NCERT Class 10 Mathematics — Statistics, Exercise 13.1. All 9 questions solved.

Every question here asks for the mean of grouped data. Once data is grouped we no longer know the individual observations, so each class is represented by its class mark — its midpoint:

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

The book gives three ways to finish the calculation, and they always agree:

$$\text{Direct:}\quad \bar x = \frac{\sum f_ix_i}{\sum f_i}$$

$$\text{Assumed mean:}\quad d_i = x_i – a, \qquad \bar x = a + \frac{\sum f_id_i}{\sum f_i}$$

$$\text{Step-deviation:}\quad u_i = \frac{x_i – a}{h}, \qquad \bar x = a + h\,\frac{\sum f_iu_i}{\sum f_i}$$

Here $a$ is an assumed mean (a class mark near the middle of the table) and $h$ is the class size.

Key insight. The assumed mean and step-deviation methods are not different answers — they are the direct method with the arithmetic shrunk. Subtracting $a$ and dividing by $h$ turns awkward class marks like $510, 530, 550$ into $-2, -1, 0, 1, 2$, and the formula adds back exactly what was taken away. So the choice of method is only about convenience: the book’s own advice is to use the direct method when $x_i$ and $f_i$ are small, and one of the other two when they are large. Question 8 shows the limit of that advice — with unequal classes there is no common $h$ to divide by.

Question 1

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in $20$ houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean, and why?

Solution. The class marks are the odd numbers $1, 3, \ldots, 13$ and the frequencies are all single digits, so every product $f_ix_i$ can be done in the head. There is nothing for the assumed mean to simplify, which makes the direct method the natural choice.

The frequencies add to $20$, the number of houses, as they should.

$$\bar x = \frac{\sum f_ix_i}{\sum f_i} = \frac{162}{20} = 8.1$$

Question 2

Consider the following distribution of daily wages of $50$ workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution. The class marks $510, 530, \ldots, 590$ are large, and they are equally spaced $20$ apart. That is exactly the situation the step-deviation method is built for: take $a = 550$ (the middle class mark) and $h = 20$, and every $u_i$ becomes a small whole number.

$$\bar u = \frac{-12}{50} = -0.24, \qquad \bar x = a + h\bar u = 550 + 20(-0.24) = 550 – 4.8 = 545.2$$

Question 3

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ $18$. Find the missing frequency $f$.

Solution. The unknown $f$ appears twice in the mean — once in $\sum f_ix_i$ and once in $\sum f_i$ — so we write both totals in terms of $f$ and set their ratio equal to $18$. The direct method keeps this cleanest, because it leaves a single linear equation in $f$.

$$\frac{752 + 20f}{44 + f} = 18 \;\Longrightarrow\; 752 + 20f = 792 + 18f \;\Longrightarrow\; 2f = 40 \;\Longrightarrow\; f = 20$$

A check: with $f = 20$ there are $64$ children and $\sum f_ix_i = 1152$, and $\tfrac{1152}{64} = 18$.

Question 4

Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Solution. The class marks are $66.5, 69.5, \ldots, 84.5$ — decimals, and awkward to multiply directly. They are all $3$ apart, so the step-deviation method with $a = 75.5$ and $h = 3$ clears the decimals completely.

$$\bar x = 75.5 + 3 \times \frac{4}{30} = 75.5 + 0.4 = 75.9$$

Question 5

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Solution. Two things to notice. First, the classes have gaps — $50$–$52$ is followed by $53$–$55$ — because the number of mangoes is a whole number. For the mean this does not matter: closing the gaps to $49.5$–$52.5$, $52.5$–$55.5$, $\ldots$ leaves every class mark where it was, at $51, 54, 57, 60, 63$. Those class marks are $3$ apart, so $h = 3$, not $2$.

Second, the frequencies are large ($110$, $135$, $115$), so multiplying them by class marks in the fifties is laborious. The step-deviation method with $a = 57$ and $h = 3$ reduces every $u_i$ to $-2, \ldots, 2$.

$$\bar x = 57 + 3 \times \frac{25}{400} = 57 + 0.1875 = 57.1875 \approx 57.19$$

Question 6

The table below shows the daily expenditure on food of $25$ households in a locality.

