NCERT Exemplar Solutions Class 10 Maths Statistics and Probability — Exercise 13.1

NCERT Exemplar Problems, Class 10 Mathematics — Statistics and Probability, Exercise 13.1. All 26 multiple-choice questions solved.

Twenty-six questions in two halves: Questions 1 to 11 on grouped data, and 12 to 26 on probability.

For the statistics half, three formulae and one habit. The formulae:

$$\bar{x} = a + h\left(\frac{\sum f_i u_i}{\sum f_i}\right) \ \text{ where } u_i = \frac{x_i – a}{h}, \qquad \text{Median} = l + \left(\frac{\frac n2 – cf}{f}\right)h$$

$$\text{Mode} = l + \left(\frac{f_1 – f_0}{2f_1 – f_0 – f_2}\right)h$$

The habit is to build the cumulative frequency column before answering anything. Five of these eleven questions are decided by it, and three of them hand you a cumulative table that has to be differenced back into ordinary frequencies first.

Two class definitions worth keeping straight:

• the median class is the first class whose cumulative frequency reaches $\frac n2$
• the modal class is simply the class with the highest frequency

They are often different classes, and Question 9 is the case where they happen to be the same.

For the probability half, everything follows from

$$P(E) = \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}, \qquad 0 \le P(E) \le 1, \qquad P(\text{not } E) = 1 – P(E)$$

A note on the current syllabus. Chapter 13 keeps mean, mode and median, but the section on drawing ogives — “Graphical Representation of Cumulative Frequency Distribution” — has been removed. Question 5 here is about the intersection of the two ogives, so it is exemplar-only material now.

It is a leftover worth knowing about, because the chapter has not been tidied. Its Introduction still promises “how to draw cumulative frequency curves, called ogives“, and its closing Note to the Reader still advises on “construction of an ogive” and on choosing scales “in case of ogives” — for a construction the chapter no longer contains.

Key insight. When a table is headed “Below 10, Below 20, …” or “More than or equal to 0, 10, …”, it is cumulative, not a frequency table. Difference consecutive entries before doing anything else. Questions 8 and 11 are decided entirely by that step, and both offer the undifferenced number as a wrong option.

Choose the correct answer from the given four options:

Question 1

In the formula $\bar{x} = a + \dfrac{\sum f_i d_i}{\sum f_i}$, for finding the mean of grouped data $d_i$’s are deviations from $a$ of

Solution. This is the assumed mean method, and in it each class is represented by a single number — its class mark, the mid-point of the interval:

$$x_i = \frac{\text{lower limit} + \text{upper limit}}{2}$$

The deviation is measured from the assumed mean $a$ to that representative value:

$$d_i = x_i – a$$

So the $d_i$ are deviations of the mid points.

The whole of grouped-data statistics rests on treating every observation in a class as though it sat at the class mark. That assumption is what Question 2 states directly, and it is why a grouped mean is only an estimate.

Question 2

While computing mean of grouped data, we assume that the frequencies are

Solution. The assumption behind every mean formula in this chapter is that all the observations in a class sit at its class mark — the mid-point.

That is exactly what makes $\sum f_i x_i$ a sensible stand-in for the sum of the original data: each of the $f_i$ observations in a class is counted as though it equalled $x_i$.

Option (A) describes a different assumption, and one the chapter does use elsewhere — the median formula assumes the observations are spread evenly across the median class, which is why it interpolates. So the two assumptions coexist in the same chapter for different purposes, and this question is about the one behind the mean.

Choosing an upper or lower limit would bias every class systematically upward or downward; the mid-point is the only choice that does not.

