NCERT Exemplar Solutions Class 10 Maths Statistics and Probability — Exercise 13.3 Q1–18

NCERT Exemplar Problems, Class 10 Mathematics — Statistics and Probability, Exercise 13.3, Questions 1 to 18. All the grouped-data questions solved.

Exercise 13.3 runs to forty-two questions. The first eighteen are grouped data and are solved here; Questions 19 to 42 are probability and are on the second part.

Three formulae carry all eighteen:

$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}, \qquad \text{Median} = l + \left(\frac{\frac n2 – cf}{f}\right)h, \qquad \text{Mode} = l + \left(\frac{f_1 – f_0}{2f_1 – f_0 – f_2}\right)h$$

Two habits decide most of the marks.

Get the class marks right. For a class $a$–$b$ the class mark is $\frac{a+b}{2}$, and that holds whether the classes are continuous ($10$–$20$) or inclusive ($4$–$7$). Questions 3, 4 and 5 are all in inclusive form, and their class marks are $5.5$, $17$ and $100.5$ — half-integers, which look wrong and are not.

Build the cumulative column before looking for a median. Questions 15 and 16 need it, and Questions 9 to 14 are about nothing else: converting between a frequency table, a “less than” table and a “more than” table.

One printed answer here is wrong. Question 13’s “more than or equal to $60$” entry is given as $60$; it should be $20$, and the key’s own “less than” column proves it. The full argument is on that question.

A note on the current syllabus. Chapter 13 keeps mean, mode and median. Cumulative frequency survives too — it is point 3 of the Summary — so Questions 9 to 14 remain on syllabus. What has gone is the section on drawing cumulative frequency curves, which affects none of these eighteen.

Key insight. For an inclusive class like $4$–$7$, converting to the continuous form $3.5$–$7.5$ moves the boundaries but leaves the class mark at $5.5$. So convert when you need a median or a mode, where the boundaries enter the formula — and don’t bother when you only need a mean, where they do not.

Question 1

Find the mean of the distribution:

Class	1–3	3–5	5–7	7–10
Frequency	9	22	27	17

Solution. The classes are not of equal width — the first three span $2$ and the last spans $3$. That rules out the step-deviation method, which assumes a common $h$, so use the direct method.

Class	$f_i$	Class mark $x_i$	$f_i x_i$
1–3	9	2	18
3–5	22	4	88
5–7	27	6	162
7–10	17	8.5	144.5
Total	75		412.5

$$\bar{x} = \frac{412.5}{75} = 5.5$$

The last class mark is $\frac{7+10}{2} = 8.5$, not $9$ — a class mark is always the midpoint of that particular class, and unequal widths make each one its own calculation.

Unequal classes affect the mode and median badly (both formulae use $h$), but the mean is untroubled: it only ever needs $x_i$ and $f_i$.

Question 2

Calculate the mean of the scores of $20$ students in a mathematics test:

Marks	10–20	20–30	30–40	40–50	50–60
Number of students	2	4	7	6	1

Solution. Equal widths of $10$ here, so any method works. Direct:

Class	$f_i$	$x_i$	$f_i x_i$
10–20	2	15	30
20–30	4	25	100
30–40	7	35	245
40–50	6	45	270
50–60	1	55	55
Total	20		700

$$\bar{x} = \frac{700}{20} = 35$$

By step deviation, as a check. Take $a = 35$ and $h = 10$, so $u_i = -2, -1, 0, 1, 2$:

$$\sum f_i u_i = 2(-2) + 4(-1) + 7(0) + 6(1) + 1(2) = -4 – 4 + 0 + 6 + 2 = 0$$

$$\bar{x} = 35 + 10\left(\frac{0}{20}\right) = 35$$

The $\sum f_i u_i$ coming out as zero says the data balances exactly at $35$ — which is what Question 3 of Exercise 13.1 was about.

Question 3

Calculate the mean of the following data:

Class	4–7	8–11	12–15	16–19
Frequency	5	4	9	10

Solution. These classes are inclusive — there is a gap between $7$ and $8$. For a mean that does not matter, because the class mark of $4$–$7$ is

$$\frac{4 + 7}{2} = 5.5$$

whether or not you convert to the continuous form $3.5$–$7.5$ (whose midpoint is also $5.5$).

