Chapter 13 · Statistics

13.1 Measures of Dispersion: Range and Mean Deviation

The mean tells you where data are centred; a measure of dispersion tells you how widely they scatter. Start with the range, then average the distances $|x_i-a|$ from a central value to get the mean deviation.

A bottling plant fills "$500\ \text{ml}$" bottles. Two machines are tested on eight bottles each, and both give a mean volume of exactly $500\ \text{ml}$. Machine P's bottles measure $498, 503, 500, 497, 502, 501, 499, 500$; machine Q's measure $492, 507, 500, 495, 510, 488, 503, 505$. Q regularly gives customers $10\ \text{ml}$ too little and wastes $10\ \text{ml}$ on others. The mean cannot see this, because it was built to describe the centre of the data; what the quality inspector needs is a number for the scatter around that centre.

Numbers that describe scatter are called measures of dispersion. In this section we meet the two simplest. The range just compares the largest and smallest values. The mean deviation looks at every observation: it measures how far each one is from a central value — the mean or the median — and averages those distances. Along the way we discover why the obvious idea, averaging the deviations themselves, always gives $0$, and why the median turns out to be the best centre to measure distances from.

Dispersion and the Range

The mean, median and mode from Class 10 are measures of central tendency: each gives one value that represents the whole data set. Two data sets can share a centre and still look completely different, as the bottles show. So a full description of data needs two kinds of number — one for the centre, one for the spread.

📘 Definition 13.1 — Dispersion; range

The dispersion (or scatter) of a set of observations is the extent to which they differ from one another or from a central value. A number that describes it is a measure of dispersion; it is $0$ when all observations are equal and grows as they spread out.

The simplest measure of dispersion is the range: $$\text{Range}=\text{largest observation}-\text{smallest observation}.$$ It has the same unit as the data.

The range is quick to find and easy to understand — weather reports quote the day's temperature range, and a teacher may say "marks ranged from $31$ to $96$". But it uses only two observations and ignores everything in between. A single unusual value, an outlier, can make it huge, and two very different data sets can have the same range.

✏️ Example 1 — The range, and what it misses

(a) The maximum daily temperatures in a town over one week were $31, 34, 29, 36, 33, 30, 35$ (in $^\circ\text{C}$). Find the range.
(b) Find the ranges of the data sets $\{10, 50, 50, 50, 50, 50, 90\}$ and $\{10, 20, 30, 50, 70, 80, 90\}$. Do they describe the spread well?
(c) Find the range of the bottle volumes of machines P and Q above.

💡 Show solution▶

(a) Largest $36$, smallest $29$: range $=36-29=7\ ^\circ\text{C}$.

(b) Both ranges are $90-10=80$. Yet in the first set five of the seven values sit exactly on $50$, while in the second they are spread evenly from $10$ to $90$. The range sees only the two end values, so it cannot tell these sets apart.

Lesson The range is a useful first look, but a better measure should use every observation.

Deviations and the Mean Deviation

To use every observation, measure how far each one is from a fixed central value $a$. The number $x_i-a$ is the deviation of $x_i$ from $a$: positive when $x_i$ lies above $a$, negative when it lies below. A natural first idea is to average the deviations from the mean. It fails completely, and the reason is worth knowing.

📐 Theorem 13.1 — Deviations from the mean add up to zero

For any observations $x_1,x_2,\dots,x_n$ with mean $\bar x=\dfrac1n\displaystyle\sum_{i=1}^{n}x_i$, $$\sum_{i=1}^{n}(x_i-\bar x)=0.$$ More generally $\displaystyle\sum_{i=1}^{n}(x_i-a)=n(\bar x-a)$, which is $0$ only when $a=\bar x$.

Proof. Split the sum: $\displaystyle\sum(x_i-a)=\sum x_i-\sum a=n\bar x-na=n(\bar x-a)$, because $\sum x_i=n\bar x$ by the definition of the mean and $a$ is added $n$ times. Putting $a=\bar x$ gives $0$. $\blacksquare$

So the positive deviations always exactly cancel the negative ones — that is what makes the mean the "balance point" of the data. The average signed deviation from the mean is $0$ for every data set, whether it is tightly packed or wildly scattered, so it measures nothing. The cure is to ignore the signs and use the distances $|x_i-a|$ (the modulus of Section 2.3). Section 13.3 uses a second cure, squaring.