Question 3

If $x_i$’s are the mid points of the class intervals of grouped data, $f_i$’s are the corresponding frequencies and $\bar{x}$ is the mean, then $\sum\left(f_i x_i – \bar{x}\right)$ is equal to

Solution. The intended quantity is the sum of the weighted deviations from the mean:

$$\sum f_i\left(x_i – \bar{x}\right)$$

Expand it:

$$\sum f_i x_i – \bar{x}\sum f_i$$

Now the definition of the mean is $\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$, so $\bar{x} \sum f_i = \sum f_i x_i$. Substituting:

$$\sum f_i x_i – \sum f_i x_i = 0$$

This is the defining property of the mean, not an accident of the algebra. The mean is precisely the value about which the deviations balance — the amounts by which observations fall short of it exactly cancel the amounts by which others exceed it.

It is also why the mean is the point where the data would balance if the observations were weights on a beam, and why a single extreme value drags it so far: to keep the balance, a large positive deviation needs many small negative ones.

Question 4

In the formula $\bar{x} = a + h\left(\dfrac{\sum f_i u_i}{\sum f_i}\right)$, for finding the mean of grouped frequency distribution, $u_i =$

Solution. This is the step-deviation method, which is the assumed mean method with one extra simplification: after subtracting $a$, divide by the common class width $h$.

$$u_i = \frac{x_i – a}{h}$$

The point is to make the numbers small. If the classes are $100$–$120$, $120$–$140$ and so on with $a = 130$, the deviations $x_i – a$ are $-20$, $0$, $20$, $40$; dividing by $h = 20$ turns them into $-1$, $0$, $1$, $2$ — integers you can multiply in your head.

The $h$ outside the bracket in the formula undoes the division at the end, which is why it is there.

Option (D) reverses the subtraction and would give a mean on the wrong side of $a$. Option (B) multiplies by $h$ instead of dividing, defeating the purpose by making the numbers larger.

Question 5

The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its

Solution. The two ogives cross at the point whose $x$-coordinate is the median.

Why. The “less than” curve at a value $x$ gives the number of observations at most $x$; the “more than” curve gives the number at least $x$. At the crossing these two counts are equal, and two counts that are equal and together account for all $n$ observations must each be $\frac n2$.

A value with half the data below it and half above is precisely the median.

So the construction is a graphical way of reading off the median without the interpolation formula — and the abscissa is the answer, not the ordinate. The ordinate at the crossing is always $\frac n2$, which tells you nothing about the data.

This question is about drawing ogives, a section that has been removed from the current Class 10 textbook — though the chapter’s Introduction and its closing Note to the Reader both still refer to them.

Question 6

For the following distribution:

Class	0–5	5–10	10–15	15–20	20–25
Frequency	10	15	12	20	9

the sum of lower limits of the median class and modal class is

Solution. Build the cumulative frequency column first.

Class	Frequency	Cumulative frequency
0–5	10	10
5–10	15	25
10–15	12	37
15–20	20	57
20–25	9	66

The median class. Here $n = 66$, so $\frac n2 = 33$. The first class whose cumulative frequency reaches $33$ is 10–15 (its $cf$ is $37$, the one before is only $25$). Lower limit: $10$.

The modal class. The largest frequency is $20$, in the class 15–20. Lower limit: $15$.

The sum:

$$10 + 15 = 25$$

Note that the two classes are different — the median class is 10–15 and the modal class 15–20. That is the usual situation, and the question is built on keeping them apart. Taking both as 15–20 gives $30$, option (C).

Question 7

Consider the following frequency distribution:

Class	0–5	6–11	12–17	18–23	24–29
Frequency	13	10	15	8	11

The upper limit of the median class is

Solution. These classes are not continuous, and that is the whole question. There is a gap between $5$ and $6$, between $11$ and $12$, and so on — the classes are in inclusive form.

The median formula requires continuous classes, so convert first. The gap is $1$, so subtract $\frac12$ from every lower limit and add $\frac12$ to every upper limit:

Original	Continuous	Frequency	Cumulative frequency
0–5	$-0.5$–$5.5$	13	13
6–11	$5.5$–$11.5$	10	23
12–17	$11.5$–$17.5$	15	38
18–23	$17.5$–$23.5$	8	46
24–29	$23.5$–$29.5$	11	57

Here $n = 57$, so $\frac n2 = 28.5$. The first cumulative frequency to reach it is $38$, in the class $11.5$–$17.5$.

