Mean of Grouped Data Class 10: Three Methods Explained

The class mark#

Because we do not know exactly where each obser­va­tion lies inside its class, we assume that every obser­va­tion in a class sits at the mid­dle of the class. That mid­dle value is called the class mark (or mid-point):

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

For the class 20–30 the class mark is $\displaystyle \frac{20 + 30}{2} = 25$. The class size (or width) is $h = \text{upper limit} - \text{lower limit}$, so here $h = 10$. This "cen­tre" assump­tion is why the mean of grouped data is an esti­mate; it is usu­ally very close to the true mean.

The three for­mu­las#

Direct method#

Treat each class mark $x_i$ as if it occurred $f_i$ times. Then the total of all obser­va­tions is about $\displaystyle \sum f_i x_i$, and

$$\displaystyle \bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$$

Assumed mean method#

Choose a con­ve­nient value $a$, usu­ally a class mark near the mid­dle of the table, and find the devi­a­tions $d_i = x_i - a$. Then

$$\displaystyle \bar{x} = a + \dfrac{\sum f_i d_i}{\sum f_i}$$

Step-devi­a­tion method#

When all the classes have the same width $h$, every devi­a­tion $d_i$ is a mul­ti­ple of $h$. Divide by $h$ to get small whole num­bers $\displaystyle u_i = \frac{x_i - a}{h}$. Then

$$\displaystyle \bar{x} = a + h \times \dfrac{\sum f_i u_i}{\sum f_i}$$

Why the three for­mu­las agree#

Write $\displaystyle n = \sum f_i$. Since $d_i = x_i - a$,

$$\displaystyle \begin{aligned} \sum f_i d_i &= \sum f_i (x_i - a) = \sum f_i x_i - a \sum f_i = \sum f_i x_i - a n \\ \dfrac{\sum f_i d_i}{n} &= \dfrac{\sum f_i x_i}{n} - a \\ a + \dfrac{\sum f_i d_i}{n} &= \dfrac{\sum f_i x_i}{n} \end{aligned}$$

So the assumed mean for­mula is exactly the direct for­mula, what­ever value of $a$ you choose. Next, $d_i = h u_i$, so $\displaystyle \sum f_i d_i = h \sum f_i u_i$, and sub­sti­tut­ing gives the step-devi­a­tion for­mula. The meth­ods dif­fer only in the size of the num­bers you mul­ti­ply, never in the answer.

Method in steps#

1. Write the class marks $x_i$ in a new col­umn.
2. Decide on the method. If the $x_i$ are small, use the direct method. If they are large, pick an assumed mean $a$ from the mid­dle of the $x_i$ col­umn.
3. For the assumed mean method, fill the col­umn $d_i = x_i - a$. For step devi­a­tion, also divide by the com­mon width $h$ to get $u_i$.
4. Mul­ti­ply across to fill the $f_i x_i$, $f_i d_i$ or $f_i u_i$ col­umn, keep­ing the signs.
5. Add the $f_i$ col­umn and the prod­uct col­umn.
6. Sub­sti­tute in the for­mula and sim­plify. State the answer with its units.

Worked exam­ples#

Exam­ple 1: one table, three meth­ods#

The marks of 50 stu­dents in a class test (our own data) are grouped below. Find the mean by all three meth­ods.

Direct method:

$$\displaystyle \bar{x} = \dfrac{\sum f_i x_i}{\sum f_i} = \dfrac{2020}{50} = 40.4$$

Assumed mean method with $a = 45$:

$$\displaystyle \bar{x} = 45 + \dfrac{-230}{50} = 45 - 4.6 = 40.4$$

Step-devi­a­tion method with $a = 45$, $h = 10$:

$$\displaystyle \bar{x} = 45 + 10 \times \dfrac{-23}{50} = 45 - 4.6 = 40.4$$

All three give a mean of 40.4 marks. Notice how much smaller the num­bers are in the last col­umn.

Exam­ple 2: large val­ues, step devi­a­tion#

The daily wages of 50 work­ers at a site (our own data) are shown. Find the mean daily wage.

$$\displaystyle \bar{x} = a + h \times \dfrac{\sum f_i u_i}{\sum f_i} = 525 + 50 \times \dfrac{-2}{50} = 525 - 2 = 523$$

The mean daily wage is ₹523. The direct method would have needed prod­ucts like $18 \times 525$; step devi­a­tion avoided them entirely.

Exam­ple 3: one miss­ing fre­quency#

The mean of the dis­tri­b­u­tion below is 52.4. Find the miss­ing fre­quency $p$.

The class marks are 10, 30, 50, 70 and 90. So

$$\displaystyle \begin{aligned} \sum f_i &= 6 + 10 + p + 12 + 8 = 36 + p \\ \sum f_i x_i &= 60 + 300 + 50p + 840 + 720 = 1920 + 50p \end{aligned}$$

Using the direct for­mula,

$$\displaystyle \begin{aligned} \dfrac{1920 + 50p}{36 + p} &= 52.4 \\ 1920 + 50p &= 1886.4 + 52.4p \\ 33.6 &= 2.4p \\ p &= 14 \end{aligned}$$

Check: with $p = 14$, $\displaystyle \sum f_i = 50$ and $\displaystyle \sum f_i x_i = 2620$, and $\displaystyle \frac{2620}{50} = 52.4$. So $p = 14$.

