Exercises — Range — definition and calculation

Step 1 — biggest. Scan the list: the largest number is $x_{\max }=12$. Why: range measures the top of the spread, so we need the highest value.

Step 2 — smallest. The smallest number is $x_{\min }=3$. Why: this pins the bottom of the spread.

Step 3 — gap. $\text{Range}=12-3=9$ Why subtract: the gap between top and bottom is the spread. Answer: $9$.

Max $=60$, min $=40$. The repeated $55$ changes nothing — range only looks at the two extremes. $\text{Range}=60-40=20\text{ units}$ Answer: $20$ units.

With one value, it is both the maximum and the minimum: $x_{\max }=x_{\min }=17$. $\text{Range}=17-17=0$ Answer: $0$. A dataset with nothing to spread across has no spread. (Note: this is $0$, unlike the empty dataset which is undefined — here we do have a max and min, they simply coincide.)

Order them on a number line — see figure s01 below, where the amber dots mark the two extremes we care about. The further left a point sits, the smaller it is. $-12<-8<-3<-1<0$ So $x_{\max }=0$ and $x_{\min }=-12$. $\text{Range}=0-(-12)=0+12=12^{\circ }\text{C}$ Why the sign flip: subtracting a negative means walking rightward on the line, i.e. adding. Answer: $12^{\circ }\text{C}$.

Step 1 — extremes. Max $=0.31$, min $=0.19$. Why: only the largest and smallest touch the ends of the spread; the three middle times don't reach either end.

Step 2 — gap. $\text{Range}=0.31-0.19=0.12\text{ s}$ Why subtract: the difference is the width of reaction-time variation.

Step 3 — state the unit, concretely. Write "$0.12$ seconds", not a bare "$0.12$". A stripped "$0.12$" is ambiguous — it could be read as $0.12$ minutes, or as $12\%$ of something. Because our data were measured in seconds and range keeps the data's units unchanged, the answer is $0.12$ s. Answer: $0.12$ seconds. (This is the units mistake made concrete — always attach the unit.)

Step 1 — extremes. Max $=100.4$, min $=99.7$.

Step 2 — range. $\text{Range}=100.4-99.7=0.7\text{ mm}$

Step 3 — compare to the rule. The rule allows anything up to $0.5$ mm. Here $0.7\text{ mm}>0.5\text{ mm}$, so the range is outside the allowed limit and the batch fails. Answer: fails (range $=0.7$ mm exceeds the $0.5$ mm limit).

Ranges. $\text{Range}(A)=100-1=99,\text{Range}(B)=100-1=99$ Why the same numbers: both sets share the same endpoints ($1$ and $100$), and range reads only the endpoints — so it must return the same value regardless of what sits between. Identical! Answer: both $99$.

The lesson. Range is blind to everything between the extremes. $A$ is tightly bunched at $1$–$4$ with one far outlier; $B$ is genuinely spread. Same range, totally different shapes. This is exactly why we also study Interquartile Range and Standard Deviation, which do look at the middle of the data.

With the error: max $=200$, min $=18$, so $\text{Range}=200-18=182\text{ min}.$ This is dominated by one bad value and misrepresents a normal delivery.

Without the outlier (remove $200$, new max $=21$): $\text{Range}=21-18=3\text{ min}.$ Honest report: "Range is $182$ min, but this is driven entirely by one erroneous value ($200$); excluding it, the range is $3$ min." See Outliers for why we flag rather than silently delete.

Rearrange the formula. Starting from $\text{Range}=x_{\max }-x_{\min }$, add $x_{\min }$ to both sides: $x_{\max }=\text{Range}+x_{\min }=37+42=79$ Why this works: the same equation, read backwards. If the gap is $37$ and the bottom is $42$, the top sits $37$ above it. Answer: $79$.

Fix the endpoints first. Min must be $3$. Since range $=x_{\max }-3=10$, we need $x_{\max }=13$.

Fill the middle freely with any values in $[3,13]$ — they can't change the range. One valid answer: $\{3,5,8,11,13\}$ Check: $13-3=10$. ✓ Answer (one of many): $\{3,5,8,11,13\}$. The middle three are your choice as long as they stay between $3$ and $13$.

The new value $v$ can enlarge the range only by becoming the new max or new min. Figure s02 below draws both moves — the amber dots are the two candidate points that stretch the bar to length $15$.