Its upper limit is $\mathbf{17.5}$.

Option (A), $17$, is the upper limit of the unconverted class — the answer you get by skipping the conversion. The textbook’s own Note to the Reader requires it: “it should be ensured that the class intervals are continuous before applying the formulae.”

Question 8

For the following distribution:

Marks	Number of students
Below 10	3
Below 20	12
Below 30	27
Below 40	57
Below 50	75
Below 60	80

the modal class is

Solution. This is a cumulative table, not a frequency table. “Below 40: 57” means $57$ students scored under $40$ — it counts everyone below, not the number in the class $30$–$40$.

Difference consecutive entries to recover the actual frequencies:

Class	Working	Frequency
0–10	$3$	3
10–20	$12 – 3$	9
20–30	$27 – 12$	15
30–40	$57 – 27$	30
40–50	$75 – 57$	18
50–60	$80 – 75$	5

The largest frequency is $30$, so the modal class is 30–40.

A check: the frequencies sum to $3 + 9 + 15 + 30 + 18 + 5 = 80$, which is the last cumulative entry ✓. That is always worth doing — it catches a subtraction slip immediately.

Reading the table as though it were a frequency table would make $80$ the largest “frequency” and 50–60 the modal class, which is option (D). A cumulative column always increases, so its last row is always the biggest — that alone should signal what kind of table you are looking at.

Question 9

Consider the data:

Class	65–85	85–105	105–125	125–145	145–165	165–185	185–205
Frequency	4	5	13	20	14	7	4

The difference of the upper limit of the median class and the lower limit of the modal class is

Solution. Cumulative frequencies first:

Class	Frequency	Cumulative frequency
65–85	4	4
85–105	5	9
105–125	13	22
125–145	20	42
145–165	14	56
165–185	7	63
185–205	4	67

The median class. $n = 67$, so $\frac n2 = 33.5$. The first cumulative frequency to reach it is $42$, in 125–145. Its upper limit is $145$.

The modal class. The largest frequency is $20$, also in 125–145. Its lower limit is $125$.

The difference:

$$145 – 125 = 20$$

Here the median class and the modal class are the same class — which happens when the data is roughly symmetric and peaked in the middle, as this is. So the question reduces to the class’s own width, $20$.

Option (A), $0$, is the trap: noticing that the two classes coincide and concluding the difference must be zero. But the question asks for the difference between the class’s upper limit and its lower limit, which is its width, not zero.

Question 10

The times, in seconds, taken by $150$ athletes to run a $110$ m hurdle race are tabulated below:

Class	13.8–14	14–14.2	14.2–14.4	14.4–14.6	14.6–14.8	14.8–15
Frequency	2	4	5	71	48	20

The number of athletes who completed the race in less than $14.6$ seconds is

Solution. “Less than $14.6$ seconds” means every class ending at or before $14.6$:

$$2 + 4 + 5 + 71 = 82$$

So $82$ athletes.

Each wrong option is a specific misreading, which makes this a good check on whether the table is being read correctly:

• (A) $11$ is $2 + 4 + 5$ — the classes below $14.4$, stopping one class too early.
• (B) $71$ is the single class $14.4$–$14.6$, treating the question as “between $14.4$ and $14.6$”.
• (D) $130$ is $2+4+5+71+48$ — the classes below $14.8$, going one class too far.

The boundary $14.6$ is the upper limit of the class $14.4$–$14.6$, so that class is entirely included. Had the question asked for “less than $14.4$”, the answer would have been $11$.