Exam­ple 4: two miss­ing fre­quen­cies#

The dis­tri­b­u­tion below has 40 obser­va­tions and a mean of 24. Find $f_1$ and $f_2$.

From the total: $5 + f_1 + 10 + f_2 + 7 = 40$, so $f_1 + f_2 = 18$.

From the mean, with class marks 5, 15, 25, 35, 45:

$$\begin{aligned} 25 + 15f_1 + 250 + 35f_2 + 315 &= 24 \times 40 \\ 590 + 15f_1 + 35f_2 &= 960 \\ 15f_1 + 35f_2 &= 370 \\ 3f_1 + 7f_2 &= 74 \end{aligned}$$

Put $f_1 = 18 - f_2$:

$$3(18 - f_2) + 7f_2 = 74 \implies 54 + 4f_2 = 74 \implies f_2 = 5$$

so $f_1 = 13$. Check: $\displaystyle \sum f_i x_i = 25 + 195 + 250 + 175 + 315 = 960$ and $\displaystyle \frac{960}{40} = 24$. So $f_1 = 13$, $f_2 = 5$.

Which method should I use?#

• Direct method: when the class marks and fre­quen­cies are small, or when a ques­tion asks you to find a miss­ing fre­quency (the alge­bra is sim­plest here).
• Assumed mean method: when the class marks are large, or the class widths are not all equal.
• Step-devi­a­tion method: when the class marks are large and all classes have the same width. It gives the small­est num­bers and the fewest slips.
• If a ques­tion names a method, you must use that method, even if another is quicker.

Com­mon mis­takes#

• Mul­ti­ply­ing the fre­quency by the lower limit instead of the class mark.
• For­get­ting the neg­a­tive signs in the $d_i$ and $u_i$ columns, or adding them as pos­i­tives.
• Using step devi­a­tion when class widths are unequal; then $u_i$ are not whole num­bers and the short­cut fails.
• Divid­ing by the num­ber of classes instead of by $\displaystyle \sum f_i$.
• In step devi­a­tion, for­get­ting to mul­ti­ply by $h$ at the end: $\displaystyle a + \frac{\sum f_i u_i}{\sum f_i}$ is wrong.
• Not check­ing that the final mean lies inside the range of the data. A mean of 4.04 for Exam­ple 1 would be an obvi­ous error.

Try these#

1. Find the mean of: classes 0–10, 10–20, 20–30, 30–40 with fre­quen­cies 3, 5, 8, 4. Answer: 21.5
2. Find the mean by step devi­a­tion: classes 100–120, 120–140, 140–160, 160–180, 180–200 with fre­quen­cies 4, 6, 10, 6, 4. Answer: 150
3. The mean of classes 10–20, 20–30, 30–40, 40–50 with fre­quen­cies 2, $k$, 6, 4 is 32.5. Find $k$. Answer: $k = 4$
4. Find the mean of: classes 5–15, 15–25, 25–35, 35–45, 45–55 with fre­quen­cies 6, 11, 15, 12, 6. Answer: 30.2
5. Rework Exam­ple 3 with $p = 14$ by the step-devi­a­tion method using $a = 50$, $h = 20$. Answer: 52.4

Key terms#

Class inter­val

A range of val­ues, such as 20–30, into which obser­va­tions are grouped.

Fre­quency

The num­ber of obser­va­tions falling in a class inter­val.

Class mark

The mid-point of a class, found by halv­ing the sum of its lower and upper lim­its.

Class size

The width $h$ of a class, the dif­fer­ence between its upper and lower lim­its.

Assumed mean

A con­ve­nient value $a$, usu­ally a cen­tral class mark, from which devi­a­tions are mea­sured.

Devi­a­tion

The dif­fer­ence $d_i = x_i - a$ between a class mark and the assumed mean.

Step devi­a­tion

The devi­a­tion divided by the class size, $\displaystyle u_i = \frac{x_i - a}{h}$.

Arith­metic mean

The sum of all obser­va­tions divided by their num­ber; for grouped data, $\displaystyle \frac{\sum f_i x_i}{\sum f_i}$.

Com­mon ques­tions#

Does the choice of assumed mean change the answer?#

No. The proof above shows that $a$ can­cels out, so any value gives the same mean. A cen­tral class mark sim­ply keeps the num­bers small.

Why is the mean of grouped data only an esti­mate?#

Because each obser­va­tion is replaced by its class mark. The real val­ues inside a class may lean towards one end, so the true mean can dif­fer slightly from the grouped mean.

Can the assumed mean be a value that is not a class mark?#

Yes, the for­mula works for any $a$. But if $a$ is not a class mark, the $u_i$ will not be whole num­bers, so a class mark is the sen­si­ble choice.

What if the classes are 10–19, 20–29 and so on?#

For the mean, the class mark is still the aver­age of the lim­its, for exam­ple $\displaystyle \frac{10 + 19}{2} = 14.5$. You may con­vert to con­tin­u­ous classes first; the class marks do not change.

Which method is safest in an exam?#

Step devi­a­tion for equal widths and large val­ues, because it has the fewest large prod­ucts. Always add a quick check that the mean lies near the tallest classes.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.
3. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.