Find the mean daily expenditure on food by a suitable method.

Solution. The class marks $125, 175, \ldots, 325$ are large and $50$ apart, so the step-deviation method with $a = 225$ and $h = 50$ is the efficient choice.

$$\bar x = 225 + 50 \times \frac{-7}{25} = 225 – 14 = 211$$

Question 7

To find out the concentration of $\mathrm{SO_2}$ in the air (in parts per million, i.e., ppm), the data was collected for $30$ localities in a certain city and is presented below:

Find the mean concentration of $\mathrm{SO_2}$ in the air.

Solution. Here the difficulty is not size but decimals. The class marks $0.02, 0.06, \ldots, 0.22$ are all $0.04$ apart, so the step-deviation method with $a = 0.10$ and $h = 0.04$ turns them into whole numbers.

$$\bar x = 0.10 + 0.04 \times \frac{-1}{30} = 0.10 – 0.00133\ldots = 0.09866\ldots \approx 0.099$$

The direct method agrees: $\sum f_ix_i = 2.96$, and $\tfrac{2.96}{30} = 0.0987$.

Question 8

A class teacher has the following absentee record of $40$ students of a class for the whole term. Find the mean number of days a student was absent.

Solution. Look at the class sizes: $6, 4, 4, 6, 8, 10, 2$. They are unequal, so each class mark has to be worked out on its own — $20$–$28$ has mark $24$, $38$–$40$ has mark $39$ — and they are not evenly spaced. Taking any class mark as $a$, the deviations share no common factor to use as $h$, so step-deviation would buy nothing. The class marks and frequencies are small, though, so the direct method is easy.

$$\bar x = \frac{499}{40} = 12.475 \approx 12.48$$

Question 9

The following table gives the literacy rate (in percentage) of $35$ cities. Find the mean literacy rate.

Solution. Equal classes of size $10$ with class marks $50, 60, \ldots, 90$: the step-deviation method with $a = 70$ and $h = 10$.

$$\bar x = 70 + 10 \times \frac{-2}{35} = 70 – \frac{4}{7} = \frac{486}{7} \approx 69.43$$

Common mistakes

• Using class limits instead of class marks. In every question each class stands for its midpoint. Multiplying $f_i$ by the lower limit (or the upper) shifts the whole mean by half a class width — in question 6 that is ₹ $25$.
• Question 2 and others, forgetting the final multiplication by $h$. The step-deviation table gives $\bar u$, not $\bar x$. Writing $\bar x = 550 – 0.24$ instead of $550 + 20(-0.24)$ is the commonest slip, and it produces an answer that still looks plausible.
• Question 3, putting $f$ in only one place. The unknown frequency changes the number of children too. Setting $752 + 20f = 18 \times 44$ ignores that and gives a non-integer $f$ — a sure sign something is wrong.
• Question 5, taking $h = 2$. The class $50$–$52$ looks two wide, but it contains three whole numbers, and successive class marks $51, 54, 57$ are $3$ apart. The class size is the gap between class marks.
• Question 8, forcing step-deviation onto unequal classes. With class sizes from $2$ to $10$ there is no single $h$ that makes every $u_i$ a whole number. Dividing by, say, $4$ anyway produces fractions such as $-\tfrac{7}{2}$ and undoes the point of the method.
• Sign errors below the assumed mean. Every class below $a$ has a negative $u_i$, and so a negative $f_iu_i$. In question 9 the totals $-16$ and $+14$ nearly cancel; one lost minus sign changes the answer by several per cent.

Practise next

• Exercise 13.2 — the mode of grouped data. Three of its six questions also ask for the mean, so the methods practised here are used again straight away.
• Exercise 13.3 — the median, completing the three measures of central tendency; its first question asks for all three and compares them.
• Class 11 Statistics, Exercise 13.1 — mean deviation about the mean, which begins with exactly the grouped mean calculated on this page.