Case A — new maximum. Keep min $=4$, need $v-4=15\Rightarrow v=19$.

Case B — new minimum. Keep max $=11$, need $11-v=15\Rightarrow v=-4$.

Any value strictly inside $[4,11]$ leaves the range at $7$ — no good. So exactly two answers: $v=19\text{or}v=-4$ Answer: $v=19$ or $v=-4$.

Let X span $[a,a+12]$ and Y span $[b,b+20]$. The merged range is (overall max) $-$ (overall min).

Smallest merged range. Slide X entirely inside Y (choose $b\le a$ and $a+12\le b+20$). Then Y's endpoints are the overall extremes, so the merged range equals Y's range $=20$. It can never drop below $20$, because the merged spread must be at least the larger of the two individual spreads (Y's extremes are both present in the merged set). Smallest $=20$.

Largest merged range. The problem places no limit on where the two classes sit, so we may push them arbitrarily far apart: put all of X far below all of Y with a huge gap between them. The overall min comes from X, the overall max from Y, and the gap can be as large as we like — so the merged range is unbounded (no largest value). If the problem additionally required the two ranges to touch end-to-end (X's top equal to Y's bottom, no gap), then the merged range would be $12+20=32$ — but that extra condition is not given here. Answer: smallest $=20$; no largest (unbounded) unless a "no-gap" condition is imposed, giving $32$.

Setup. Let the original max and min be $M$ and $m$, so ${\text{Range}}_{\text{old}}=M-m\ge 0$.

Adding $c$. Every value shifts by the same amount, so the largest stays largest and the smallest stays smallest: new max $=M+c$, new min $=m+c$. ${\text{Range}}_{\text{new}}=(M+c)-(m+c)=M-m$ The $+c$ terms cancel — the whole dataset slides along the line but its width is untouched. Range is unchanged. ✓

Multiplying by $k>0$. Multiplying by a positive number keeps order (bigger stays bigger), so new max $=kM$, new min $=km$. ${\text{Range}}_{\text{new}}=kM-km=k(M-m)=k\cdot {\text{Range}}_{\text{old}}$ Range scales by $k$. ✓ (This is why converting cm→mm multiplies a range by $10$.)

Multiplying by $k<0$ (the flipped case). A negative multiplier reverses order: the value that was largest, $M$, becomes the most negative ($kM$), and the old min $m$ becomes the largest ($km$). So new max $=km$ and new min $=kM$ — they swap roles. Then ${\text{Range}}_{\text{new}}=km-kM=k(m-M)=-k(M-m)=|k|(M-m).$ Since $k<0$, $-k=|k|>0$, so the result is positive, as a range must be. Range scales by $|k|$. ✓ (E.g. multiplying every value by $-1$ mirrors the data but keeps the same width, so the range is unchanged: $|-1|(M-m)=M-m$.)

Combined rule. For any linear change $x\mapsto kx+c$: shift $c$ does nothing to the range, and scale $k$ multiplies it by $|k|$.

False as stated — it depends on sort direction. Sorted ascending: $1,2,4,9$. "Last minus first" $=9-1=8$ ✓ (this happens to work). Sorted descending: $9,4,2,1$. "Last minus first" $=1-9=-8$ ✗ (negative, wrong).

The safe rule never mentions position; it names the values: $\text{Range}=x_{\max }-x_{\min }=9-1=8.$ Answer: false in general — "last minus first" only works if you promise to sort ascending. Always anchor to max and min, not to array positions.

Upper bound. Since all values sit in $[6,16]$, the max is at most $16$ and the min is at least $6$, so $\text{Range}=x_{\max }-x_{\min }\le 16-6=10.$ No dataset inside this interval can beat $10$.

Can we reach $10$ while keeping mean $=10$? We need a $16$ and a $6$ to be present (to hit both ends), and the five values must sum to $5\times 10=50$ (that is what "mean $=10$" means). Take $6$ and $16$ (their sum is $22$); the remaining three must total $50-22=28$ while staying inside $[6,16]$. One clean choice is $6,16,6$ again: $\{6,16,6,16,6\}.$ Check: sum $=6+16+6+16+6=50$ so mean $=10$ ✓; all values in $[6,16]$ ✓; range $=16-6=10$ ✓. Answer: largest possible range $=10$, achieved e.g. by $\{6,16,6,16,6\}$.