📘 Definition 13.2 — Mean deviation about a point

The mean deviation of the observations $x_1,x_2,\dots,x_n$ about a number $a$ is the mean of their absolute deviations from $a$: $$\text{M.D.}(a)=\frac1n\sum_{i=1}^{n}|x_i-a|.$$ It is almost always taken about the mean or the median: $$\text{M.D.}(\bar x)=\frac1n\sum_{i=1}^{n}|x_i-\bar x|,\qquad \text{M.D.}(M)=\frac1n\sum_{i=1}^{n}|x_i-M|.$$ It has the same unit as the data, and it is $0$ exactly when every observation equals $a$.

🧭 Finding a mean deviation (ungrouped data)

Find the centre: the mean $\bar x=\tfrac1n\sum x_i$, or the median $M$ (arrange the data in order first; for odd $n$ it is the middle value, for even $n$ the average of the two middle values).
Write down each deviation $x_i-a$ and drop its sign to get $|x_i-a|$.
Add the absolute deviations and divide by $n$, the number of observations. Give the unit.
Check: before dropping the signs, the deviations from the mean must add up to $0$ (Theorem 13.1).

✏️ Example 2 — Mean deviation about the mean

Find the mean deviation about the mean of $12, 5, 9, 14, 7, 11, 8, 6$.

💡 Show solution▶

Mean $\sum x_i=72$ and $n=8$, so $\bar x=\dfrac{72}{8}=9$.

Table

$x_i$	12	5	9	14	7	11	8	6
$x_i-\bar x$	3	−4	0	5	−2	2	−1	−3
$\|x_i-\bar x\|$	3	4	0	5	2	2	1	3

Answer $\sum|x_i-\bar x|=20$, so $\text{M.D.}(\bar x)=\dfrac{20}{8}=2.5$.

Check The signed deviations add to $3-4+0+5-2+2-1-3=0$ ✓ (Theorem 13.1). On average an observation is $2.5$ away from the mean.

✏️ Example 3 — Mean deviation about the median

Find the mean deviation about the median of
(a) $34, 23, 40, 19, 28, 45, 31, 26, 37$;
(b) $14, 9, 21, 17, 11, 25, 13, 18$.

💡 Show solution▶

(a) Median In order: $19, 23, 26, 28, \mathbf{31}, 34, 37, 40, 45$. With $n=9$ (odd) the median is the $\tfrac{9+1}{2}=5$th value, $M=31$.

(a) Distances $|x_i-31|$: $12, 8, 5, 3, 0, 3, 6, 9, 14$, with sum $60$. So $\text{M.D.}(M)=\dfrac{60}{9}=6.67$ (to $2$ decimal places).

(b) Median In order: $9, 11, 13, \mathbf{14}, \mathbf{17}, 18, 21, 25$. With $n=8$ (even) the median is the mean of the $4$th and $5$th values: $M=\dfrac{14+17}{2}=15.5$.

(b) Distances $|x_i-15.5|$: $6.5, 4.5, 2.5, 1.5, 1.5, 2.5, 5.5, 9.5$, with sum $34$. So $\text{M.D.}(M)=\dfrac{34}{8}=4.25$.

✏️ Example 4 — Which machine is more consistent?

Find the mean deviation about the mean for the volumes (in ml) filled by machines P and Q at the start of this section, and decide which machine is more consistent.

💡 Show solution▶

Means Both totals are $4000$, so both means are $\dfrac{4000}{8}=500\ \text{ml}$.

P $|x_i-500|$: $2, 3, 0, 3, 2, 1, 1, 0$; sum $12$; $\text{M.D.}(\bar x)=\dfrac{12}{8}=1.5\ \text{ml}$.

Q $|x_i-500|$: $8, 7, 0, 5, 10, 12, 3, 5$; sum $50$; $\text{M.D.}(\bar x)=\dfrac{50}{8}=6.25\ \text{ml}$.

Conclusion A typical bottle from Q is about four times as far from $500\ \text{ml}$ as one from P. Machine P is more consistent — the same verdict as the ranges ($6$ ml against $22$ ml), but now based on all sixteen bottles.