Question 11

Consider the following distribution:

Marks obtained	Number of students
More than or equal to 0	63
More than or equal to 10	58
More than or equal to 20	55
More than or equal to 30	51
More than or equal to 40	48
More than or equal to 50	42

the frequency of the class 30–40 is

Solution. This is a “more than” cumulative table, so the numbers decrease as the mark rises — each entry counts everyone at or above that mark.

To get the frequency of a single class, subtract the next entry from the current one:

$$f(30\text{–}40) = (\text{at least } 30) – (\text{at least } 40) = 51 – 48 = 3$$

The students counted in “at least $30$” but not in “at least $40$” are exactly those scoring between $30$ and $40$.

The full frequency table, for completeness:

Class	Working	Frequency
0–10	$63 – 58$	5
10–20	$58 – 55$	3
20–30	$55 – 51$	4
30–40	$51 – 48$	3
40–50	$48 – 42$	6
50 and above	—	42

Note the direction of the subtraction. For a “less than” table (Question 8) you subtract the previous entry from the current one; for a “more than” table you subtract the next from the current. Getting it backwards here gives $48 – 51 = -3$, and a negative frequency is the signal that the table has been read the wrong way.

Options (C) and (D) are the two cumulative entries themselves, offered undifferenced.

Question 12

If an event cannot occur, then its probability is

Solution. An event that cannot occur is an impossible event, and no outcome favours it:

$$P(E) = \frac{0}{\text{total outcomes}} = 0$$

At the other end, an event certain to occur is favoured by every outcome, so its probability is $1$. Those two are the extremes of the scale, and every other probability lies between them.

Examples of impossible events for a single throw of an ordinary die: getting a $7$, getting a negative number, getting two numbers at once. Each has probability $0$.

Question 13

Which of the following cannot be the probability of an event?

Solution. A probability must satisfy

$$0 \le P(E) \le 1$$

Check each option against that:

• (A) $\frac13 \approx 0.333$ — between $0$ and $1$ ✓
• (B) $0.1$ ✓
• (C) $3\% = 0.03$ ✓
• (D) $\frac{17}{16} = 1.0625 > 1$ ✗

So $\frac{17}{16}$ cannot be a probability.

Option (C) is the one that catches people, because a percentage looks like the wrong kind of object. But $3\%$ is just $0.03$, comfortably inside the range — and probabilities are often quoted as percentages in everyday use (“a $30\%$ chance of rain”). What matters is the value, not the notation.

The spot check for any fraction: if the numerator exceeds the denominator, it is more than $1$ and cannot be a probability. $\frac{17}{16}$ fails that at a glance.

Question 14

An event is very unlikely to happen. Its probability is closest to

Solution. “Very unlikely” means the probability is near $0$, so choose the smallest of the four.

Line them up by size:

$$0.0001 < 0.001 < 0.01 < 0.1$$

The smallest is $0.0001$, which is one chance in ten thousand.

The decimals are easy to misrank because the digits are identical and only the number of zeros differs. Counting decimal places is the reliable way: $0.0001$ has four, $0.1$ has one, and more decimal places before the $1$ means a smaller number.

For a sense of scale, $0.1$ is one chance in ten — that describes something that happens fairly often, not something very unlikely.

Question 15

If the probability of an event is $p$, the probability of its complementary event will be

Solution. The complementary event is “$E$ does not happen”. Exactly one of $E$ and its complement occurs, and between them they cover every outcome, so their probabilities sum to $1$:

$$P(E) + P(\text{not } E) = 1 \implies P(\text{not } E) = 1 – p$$

Option (A), $p – 1$, is the same subtraction reversed and is negative whenever $p < 1$ — which is to say almost always. A negative probability is impossible, so (A) can be ruled out without any reasoning about complements at all.

Option (D) is worse: with $p = \frac12$ it gives $1 – 2 = -1$.

The relation is most useful in the direction $P(E) = 1 – P(\text{not } E)$, when the complement is easier to count. Question 21 is an example — counting the cards that are the ace of hearts is much quicker than counting the $51$ that are not.