Class	$f_i$	$x_i$	$f_i x_i$
4–7	5	5.5	27.5
8–11	4	9.5	38
12–15	9	13.5	121.5
16–19	10	17.5	175
Total	28		362

$$\bar{x} = \frac{362}{28} = 12.928\ldots \approx 12.93$$

The half-integer class marks look suspicious the first time you meet them, and they are correct. The mistake to avoid is taking the midpoint between classes — $\frac{4+8}{2} = 6$ — which uses the next class’s lower limit instead of this class’s upper one.

Question 4

The following table gives the number of pages written by Sarika for completing her own book for $30$ days:

Number of pages written per day	16–18	19–21	22–24	25–27	28–30
Number of days	1	3	4	9	13

Find the mean number of pages written per day.

Solution. Inclusive classes again, and here the class marks come out whole: $\frac{16+18}{2} = 17$, and so on.

Class	$f_i$	$x_i$	$f_i x_i$
16–18	1	17	17
19–21	3	20	60
22–24	4	23	92
25–27	9	26	234
28–30	13	29	377
Total	30		780

$$\bar{x} = \frac{780}{30} = 26$$

A check on the size. The frequencies rise steadily towards the last class, so the mean should sit in the upper half of the range $16$ to $30$. It does — $26$ is well up, pulled there by the $13$ days in the $28$–$30$ class.

Note that the class marks here are $17, 20, 23, 26, 29$, three apart, so the step-deviation method with $a = 26$, $h = 3$ would also work and give $u_i = -3, -2, -1, 0, 1$.

Question 5

The daily income of a sample of $50$ employees are tabulated as follows:

Income (in ₹)	1–200	201–400	401–600	601–800
Number of employees	14	15	14	7

Find the mean daily income of employees.

Solution. Inclusive classes with a gap of $1$, so the class marks are half-integers:

$$\frac{1 + 200}{2} = 100.5, \qquad \frac{201+400}{2} = 300.5, \qquad 500.5, \qquad 700.5$$

Class	$f_i$	$x_i$	$f_i x_i$
1–200	14	100.5	1407
201–400	15	300.5	4507.5
401–600	14	500.5	7007
601–800	7	700.5	4903.5
Total	50		17825

$$\bar{x} = \frac{17825}{50} = 356.5$$

So the mean daily income is ₹356.50.

Taking the class marks as $100$, $300$, $500$, $700$ — ignoring the half — would give $356$, only fifty paise out here but wrong in principle. Each class genuinely spans $200$ values ($1$ to $200$ inclusive), and its centre is $100.5$.

Question 6

An aircraft has $120$ passenger seats. The number of seats occupied during $100$ flights is given in the following table:

Number of seats	100–104	104–108	108–112	112–116	116–120
Frequency	15	20	32	18	15

Determine the mean number of seats occupied over the flights.

Solution. Continuous classes of width $4$.

Class	$f_i$	$x_i$	$f_i x_i$
100–104	15	102	1530
104–108	20	106	2120
108–112	32	110	3520
112–116	18	114	2052
116–120	15	118	1770
Total	100		10992

$$\bar{x} = \frac{10992}{100} = 109.92$$

So on average about $110$ seats were occupied.

In context: the aircraft has $120$ seats, so it flew at roughly $92\%$ capacity. That is a sensible load factor, and the kind of check worth making — an answer above $120$ or below $100$ would contradict the data’s own range.

Seats come in whole numbers, so $109.92$ cannot be an actual observation. A mean need not be attainable; it is a balance point, not a value from the data.

Question 7

The weights (in kg) of $50$ wrestlers are recorded in the following table:

Weight (in kg)	100–110	110–120	120–130	130–140	140–150
Number of wrestlers	4	14	21	8	3

Find the mean weight of the wrestlers.

Solution.

Class	$f_i$	$x_i$	$f_i x_i$
100–110	4	105	420
110–120	14	115	1610
120–130	21	125	2625
130–140	8	135	1080
140–150	3	145	435
Total	50		6170

$$\bar{x} = \frac{6170}{50} = 123.4 \text{ kg}$$

By step deviation, which is quicker here. Take $a = 125$ (the middle class mark) and $h = 10$, giving $u_i = -2, -1, 0, 1, 2$:

$$\sum f_i u_i = 4(-2) + 14(-1) + 21(0) + 8(1) + 3(2) = -8 – 14 + 0 + 8 + 6 = -8$$

$$\bar{x} = 125 + 10\left(\frac{-8}{50}\right) = 125 – 1.6 = 123.4 \text{ kg}$$

The negative $\sum f_i u_i$ says the data leans below $125$, which matches the table — more wrestlers sit in the two lighter classes than the two heavier ones.