Each dot is one observation, on its own row so that you can see its deviation as a bar from the chosen centre (red if the dot is below the centre, blue if above). Drag any dot sideways. The readouts recompute the mean, median, range and mean deviation at once. Notice the last readout: the signed deviations from the mean always add to $0$, however you move the dots. Press Play to stretch and squeeze the whole data set about its mean.

measure from

data—

mean $\bar x$ · median $M$—

range—

M.D. about the centre—

$\sum(x_i-\bar x)$—

Why the Median Is the Best Centre for Distances

In Example 3 we measured distances from the median, in Example 2 from the mean. Does the choice matter? Think of the observations as houses along a straight road and $a$ as the position of a bus stop; $n\cdot\text{M.D.}(a)=\sum|x_i-a|$ is the total distance everyone walks to the stop. Where should the stop go?

📐 Theorem 13.2 — The mean deviation is least about the median

Let $M$ be the median of $x_1,x_2,\dots,x_n$. For every real number $a$, $$\text{M.D.}(a)\ \ge\ \text{M.D.}(M).$$ When $n$ is odd, equality holds only for $a=M$. When $n$ is even, every $a$ between the two middle observations (inclusive) gives the same least value. In particular $\text{M.D.}(M)\le\text{M.D.}(\bar x)$.

Proof. Arrange the data in increasing order and look at $S(a)=\sum|x_i-a|$. Suppose $a$ moves a small distance $h\gt0$ to the right without passing any observation. Every observation to the left of $a$ is now $h$ farther away, and every observation to the right is $h$ nearer. If $L$ observations lie to the left of $a$ and $R$ to the right, $S$ changes by $h(L-R)$. So $S$ decreases as $a$ moves right while $L\lt R$ (fewer than half the observations are to the left), and increases once $L\gt R$. The switch happens where half the observations lie on each side — at the median. For odd $n$ that is the single middle value; for even $n$, $L=R=\tfrac n2$ all the way between the two middle values, so $S$ is constant (and least) there. Dividing by $n$ gives the result for $\text{M.D.}(a)$. $\blacksquare$

The mean minimises a different quantity — the sum of squared deviations — which is why it belongs with the variance of Section 13.3. When the data are roughly symmetric the mean and median are close and the two mean deviations nearly agree; an outlier pulls the mean towards it and makes $\text{M.D.}(\bar x)$ noticeably larger than $\text{M.D.}(M)$.

✏️ Example 5 — An unknown observation, then both mean deviations

The mean of $4, 7, x, 10, 14$ is $9$. Find $x$, and then the mean deviation about the mean and about the median. Which is smaller?

💡 Show solution▶

Find $x$ $\dfrac{4+7+x+10+14}{5}=9$ gives $35+x=45$, so $x=10$. The data are $4, 7, 10, 10, 14$.

About the mean $|x_i-9|$: $5, 2, 1, 1, 5$; sum $14$; $\text{M.D.}(\bar x)=\dfrac{14}{5}=2.8$.

About the median In order the middle value is $M=10$. $|x_i-10|$: $6, 3, 0, 0, 4$; sum $13$; $\text{M.D.}(M)=\dfrac{13}{5}=2.6$.

Check $2.6\lt2.8$, as Theorem 13.2 promises.

✏️ Example 6 — The bus-stop problem

Six friends live along a straight road at the points $1, 3, 4, 6, 9, 11$ (in km from a junction). (a) Where should they meet so that the total distance travelled is as small as possible? (b) What is the least total distance? (c) How much farther do they travel in total if they meet at the mean position?

💡 Show solution▶

(a) $n=6$ is even; the two middle values are $4$ and $6$. By Theorem 13.2 any meeting point $a$ with $4\le a\le6$ gives the least total; the median $M=5$ is one of them.

(b) At $a=5$: $4+2+1+1+4+6=18\ \text{km}$. (At $a=4$: $3+1+0+2+5+7=18$ km too.) So $\text{M.D.}(M)=\tfrac{18}{6}=3\ \text{km}$.

(c) The mean is $\bar x=\tfrac{34}{6}=\tfrac{17}{3}\approx5.67$, which also lies between $4$ and $6$ — so the total is again $18$ km, and they travel no farther. (Three friends are to the left and three to the right of any point in $[4,6]$, so moving the meeting point there changes nothing.)

Lesson For even $n$ the least value is shared by a whole interval. If the friend at $11$ moved to $40$, the mean would jump to $\tfrac{63}{6}=10.5$ while the median stays $5$ — and meeting at the mean would then cost far more.

