \]
Verify: notice what an equal bonus does not do
Why: Everyone is better off by the same amount, so the gap between the highest and lowest paid is exactly what it was. A flat bonus raises pay without narrowing pay differences at all — which is a real observation about how such policies work rather than an artefact of the arithmetic.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.2 Apply Transformations to Data §11.2, pp. 753-753
Fill the middle
A combined transformation.
Fill in the blanks
1.03(37.2)+1.2 \approx 39.5
Why: One point zero three times 37.2 is about 38.3, plus 1.2 gives about 39.5. The mean takes both parts of the transformation.
Worked example
Combining both transformations.
\[ \text{Every salary is raised by } 3 \text{ percent and then given a } 1.2 \text{ thousand bonus. Find the new statistics.} \]
Write the transformation
Why: Multiply by 1.03, then add 1.2.
\[ 1.03 x + 1.2 \]
Scale then shift the mean
Why: One point zero three times 37.2, plus 1.2.
\[ \text{about } 39.5 \]
Do the same for median and mode
Why: One point zero three times 38 and 28.5, each plus 1.2.
\[ \text{about } 40.3\text{ and } 30.6 \]
Scale the spreads only
Why: One point zero three times 19.5 and 6.9.
\[ \text{about } 20.1\text{ and } 7.1 \]
Figure (svg): The general transformation multiply then add, with its effect on each statistic
\[ a\bar{x}+b = 1.03(37.2)+1.2 \approx 39.5 \]
Verify: compare the two policies
Why: The flat bonus alone left the spread at 19.5; adding the percentage raise pushed it to 20.1. A percentage raise widens pay gaps because it gives more to those already paid more, while a flat bonus does not. The two rules together make that difference calculable rather than merely arguable.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.2 Apply Transformations to Data §11.2, pp. 752-753
Error analysis
A student transforms data by 1.03x plus 1.2 and reports the new standard deviation.
Annotate
On: \( s = 1.03(6.9)+1.2 \approx 8.3 \)
In a combined transformation the multiplier reaches every statistic and the constant reaches only the three centres.
Sorting
For the transformation ax plus b.
Sort into buckets
Sort each statistic by what affects it.
The split is the same one that has run through the whole lesson: centres take everything, spreads take only the multiplier.
Two truths and a lie
All three are about transforming data.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. The squared deviations do scale by a squared, but the standard deviation takes a square root at the end, which brings the factor back to a. Checking the units settles it: a standard deviation is in the same units as the data, so a conversion factor can only appear once.
Prediction
Commit before reasoning.
Predict first
A company can give everyone a flat bonus or the same percentage raise. Which widens the pay gaps?
Correct: The percentage raise, since it multiplies the spread while the bonus leaves it alone.
\[ \text{bonus: } s \to s; \qquad \text{raise: } s \to 1.03s \]
Why: A flat bonus adds the same amount to every salary, so the range and standard deviation are unchanged — the gaps are exactly as they were. A percentage raise multiplies every salary, so it multiplies the gaps too: a 3 percent raise widens a 19.5 thousand range to 20.1. The two rules of this lesson turn an argument about fairness into a calculation, which is what makes them worth knowing.
Comparison
Fill the blanks. Centres and spreads behave differently.
Comparison matrix
| Statistic | Add b | Multiply by a |
|---|---|---|
| Mean, median, mode | each increases by b | each multiplies by a |
| Range | unchanged | multiplies by a |
| Standard deviation | unchanged | multiplies by a |
| Why | b cancels in every difference | a factors out of every difference |
The last row explains the other three: spread measures are differences, and a shift disappears from a difference while a multiplier does not.
Pattern
Compute once, then transform.
Never recompute from the transformed data unless you want a check; the rules give the answers directly.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.2 Apply Transformations to Data §11.2, pp. 751-753
Check
Adding a constant.
Check your understanding
A data set has standard deviation 10. Three is added to every value. What is the new standard deviation?
Answer: A
Why: A shift moves every value equally, so no deviation changes.
Check
Multiplying by a constant.
Check your understanding
The range of a data set is 21 and every value is multiplied by 3. What is the new range?