Question 16

The probability expressed as a percentage of a particular occurrence can never be

Solution. As a percentage, the range $0 \le P \le 1$ becomes

$$0\% \le P \le 100\%$$

So a probability expressed as a percentage can never be less than $0$.

Check why each of the others is wrong:

• (A) Probabilities less than $100\%$ are the ordinary case — anything not certain.
• (C) A percentage can exceed $1$: $50\%$ is a perfectly good probability and $50 > 1$. This option confuses the two scales, and it is the trap the question is built on. The bound of $1$ applies to the decimal form, not the percentage form.
• (D) A probability need not be a whole number of per cent: $\frac13$ is $33.33\ldots\%$.

The careless reading of (C) comes from carrying “probability cannot exceed $1$” across to a question that has just changed the units. When the scale changes, the bound changes with it.

Question 17

If $P(A)$ denotes the probability of an event $A$, then

Solution. The probability of any event lies between $0$ and $1$ inclusive:

$$0 \le P(A) \le 1$$

The two endpoints are attainable, which is why the inequalities are not strict: an impossible event has $P = 0$ (Question 12) and a certain event has $P = 1$.

Options (A) and (B) assert that every probability is negative, or that every probability exceeds $1$ — both impossible for any event at all.

Option (D) is the near-miss. Its upper bound is right and its lower bound is wrong: no probability is ever negative, because a count of favourable outcomes cannot be negative and neither can a count of total outcomes.

Question 18

A card is selected from a deck of $52$ cards. The probability of its being a red face card is

Solution. Count the red face cards carefully; both words matter.

Face cards are the Jack, Queen and King — three in each suit. Red suits are hearts and diamonds — two of the four.

$$\text{red face cards} = 2 \text{ suits} \times 3 \text{ face cards} = 6$$

Specifically: Jack, Queen, King of hearts and Jack, Queen, King of diamonds.

$$P = \frac{6}{52} = \frac{3}{26}$$

Option (B), $\frac{3}{13}$, is $\frac{12}{52}$ — all the face cards, red and black. That is the answer to a question that dropped the word “red”, and it is why the count has to be done in two steps.

Note that the Ace is not a face card in the convention this book uses; if it were, there would be $8$ red ones.

Question 19

The probability that a non leap year selected at random will contain $53$ Sundays is

Solution. A non-leap year has $365$ days, and

$$365 = 52 \times 7 + 1$$

So the year is exactly $52$ complete weeks plus one extra day.

Those $52$ complete weeks already supply $52$ Sundays, guaranteed. A $53$rd Sunday happens if and only if the one leftover day is itself a Sunday.

The leftover day can be any of the seven days of the week, each equally likely, so

$$P = \frac17$$

For a leap year the answer differs, and the contrast is the point of the question. A leap year has $366 = 52 \times 7 + 2$ days, so there are two leftover days, and they are consecutive: (Sun, Mon), (Mon, Tue), …, (Sat, Sun) — seven equally likely pairs. Two of those seven contain a Sunday, so the probability would be $\frac27$.

That is where option (B) comes from: it is the right answer to the leap-year version.

Question 20

When a die is thrown, the probability of getting an odd number less than $3$ is

Solution. Both conditions have to hold at once — odd and less than $3$.

The numbers less than $3$ are $1$ and $2$. Of those, the odd one is just $\mathbf{1}$.

So exactly one of the six outcomes is favourable:

$$P = \frac16$$

Each wrong option answers a different question:

• (B) $\frac13$ counts $\{1, 2\}$ — everything less than $3$, dropping “odd”.
• (C) $\frac12$ counts $\{1, 3, 5\}$ — all the odd numbers, dropping “less than $3$”.
• (D) $0$ treats the two conditions as incompatible, which they are not.

Listing the outcomes that satisfy both conditions, rather than reasoning about them separately, is what keeps this straight.