The curve shows $\text{M.D.}(a)$ for every centre $a$; the data are the small triangles on the axis. Move $a$ (or press Play to sweep it). The curve is made of straight pieces that bend only at data points: it slopes down while more observations lie to the right of $a$ than to the left, and up once more lie to the left. The bottom is at the median (green) — a single point for odd $n$, a flat stretch for even $n$. The mean (purple) is marked for comparison.

data$a$9

left · right of $a$—

$\text{M.D.}(a)$—

$\text{M.D.}(M)$—

$\text{M.D.}(\bar x)$—

⚠️ Common mistakes with the range and mean deviation

1. Forgetting the modulus. Without $|\ \ |$ the deviations from the mean add to $0$ and you will "find" a mean deviation of $0$ for any data.

2. Finding the median without sorting. The median is the middle of the ordered list. For $14, 9, 21, 17$ it is $\tfrac{14+17}{2}=15.5$, not $\tfrac{9+21}{2}$.

3. Dividing by the wrong number. Divide the sum of the absolute deviations by $n$, the number of observations — not by the number of distinct values, and not by $n-1$.

4. Range as "smallest to largest". The range is one number, the difference $\text{max}-\text{min}$; "from $29$ to $36$" is the spread written out, and its range is $7$.

5. Mixing the centres. If the question says "about the median", measure distances from $M$, even when you have already worked out $\bar x$.

Exercises

Range and deviations.

1 Find the range of $23, 17, 35, 41, 29, 12, 38$.
2 The minimum temperatures (in $^\circ\text{C}$) in a hill town over a week were $-2, 3, 5, 1, -4, 0, 2$. Find the range.
3 Find the deviations of $3, 8, 6, 11, 7$ from their mean, and check that they add to $0$.
4 The range of a data set is $18$ and its smallest value is $27$. Find its largest value.
5 Show that $\sum_{i=1}^{n}(x_i-a)=0$ if and only if $a=\bar x$.
6 The deviations of five observations from $20$ are $-3, 5, 0, -6$ and $k$. If the mean of the observations is $21$, find $k$ and the five observations.

💡 Show solutions▶

1 $41-12=29$.

2 Largest $5$, smallest $-4$: range $=5-(-4)=9\ ^\circ\text{C}$.

3 $\bar x=\tfrac{35}{5}=7$; deviations $-4, 1, -1, 4, 0$; sum $0$ ✓.

4 $27+18=45$.

5 $\sum(x_i-a)=n\bar x-na=n(\bar x-a)$, which is $0$ exactly when $a=\bar x$ (as $n\ne0$).

6 $\sum(x_i-20)=n(\bar x-20)=5(21-20)=5$, so $-3+5+0-6+k=5$ and $k=9$. Observations: $17, 25, 20, 14, 29$ (mean $\tfrac{105}{5}=21$ ✓).

Mean deviation about the mean. (Give answers to $2$ decimal places where needed.)

1 $5, 8, 9, 11, 12, 14, 15, 22$
2 $36, 72, 45, 41, 47, 58, 61, 49, 52, 39$
3 the first eight even natural numbers
4 $2.5, 3.1, 1.8, 4.2, 3.4$
5 $-5, -2, 0, 3, 4$
6 $a, a+d, a+2d, a+3d, a+4d$, where $d\gt0$
7 the ages (in years) of six people in a family: $8, 12, 38, 41, 65, 70$

Check: mean deviation of your own list

dataabout

$n$—

centre—

$\sum|x_i-a|$—

mean deviation—

💡 Show solutions▶

1 $\bar x=\tfrac{96}{8}=12$; $|x_i-12|$: $7,4,3,1,0,2,3,10$; sum $30$; M.D. $=\tfrac{30}{8}=3.75$.

2 $\bar x=\tfrac{500}{10}=50$; $|x_i-50|$: $14,22,5,9,3,8,11,1,2,11$; sum $86$; M.D. $=8.6$.

3 $2,4,\dots,16$: $\bar x=\tfrac{72}{8}=9$; distances $7,5,3,1,1,3,5,7$; sum $32$; M.D. $=4$.

4 $\bar x=\tfrac{15}{5}=3$; distances $0.5,0.1,1.2,1.2,0.4$; sum $3.4$; M.D. $=0.68$.

5 $\bar x=0$; distances $5,2,0,3,4$; sum $14$; M.D. $=2.8$.

6 $\bar x=a+2d$; distances $2d,d,0,d,2d$; sum $6d$; M.D. $=\tfrac{6d}{5}$ — it does not depend on $a$.