Question 8

The mileage (km per litre) of $50$ cars of the same model was tested by a manufacturer and details are tabulated as given below:

Mileage (km/$l$)	10–12	12–14	14–16	16–18
Number of cars	7	12	18	13

Find the mean mileage. The manufacturer claimed that the mileage of the model was $16$ km/litre. Do you agree with this claim?

Solution.

Class	$f_i$	$x_i$	$f_i x_i$
10–12	7	11	77
12–14	12	13	156
14–16	18	15	270
16–18	13	17	221
Total	50		724

$$\bar{x} = \frac{724}{50} = 14.48 \text{ km/litre}$$

The claim does not hold. The manufacturer says $16$ km/litre; the tested average is $14.48$, so the claim overstates the mileage by

$$16 – 14.48 = 1.52 \text{ km/litre}$$

— about $10.5\%$ too high.

Why the claim is not simply a lie. Thirteen of the fifty cars did fall in the $16$–$18$ class, so $16$ km/litre is achievable. But it is close to the top of the range, not the middle. A figure quoted as “the mileage of the model” should be the average performance, and only $26\%$ of the cars reached it.

That distinction — between what is attainable and what is typical — is the point of asking the question.

Question 9

The following is the distribution of weights (in kg) of $40$ persons:

Weight (in kg)	40–45	45–50	50–55	55–60	60–65	65–70	70–75	75–80
Number of persons	4	4	13	5	6	5	2	1

Construct a cumulative frequency distribution (of the less than type) table for the data above.

Solution. A “less than” cumulative frequency is a running total: each entry counts everyone in that class and all the classes before it.

Weight (in kg)	Number of persons
Less than 45	4
Less than 50	$4 + 4 = 8$
Less than 55	$8 + 13 = 21$
Less than 60	$21 + 5 = 26$
Less than 65	$26 + 6 = 32$
Less than 70	$32 + 5 = 37$
Less than 75	$37 + 2 = 39$
Less than 80	$39 + 1 = 40$

Two checks that catch every slip.

The entries must never decrease — a running total of non-negative frequencies can only rise or stay level. And the last entry must equal the total, here $40$, because by the end everyone has been counted.

Note that the labels use the upper limit of each class. “Less than $45$” covers the class $40$–$45$, since every weight in it is below $45$. Starting the table at “less than $40$” with a value of $0$ would also be correct and is sometimes printed.

Question 10

The following table shows the cumulative frequency distribution of marks of $800$ students in an examination:

Marks	Number of students
Below 10	10
Below 20	50
Below 30	130
Below 40	270
Below 50	440
Below 60	570
Below 70	670
Below 80	740
Below 90	780
Below 100	800

Construct a frequency distribution table for the data above.

Solution. This is Question 9 run backwards: difference consecutive entries to recover the individual class frequencies.

Class	Working	Frequency
0–10	$10$	10
10–20	$50 – 10$	40
20–30	$130 – 50$	80
30–40	$270 – 130$	140
40–50	$440 – 270$	170
50–60	$570 – 440$	130
60–70	$670 – 570$	100
70–80	$740 – 670$	70
80–90	$780 – 740$	40
90–100	$800 – 780$	20
Total		800

The check is the total. The frequencies must sum to the last cumulative entry, $800$, and they do. Any single subtraction slip would show up there.

The shape is worth reading: the frequencies rise to a peak of $170$ in $40$–$50$ and fall away either side, so the distribution is single-peaked and roughly symmetric. That makes $40$–$50$ the modal class, and with $\frac n2 = 400$ falling inside it (the cumulative reaches $440$ there against $270$ before), it is the median class too.

Question 11

Form the frequency distribution table from the following data:

Marks (out of 90)	Number of candidates
More than or equal to 80	4
More than or equal to 70	6
More than or equal to 60	11
More than or equal to 50	17
More than or equal to 40	23
More than or equal to 30	27
More than or equal to 20	30
More than or equal to 10	32
More than or equal to 0	34

Solution. A “more than” table decreases as the mark rises, so the differencing runs the opposite way to Question 10: subtract the entry below from the entry above it in the table as printed.