Answer: A
Why: Both extremes triple, so their difference triples.
Check
Combining both.
Check your understanding
Data is transformed by 1.03x + 1.2. Its standard deviation was 6.9. What is the new one?
Answer: A
Why: Only the multiplier reaches a spread measure: 1.03 times 6.9.
Real world
A class of temperatures is recorded in Celsius with mean 22 degrees and standard deviation 4 degrees. Fahrenheit is obtained by multiplying by 1.8 and adding 32.
Discussion prompt
Find the mean and standard deviation in Fahrenheit, and explain why the two are transformed differently.
Hint: This is exactly a times x plus b.
Answer:
\[ \bar{x}_F = 1.8(22)+32 = 39.6+32 = 71.6\degree F \]
\[ s_F = 1.8(4) = 7.2\degree F \]
The mean becomes 71.6 degrees and the standard deviation 7.2 degrees. The 32 reaches the mean and not the standard deviation.
That asymmetry is not a quirk of the formula but a fact about what the two numbers mean. The mean is a temperature, so it must be converted as a temperature — 32 degrees Fahrenheit is the same thing as zero Celsius, and the offset matters. The standard deviation is a temperature DIFFERENCE, and a difference of 4 Celsius degrees is 7.2 Fahrenheit degrees regardless of where the zero is placed. This is why weather reports can say a temperature rose by five degrees without specifying a scale ambiguity that would matter for the temperature itself — differences and positions convert by different rules.
Commit first
Answer, then rate your confidence honestly.
Predict first
A data set has standard deviation 10, and 3 is added to every value. Is the new standard deviation 13?
Correct: No — it stays 10, because the mean shifts too and every deviation is unchanged.
\[ (x+3)-(\bar{x}+3) = x-\bar{x} \]
Why: Adding 3 to every value moves the whole data set together, so no value ends up further from the centre than before. Algebraically the new deviation is x plus 3 minus the quantity mean plus 3, and the two threes cancel. This is lesson exercise 9's printed error. A quick picture settles it: five dots on a number line slid three units right are still exactly as far apart as they were, and both the range and the standard deviation measure exactly that.
Explain it
They think every statistic changes the same way when data changes.
Discussion prompt
In four sentences or fewer, explain why adding a constant changes the mean but not the spread.
Hint: Picture dots on a number line.
Answer:
Imagine the data as dots on a number line. Adding the same number to every value slides all the dots the same distance in the same direction.
So the middle of the group moves, which changes the mean. But the gaps between the dots are exactly what they were, and the range and standard deviation only measure those gaps.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the shift, picture the dots sliding and the gaps staying. For the scaling, remember every statistic shares the data's units. For the factor, follow the a squared through the square root. For combined transformations, apply the multiplier to all five and the constant to the three centres.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a transformations page. Top left: draw a small data set as dots on a number line, then the same data shifted by ten below it, and write which statistics moved and which did not. Top right: write out the deviation cancellation in two lines and say in one sentence why the mean shifting is the crucial step. Middle: build the astronaut table with and without suits, filling in all five statistics both ways with no recomputation. Bottom left: build the triple jump table in metres, feet and yards, showing that every statistic used the same factor. Bottom right: work out what happens to all five under a 3 percent raise plus a flat bonus, and write one sentence contrasting the two policies' effect on the spread.
If your transformed standard deviation picked up the additive constant, redo it: only the multiplier ever reaches a spread measure.
Recap
Five things, and no statistic ever needs recomputing.
| If you see | Then |
|---|---|
| Add b to every value | Add b to the mean, median and mode only |
| Multiply every value by a | Multiply all five statistics by a |
| A unit conversion | It is a multiplication, so everything scales |
| A flat bonus | It is an addition, so the spreads are untouched |
| ax + b | All five take a; only the three centres take b |
| A spread with the constant added | An error; shifts never change spreads |
Lesson 11.3 uses the mean and standard deviation together to describe a normal distribution, where the two numbers determine the whole shape.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.2 Apply Transformations to Data §11.2, pp. 751-753 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Algebra 2 — $55/session, free consultation.