7 $\bar x=\tfrac{234}{6}=39$; distances $31,27,1,2,26,31$; sum $118$; M.D. $=\tfrac{118}{6}=19.67$ years.

III

Mean deviation about the median.

1 $21, 15, 18, 24, 30, 19, 27$
2 $22, 35, 28, 41, 30, 26, 33, 38$
3 $3, 5, 5, 6, 8, 9, 12, 12, 12, 15, 20$
4 For $2, 4, 6, 8, 30$ find the mean deviation about the mean and about the median. Which is smaller, and why?
5 $1.2, 0.8, 1.5, 2.1, 0.9, 1.1$
6 the heights (in cm) of nine seedlings: $42, 51, 38, 47, 55, 40, 49, 44, 53$
7 The observations $5, 9, 11, x, 17$ are in increasing order and their mean deviation about the median is $3.6$. Find $x$.

💡 Show solutions▶

1 Ordered: $15,18,19,21,24,27,30$; $M=21$; distances $6,3,2,0,3,6,9$; sum $29$; M.D. $=\tfrac{29}{7}=4.14$.

2 Ordered: $22,26,28,30,33,35,38,41$; $M=\tfrac{30+33}{2}=31.5$; distances $9.5,5.5,3.5,1.5,1.5,3.5,6.5,9.5$; sum $41$; M.D. $=5.125$.

3 $n=11$, $M=$ the $6$th value $=9$; distances $6,4,4,3,1,0,3,3,3,6,11$; sum $44$; M.D. $=4$.

4 $\bar x=10$: distances $8,6,4,2,20$, M.D.$(\bar x)=\tfrac{40}{5}=8$. $M=6$: distances $4,2,0,2,24$, M.D.$(M)=\tfrac{32}{5}=6.4$. The median gives the smaller value (Theorem 13.2); the outlier $30$ drags the mean up, away from the other four values.

5 Ordered: $0.8,0.9,1.1,1.2,1.5,2.1$; $M=1.15$; distances $0.35,0.25,0.05,0.05,0.35,0.95$; sum $2$; M.D. $=\tfrac26=0.33$.

6 Ordered: $38,40,42,44,47,49,51,53,55$; $M=47$; distances $9,7,5,3,0,2,4,6,8$; sum $44$; M.D. $=\tfrac{44}{9}=4.89\ \text{cm}$.

7 $M=11$ (the middle value). Since $11\le x\le17$, $|x-11|=x-11$, so $6+2+0+(x-11)+6=5\times3.6=18$, giving $x+3=18$ and $x=15$. Check: distances $6,2,0,4,6$, sum $18$, and $\tfrac{18}{5}=3.6$ ✓.

Applications and reasoning.

1 The July rainfall (in mm) over six years was $210, 250, 230, 240, 220, 230$ in town A and $120, 340, 180, 300, 150, 290$ in town B. Find the mean, the range and the mean deviation about the mean for each town. Which town's July rainfall is more dependable?
2 A student's journey times to school (in minutes) on five days were $25, 32, 28, 45, 30$. Find the mean deviation about the median and about the mean.
3 Find the mean deviation about the mean of $3, 7, 11, 15$. Each observation is then doubled and $5$ is added. Find the new mean deviation directly, and explain the result.
4 Without recalculating, say how the mean deviation about the mean of a data set changes if (a) $7$ is added to every observation; (b) every observation is multiplied by $-2$. Justify your answers.
5 The monthly incomes (in thousands of rupees) of seven workers in a small office are $18, 20, 21, 22, 23, 25, 300$. Find the range with and without the last value. Why is the range a poor measure of spread here?
6 Find the mean deviation about the mean of the first $n$ natural numbers for $n=5$ and for $n=6$.
7 Seven students' times (in minutes) to finish a puzzle were recorded as $4, 6, 6, 7, 9, 10, 21$. (a) Find the mean deviation about the median. (b) It was then found that $21$ had been copied wrongly for $12$. Find the correct mean deviation about the median.

💡 Show solutions▶

1 Both totals are $1380$, so both means are $230$ mm. A: range $250-210=40$ mm; distances $20,20,0,10,10,0$; M.D. $=\tfrac{60}{6}=10$ mm. B: range $340-120=220$ mm; distances $110,110,50,70,80,60$; M.D. $=\tfrac{480}{6}=80$ mm. Town A's rainfall is far more dependable.