The total is $34$ — everyone scores at least $0$.

Class	Working	Frequency
0–10	$34 – 32$	2
10–20	$32 – 30$	2
20–30	$30 – 27$	3
30–40	$27 – 23$	4
40–50	$23 – 17$	6
50–60	$17 – 11$	6
60–70	$11 – 6$	5
70–80	$6 – 4$	2
80–90	$4$	4
Total		34

The last class takes the final entry itself: nobody scores above $90$, so “at least $80$” and “in $80$–$90$” are the same $4$ candidates.

The check: the frequencies sum to $34$ ✓.

Getting the direction right. A frequency must be positive, so if a subtraction gives a negative number you are differencing the wrong way. In a “less than” table subtract the previous row; in a “more than” table subtract the next one.

Question 12

Find the unknown entries $a$, $b$, $c$, $d$, $e$, $f$ in the following distribution of heights of students in a class:

Height (in cm)	Frequency	Cumulative frequency
150–155	12	$a$
155–160	$b$	25
160–165	10	$c$
165–170	$d$	43
170–175	$e$	48
175–180	2	$f$
Total	50

Solution. Work down the table, using the two relations that link the columns:

$$cf_n = cf_{n-1} + f_n \qquad \text{and} \qquad f_n = cf_n – cf_{n-1}$$

$a$ — the first cumulative entry is just the first frequency:

$$a = 12$$

$b$ — from $cf = 25$ after the second class:

$$b = 25 – 12 = 13$$

$c$ — add the third frequency to the running total:

$$c = 25 + 10 = 35$$

$d$ — from $cf = 43$:

$$d = 43 – 35 = 8$$

$e$ — from $cf = 48$:

$$e = 48 – 43 = 5$$

$f$ — add the last frequency:

$$f = 48 + 2 = 50$$

Two independent checks. The final cumulative frequency must equal the total, and $f = 50$ ✓. And the frequencies must sum to $50$:

$$12 + 13 + 10 + 8 + 5 + 2 = 50 \ \checkmark$$

Both hold, so every entry is confirmed. Filling in the table row by row and checking at the end is safer than trying to solve for the letters simultaneously.

Question 13

The following are the ages of $300$ patients getting medical treatment in a hospital on a particular day:

Age (in years)	10–20	20–30	30–40	40–50	50–60	60–70
Number of patients	60	42	55	70	53	20

Form: (i) Less than type cumulative frequency distribution. (ii) More than type cumulative frequency distribution.

Solution.

(i) Less than type — running totals from the top:

Age (in years)	Number of patients
Less than 10	0
Less than 20	60
Less than 30	$60 + 42 = 102$
Less than 40	$102 + 55 = 157$
Less than 50	$157 + 70 = 227$
Less than 60	$227 + 53 = 280$
Less than 70	$280 + 20 = 300$

Nobody is under $10$, so that row is $0$, and the last row reaches the total of $300$ ✓.

(ii) More than type — start at the total and subtract each class as you pass it:

Age (in years)	Number of patients
More than or equal to 10	300
More than or equal to 20	$300 – 60 = 240$
More than or equal to 30	$240 – 42 = 198$
More than or equal to 40	$198 – 55 = 143$
More than or equal to 50	$143 – 70 = 73$
More than or equal to 60	$73 – 53 = \mathbf{20}$

The printed answer for the last entry is wrong. The key gives $60$ for “more than or equal to $60$”; the correct value is $\mathbf{20}$.

Three independent confirmations:

1. From the data. Only the class $60$–$70$ lies at or above $60$, and it holds $20$ patients.
2. From the running subtraction. $73 – 53 = 20$, and the key’s own preceding entry of $73$ is correct.
3. From the key’s own other table. It prints “less than $60$” as $280$ against a total of $300$, so the number at or above $60$ must be $300 – 280 = 20$.

So the key contradicts itself — its two tables disagree — and $20$ is the value both the data and its own less-than column support.

A check worth applying to every such pair. For any value $x$,

$$(\text{less than } x) + (\text{more than or equal to } x) = \text{total}$$

Run it down the two columns: $0 + 300$, $60 + 240$, $102 + 198$, $157 + 143$, $227 + 73$, $280 + 20$ — every pair gives $300$ ✓. With the printed $60$, the last pair would give $340$.