Question 21

A card is drawn from a deck of $52$ cards. The event $E$ is that card is not an ace of hearts. The number of outcomes favourable to $E$ is

Solution. The question asks for a count, not a probability — read it twice, because every option is a plausible count.

There is exactly one ace of hearts in a deck. So the cards that are not the ace of hearts number

$$52 – 1 = 51$$

That is the count of favourable outcomes. (The probability would be $\frac{51}{52}$, but that is not what was asked.)

The wrong options come from misreading which single card is excluded:

• (A) $4$ is the number of aces.
• (B) $13$ is the number of hearts.
• (C) $48$ is $52 – 4$ — excluding all four aces, the answer to “not an ace”.

“Ace of hearts” names one specific card, not a whole rank or a whole suit. The phrase X of Y in a card question always picks out exactly one card.

Question 22

The probability of getting a bad egg in a lot of $400$ is $0.035$. The number of bad eggs in the lot is

Solution. Run the probability definition backwards. From

$$P = \frac{\text{number of bad eggs}}{\text{total eggs}}$$

we get

$$\text{number of bad eggs} = P \times \text{total} = 0.035 \times 400 = 14$$

The multiplication is easier as a fraction: $0.035 = \frac{35}{1000} = \frac{7}{200}$, so

$$\frac{7}{200} \times 400 = 7 \times 2 = 14$$

Check the answer is a whole number, as a count of eggs must be. That is a genuine constraint here: had the probability been, say, $0.037$, the product would be $14.8$ and the data would be inconsistent. A non-integer answer to a counting question means an arithmetic slip.

Option (A), $7$, is what you get from $0.035 \times 200$ — the numerator of the simplified fraction, mistaken for the answer.

Question 23

A girl calculates that the probability of her winning the first prize in a lottery is $0.08$. If $6000$ tickets are sold, how many tickets has she bought?

Solution. Her chance of winning is the share of tickets she holds:

$$P = \frac{\text{her tickets}}{\text{total tickets}} = 0.08$$

So

$$\text{her tickets} = 0.08 \times 6000 = 480$$

As a fraction, $0.08 = \frac{8}{100} = \frac{2}{25}$, and $\frac{2}{25} \times 6000 = 2 \times 240 = 480$.

A check on the size. $0.08$ is $8\%$, so she should hold about a twelfth of the tickets. A twelfth of $6000$ is $500$ — and $480$ is right beside it ✓.

Option (D), $750$, is $\frac{6000}{8}$ — dividing by $8$ instead of multiplying by $0.08$, which inverts the relationship. The two differ by a factor of over a hundred, so the plausibility check catches it.

This is the same structure as Question 22: probability times total gives the count.

Question 24

One ticket is drawn at random from a bag containing tickets numbered $1$ to $40$. The probability that the selected ticket has a number which is a multiple of $5$ is

Solution. Count the multiples of $5$ from $1$ to $40$:

$$5, 10, 15, 20, 25, 30, 35, 40$$

There are $8$ of them — or, without listing, $\left\lfloor \frac{40}{5} \right\rfloor = 8$.

$$P = \frac{8}{40} = \frac15$$

The answer is $\frac15$, and that is no coincidence: among the first $n$ multiples-of-$5$-worth of integers, exactly one in five is a multiple of $5$. Whenever the range runs from $1$ to an exact multiple of the divisor, the probability is simply $\frac{1}{\text{divisor}}$.

The shortcut fails when the range does not divide exactly. From $1$ to $42$ there would still be $8$ multiples, giving $\frac{8}{42} = \frac{4}{21}$ — close to $\frac15$ but not equal. Counting is safer than assuming.