2 Ordered $25,28,30,32,45$, $M=30$: distances $5,2,0,2,15$, M.D.$(M)=\tfrac{24}{5}=4.8$ min. $\bar x=\tfrac{160}{5}=32$: distances $7,0,4,13,2$, M.D.$(\bar x)=\tfrac{26}{5}=5.2$ min.

3 $\bar x=9$; distances $6,2,2,6$; M.D. $=4$. New data $11,19,27,35$ with mean $23$; distances $12,4,4,12$; M.D. $=8$. Adding $5$ moves every value and the mean together, so no distance changes; doubling doubles every distance.

5 With $300$: $300-18=282$ thousand rupees; without it: $25-18=7$ thousand rupees. One unusual income (an outlier) decides the range on its own, although six of the seven incomes lie within ₹7000 of each other.

6 $n=5$: $\bar x=3$, distances $2,1,0,1,2$, M.D. $=\tfrac65=1.2$. $n=6$: $\bar x=3.5$, distances $2.5,1.5,0.5,0.5,1.5,2.5$, M.D. $=\tfrac96=1.5$.

7 (a) $M=7$ (the $4$th value); distances $3,1,1,0,2,3,14$; M.D. $=\tfrac{24}{7}=3.43$ min. (b) The data $4,6,6,7,9,10,12$ still have $M=7$; distances $3,1,1,0,2,3,5$; M.D. $=\tfrac{15}{7}=2.14$ min. (The median did not change, but the mean deviation did.)

🎯 Exam corner Board · 1 markBoard · 3 marksBoard · Case studyJEE Main

How this is examined: Board papers ask for a mean deviation about the mean or the median of a short list (2–3 marks, usually set out as a table), MCQs and assertion–reason items on $\sum(x_i-\bar x)=0$, and case studies comparing the consistency of two data sets. JEE Main likes data in arithmetic progression and the "least mean deviation" property of the median.

E1. Board · 1 mark The mean deviation about the mean of $4, 8, 12, 16, 20$ is (A) $4.8$ (B) $12$ (C) $0$ (D) $24$

💡 Show solution▶

Idea $\bar x=\tfrac{60}{5}=12$; distances $8,4,0,4,8$; sum $24$; M.D. $=\tfrac{24}{5}=4.8$.

Trap (C) is the sum of the signed deviations (always $0$); (D) forgets to divide by $n$; (B) is the mean itself.

Answer (A).

E2. Board · 1 mark Assertion (A): For every data set, $\sum(x_i-\bar x)=0$. Reason (R): The mean deviation is defined using the absolute values $|x_i-\bar x|$. Choose: (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

💡 Show solution▶

A True: $\sum(x_i-\bar x)=n\bar x-n\bar x=0$ (Theorem 13.1).

R True — that is the definition. But R does not explain A; it is the other way round: because the signed deviations always cancel, the definition has to use absolute values.

Answer (B).

E3. Board · 3 marks Find the mean deviation about the median of $11, 3, 8, 7, 5, 9, 14, 6$.

💡 Show solution▶

Order $3, 5, 6, 7, 8, 9, 11, 14$; $n=8$, so $M=\dfrac{7+8}{2}=7.5$. (1 mark)

Distances $|x_i-7.5|$: $4.5, 2.5, 1.5, 0.5, 0.5, 1.5, 3.5, 6.5$; sum $21$. (1 mark)

Answer $\text{M.D.}(M)=\dfrac{21}{8}=2.625$. (1 mark)

E4. Board · Case study A health department records the waiting times (in minutes) of six patients at two clinics. Clinic X: $12, 15, 18, 15, 20, 10$. Clinic Y: $5, 30, 8, 25, 12, 10$. (i) Show that the mean waiting times are equal. (1 mark) (ii) Find the range for each clinic. (1 mark) (iii) Find the mean deviation about the mean for each clinic and say which clinic is more consistent. (2 marks)

💡 Show solution▶

(i) Both totals are $90$, so both means are $\tfrac{90}{6}=15$ min. (1 mark)

(ii) X: $20-10=10$ min; Y: $30-5=25$ min. (1 mark)

(iii) X Distances $3,0,3,0,5,5$; sum $16$; M.D. $=\tfrac{16}{6}=2.67$ min. (1 mark)

(iii) Y Distances $10,15,7,10,3,5$; sum $50$; M.D. $=\tfrac{50}{6}=8.33$ min. Clinic X is more consistent. (1 mark)

E5. JEE Main For the data $2, 3, 7, 8, 10, 15$, the set of all real $a$ for which $\displaystyle\sum_{i=1}^{6}|x_i-a|$ is least is (A) $\{7.5\}$ (B) $[7,8]$ (C) $\{7\}$ (D) $[3,10]$

💡 Show solution▶

Idea By Theorem 13.2 the sum is least about the median. With $n=6$ (even), it is least — and constant — for every $a$ between the two middle values $7$ and $8$.