Question 14

Given below is a cumulative frequency distribution showing the marks secured by $50$ students of a class:

Marks	Below 20	Below 40	Below 60	Below 80	Below 100
Number of students	17	22	29	37	50

Form the frequency distribution table for the data.

Solution. A “less than” table, so difference each entry from the one before.

Class	Working	Frequency
0–20	$17$	17
20–40	$22 – 17$	5
40–60	$29 – 22$	7
60–80	$37 – 29$	8
80–100	$50 – 37$	13
Total		50

The frequencies sum to $50$ ✓.

The shape is unusual and worth noticing. The frequencies run $17, 5, 7, 8, 13$ — high at both ends and low in the middle. That is a bimodal distribution: many students scored under $20$ and many scored over $80$, with few in between.

It matters because the mean would be misleading here. With $\sum f_i x_i = 17(10) + 5(30) + 7(50) + 8(70) + 13(90) = 2400$, the mean is $48$ — a mark almost nobody actually scored. For data like this the mean is a poor summary, and saying so is often worth more than computing it.

Question 15

Weekly income of $600$ families is tabulated below:

Weekly income (in ₹)	0–1000	1000–2000	2000–3000	3000–4000	4000–5000	5000–6000
Number of families	250	190	100	40	15	5

Compute the median income.

Solution. Build the cumulative column and find where $\frac n2$ falls.

Class	$f$	$cf$
0–1000	250	250
1000–2000	190	440
2000–3000	100	540
3000–4000	40	580
4000–5000	15	595
5000–6000	5	600

$n = 600$, so $\frac n2 = 300$. The first cumulative frequency to reach $300$ is $440$, so the median class is $1000$–$2000$, with

$$l = 1000, \quad cf = 250, \quad f = 190, \quad h = 1000$$

$$\text{Median} = l + \left(\frac{\frac n2 – cf}{f}\right)h = 1000 + \left(\frac{300 – 250}{190}\right)(1000)$$

$$= 1000 + \frac{50\,000}{190} = 1000 + 263.157\ldots = ₹1263.16$$

On the printed answer. The key gives ₹$1263.15$. The exact value is $1000 + \frac{50000}{190} = 1263.1578\ldots$, which rounds to $1263.16$ and truncates to $1263.15$. Either will be accepted; rounding is the convention for money.

Note how far the median sits below the mean. The mean here is ₹$1673$, because the few high-income families at $5000$–$6000$ drag it up while leaving the median untouched. Income data is almost always skewed this way, which is exactly why median income is the figure usually quoted.

Question 16

The maximum bowling speeds, in km per hour, of $33$ players at a cricket coaching centre are given as follows:

Speed (km/h)	85–100	100–115	115–130	130–145
Number of players	11	9	8	5

Calculate the median bowling speed.

Solution.

Class	$f$	$cf$
85–100	11	11
100–115	9	20
115–130	8	28
130–145	5	33

$n = 33$, so $\frac n2 = 16.5$. The first cumulative frequency to reach it is $20$, so the median class is $100$–$115$, with

$$l = 100, \quad cf = 11, \quad f = 9, \quad h = 15$$

$$\text{Median} = 100 + \left(\frac{16.5 – 11}{9}\right)(15) = 100 + \frac{5.5 \times 15}{9} = 100 + \frac{82.5}{9}$$

$$= 100 + 9.1\overline{6} = 109.17 \text{ km/h}$$

Use $\frac n2$, not $\frac{n+1}{2}$. For ungrouped data with $n$ odd you would take the $\frac{n+1}{2}$th observation, here the $17$th. For grouped data the formula uses $\frac n2 = 16.5$ throughout, because it interpolates within the class rather than picking out an individual value. Using $17$ instead would give $109.5$ — close, and wrong.

The class width $h = 15$ is worth checking too; it is easy to assume $10$ out of habit.

Question 17

The monthly income of $100$ families are given as below:

Income (in ₹)	Number of families
0–5000	8
5000–10000	26
10000–15000	41
15000–20000	16
20000–25000	3
25000–30000	3
30000–35000	2
35000–40000	1

Calculate the modal income.

Solution. The modal class is the one with the largest frequency: $41$, in $10000$–$15000$.