Question 25

Someone is asked to take a number from $1$ to $100$. The probability that it is a prime is

Solution. There are $25$ primes below $100$, a fact worth knowing outright:

$$2, 3, 5, 7, \quad 11, 13, 17, 19, \quad 23, 29, \quad 31, 37, \quad 41, 43, 47, \quad 53, 59, \quad 61, 67, \quad 71, 73, 79, \quad 83, 89, \quad 97$$

Grouped by decade: $4, 4, 2, 2, 3, 2, 2, 3, 2, 1$ — which sums to $25$ and is the easiest way to recount without missing any.

$$P = \frac{25}{100} = \frac14$$

Two easy miscounts. Including $1$ gives $26$ and option (D), $\frac{26}{100} = \frac{13}{50}$ — but $1$ is not prime, since a prime has exactly two distinct factors and $1$ has only one. And forgetting $2$, on the grounds that it is even, gives $24$ and option (B), $\frac{24}{100} = \frac{6}{25}$ — but $2$ is prime, and is the only even prime.

Both wrong options are exactly one prime away from the right answer, which is the question’s design.

Question 26

A school has five houses $A$, $B$, $C$, $D$ and $E$. A class has $23$ students, $4$ from house $A$, $8$ from house $B$, $5$ from house $C$, $2$ from house $D$ and rest from house $E$. A single student is selected at random to be the class monitor. The probability that the selected student is not from $A$, $B$ and $C$ is

Solution. House $E$’s count is not given and has to be recovered first:

$$E = 23 – (4 + 8 + 5 + 2) = 23 – 19 = 4$$

“Not from $A$, $B$ and $C$” means from $D$ or $E$:

$$2 + 4 = 6$$

$$P = \frac{6}{23}$$

The complement route is quicker and worth practising. Students from $A$, $B$ or $C$ number $4 + 8 + 5 = 17$, so

$$P(\text{not } A, B, C) = 1 – \frac{17}{23} = \frac{6}{23}$$

Option (D), $\frac{17}{23}$, is that complement left un-subtracted — the probability of the event you counted on the way, not the one asked for. That is the standard hazard of the complement method: having computed $P(A \text{ or } B \text{ or } C)$, it is easy to stop there.

Option (A) is house $E$ alone, and (C) is house $B$ alone.

Common mistakes

• Questions 8 and 11 — treating a cumulative table as a frequency table. “Below 40” and “More than or equal to 30” are running totals. Difference consecutive entries first, and check that the recovered frequencies sum to the total.
• Question 11 — differencing in the wrong direction. For a “less than” table subtract the previous entry; for a “more than” table subtract the next. A negative frequency means you have it backwards.
• Question 7 — using inclusive classes in the median formula. Classes like 0–5, 6–11 have gaps. Subtract $\frac12$ from each lower limit and add $\frac12$ to each upper before proceeding; the answer is $17.5$, not $17$.
• Questions 6 and 9 — confusing the median class with the modal class. The median class is the first whose cumulative frequency reaches $\frac n2$; the modal class simply has the largest frequency. They coincide only sometimes, as in Question 9.
• Question 16 — carrying the bound of $1$ across to a percentage. On the percentage scale the bounds are $0\%$ and $100\%$. A probability of $50\%$ is greater than $1$ and perfectly legitimate.
• Question 21 — answering with a probability when a count was asked for. “The number of outcomes favourable to $E$” wants $51$, not $\frac{51}{52}$.
• Questions 18 and 21 — over- or under-counting cards. “Red face card” needs both conditions ($6$ cards); “ace of hearts” names exactly one card, not the four aces or the thirteen hearts.
• Question 25 — miscounting the primes below $100$. There are $25$: $1$ is not prime, and $2$ is. Both wrong options here are exactly one prime out.

Practise next

• Exercise 13.2 — fourteen reasoning questions on why the formulae of Questions 1 to 4 work, and on when outcomes are genuinely equally likely.
• Exercise 13.3 — forty-two short-answer questions, twenty on grouped data and twenty-two on probability, where these ideas are applied rather than recognised.
• Exercise 11.4 — worth revisiting for the cost and rate reasoning, which pairs with the counting arguments here.