Check At $a=7$: $5+4+0+1+3+8=21$; at $a=8$: $6+5+1+0+2+7=21$; at $a=7.5$: $5.5+4.5+0.5+0.5+2.5+7.5=21$; at $a=6$: $4+3+1+2+4+9=23$.

Trap (A) is the median and the mean ($\tfrac{45}{6}=7.5$), but it is not the only minimiser; (D) is too wide — between $3$ and $7$ more observations lie to the right, so the sum still falls as $a$ increases.

Answer (B).

E6. JEE Main Numerical answer. Find the mean deviation about the mean of the $20$ numbers $4, 8, 12, \dots, 80$.

💡 Show solution▶

Mean The numbers are $4k$ for $k=1,\dots,20$; $\bar x=4\times\tfrac{21}{2}=42$.

Distances $|4k-42|=4|k-10.5|$. For $k=1,\dots,20$ the values $|k-10.5|$ are $9.5, 8.5,\dots,0.5$ twice, with sum $2(0.5+1.5+\cdots+9.5)=2\times50=100$.

Answer $\text{M.D.}(\bar x)=\dfrac{4\times100}{20}=20$.

⏱ Fast route For $1,2,\dots,n$ with $n$ even, M.D. $=\tfrac n4$; here $\tfrac{20}{4}=5$, and multiplying the data by $4$ multiplies the mean deviation by $4$: $20$.

🔬 Research–Teaching NexusMini project · about 40 minutes

Research question. Is your reaction time more consistent with your writing hand than with your other hand — and does the answer depend on which measure of spread you use?

Why it matters. Sprint starters, goalkeepers and drivers all depend on fast and reliable reactions. Sports scientists and road-safety researchers report not only an average reaction time but also its variability, because an occasional very slow reaction can matter more than a slightly slower average.

What to do.

Work in pairs. Your partner holds a $30\ \text{cm}$ ruler vertically with its $0$ mark level with your open thumb and finger, then lets go without warning; you catch it and read the distance $d$ (in cm) at your fingers.
Do $10$ catches with each hand, recording every value of $d$ (a smaller $d$ means a faster reaction).
For each hand find the mean, the median, the range, $\text{M.D.}(\bar x)$ and $\text{M.D.}(M)$. Use the check-lab above to confirm your arithmetic.
Compare the hands. Do the range and the mean deviation agree about which hand is more consistent? Does one bad catch change the range much more than the mean deviation?

Tools: A $30\ \text{cm}$ ruler, a partner, a notebook or spreadsheet

Hand in: Your two data lists, a dot plot for each hand on the same scale, a table of the five statistics, and a short paragraph stating which hand is more consistent and which measure you trust most, with a reason.

Going further. A ruler falling freely from rest drops $d=\tfrac12gt^2$, so the reaction time is $t=\sqrt{2d/g}$ with $g\approx9.8\ \text{m/s}^2$ (use $d$ in metres). Convert your catches to times. Is the mean deviation of the times the same multiple of the mean deviation of the distances for every catch? Why not?

Instructor note: typical catch distances are $10$–$20\ \text{cm}$ (reaction times of about $0.14$–$0.20\ \text{s}$). Expect the range to be dominated by one or two "late" catches, which makes a good discussion of outliers and of why $\text{M.D.}(M)\le\text{M.D.}(\bar x)$. In "Going further", the conversion $t=\sqrt{2d/g}$ is not linear, so mean deviations do not simply scale — unlike the linear changes in Exercise IV.3–4.

🧪 Self-check quiz

0/5 correct

Q1. The range of $7, 3, 12, 9, 5$ is

Q2. For any data set, the sum of the deviations from the mean, $\sum(x_i-\bar x)$, is

Q3. The mean deviation about the mean of $2, 4, 6, 8, 10$ is

Q4. The mean deviation about the median of $8, 3, 6, 10, 1, 7$ is

Q5. A data set has mean deviation about the mean equal to $2$. Every observation is multiplied by $3$ and then $4$ is added. The new mean deviation about the mean is

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