Read off the four quantities the formula needs:

$$l = 10\,000 \ (\text{lower limit of the modal class})$$ $$f_1 = 41 \ (\text{its frequency}), \quad f_0 = 26 \ (\text{the class before}), \quad f_2 = 16 \ (\text{the class after}), \quad h = 5000$$

$$\text{Mode} = l + \left(\frac{f_1 – f_0}{2f_1 – f_0 – f_2}\right)h = 10\,000 + \left(\frac{41 – 26}{82 – 26 – 16}\right)(5000)$$

$$= 10\,000 + \frac{15}{40}(5000) = 10\,000 + 1875 = ₹11\,875$$

What the formula is doing. It places the mode inside the modal class, shifted towards whichever neighbour has the smaller frequency. Here the class after ($16$) is lighter than the class before ($26$), so the mode is pulled upward from the midpoint of $12\,500$ — and indeed $11\,875$ is… below it.

That is worth pausing on. The shift is $\frac{f_1 – f_0}{2f_1 – f_0 – f_2}$ of the way up from the lower limit, and $\frac{15}{40} = 0.375$ is less than $\frac12$, so the mode sits in the lower part of the class, nearer the heavier neighbour at $5000$–$10000$. The formula leans towards the side with more data, which is what you would want.

Question 18

The weight of coffee in $70$ packets are shown in the following table:

Weight (in g)	Number of packets
200–201	12
201–202	26
202–203	20
203–204	9
204–205	2
205–206	1

Determine the modal weight.

Solution. The largest frequency is $26$, so the modal class is $201$–$202$, and

$$l = 201, \quad f_1 = 26, \quad f_0 = 12, \quad f_2 = 20, \quad h = 1$$

$$\text{Mode} = 201 + \left(\frac{26 – 12}{52 – 12 – 20}\right)(1) = 201 + \frac{14}{20} = 201 + 0.7 = 201.7 \text{ g}$$

The class width is $1$, which makes the fraction $\frac{14}{20} = 0.7$ the whole of the shift — a useful case for seeing what the formula does. The mode sits $70\%$ of the way up the class, leaning towards the heavier neighbour above ($20$) rather than the lighter one below ($12$).

In context. The packets are presumably meant to contain $200$ g, and the most common weight is $201.7$ g — so the filling machine is running slightly over, by not quite two grams. That is the sort of thing this calculation is for: a manufacturer wants the mode near the target, and a systematic overfill is money given away.

Common mistakes

• Question 13 — trusting the printed “more than” table. Its last entry should be $20$, not $60$. Check any such pair with $(\text{less than } x) + (\text{at least } x) = \text{total}$; here the key’s own two columns fail that test.
• Questions 3, 4 and 5 — taking the midpoint between classes. For the inclusive class $4$–$7$ the class mark is $\frac{4+7}{2} = 5.5$, not $\frac{4+8}{2} = 6$. Half-integer class marks are normal in inclusive tables.
• Question 1 — using the step-deviation method on unequal classes. The widths here are $2, 2, 2, 3$, so there is no common $h$. The direct method always works.
• Questions 10, 11 and 14 — differencing in the wrong direction. In a “less than” table subtract the previous row; in a “more than” table subtract the next. A negative frequency means it has been done backwards.
• Question 16 — using $\frac{n+1}{2}$ for grouped data. The grouped median formula uses $\frac n2$, even when $n$ is odd, because it interpolates rather than selecting an observation.
• Questions 17 and 18 — misreading $f_0$ and $f_2$. They are the frequencies of the classes immediately before and after the modal class, not the two smallest or the neighbouring class marks.
• Question 15 — truncating money instead of rounding. $1263.1578\ldots$ is ₹$1263.16$ to the nearest paisa. The key truncates; say which convention you used.
• All the mean questions — forgetting to check $\sum f_i$. Every table here states its total ($50$ wrestlers, $800$ students, $600$ families). If your column sum disagrees, a frequency has been misread before any averaging begins.

Practise next

• Exercise 13.3, Questions 19 to 42 — the probability half of this exercise, twenty-four questions on dice, cards, coins and counting.
• Exercise 13.1 — worth revisiting for Questions 6 to 11, which are these same table manipulations set as multiple choice.
• Exercise 13.2 — its Questions 1 to 4 explain why the formulae used here behave as they do, including why the grouped median is only an estimate.