Mode of Grouped Data Class 10: Formula and Examples

The for­mula#

$$\displaystyle \text{Mode} = l + \left( \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$$

where

• $l$ = lower limit of the modal class,
• $h$ = size (width) of the class inter­vals,
• $f_1$ = fre­quency of the modal class,
• $f_0$ = fre­quency of the class just before the modal class,
• $f_2$ = fre­quency of the class just after the modal class.

If the modal class is the first class, there is no class before it, so $f_0 = 0$. If it is the last class, $f_2 = 0$.

Why the for­mula makes sense#

The denom­i­na­tor can be writ­ten as $(f_1 - f_0) + (f_1 - f_2)$. So the frac­tion

$$\displaystyle \dfrac{f_1 - f_0}{(f_1 - f_0) + (f_1 - f_2)}$$

com­pares how much the modal class rises above its left neigh­bour with the total rise above both neigh­bours. If both neigh­bours are equal ($f_0 = f_2$), the frac­tion is $\displaystyle \frac{1}{2}$ and the mode is the exact mid­dle of the class. If the class after is much smaller than the class before, the frac­tion is less than $\displaystyle \frac{1}{2}$ and the mode is pulled towards the left end, towards the big­ger neigh­bour. The mode always leans towards the heav­ier side.

There is also a neat pic­ture. On the his­togram, join the top-left cor­ner of the modal bar to the top-left cor­ner of the next bar, and the top-right cor­ner of the modal bar to the top-right cor­ner of the pre­vi­ous bar. These two lines cross at a point whose hor­i­zon­tal posi­tion is exactly the mode given by the for­mula. Tak­ing the left edge of the modal class as posi­tion $0$ and writ­ing $t$ for the frac­tion of the width, the first line has height $f_1 + t(f_2 - f_1)$ and the sec­ond has height $f_0 + t(f_1 - f_0)$. Set­ting them equal:

$$\displaystyle \begin{aligned} f_1 + t(f_2 - f_1) &= f_0 + t(f_1 - f_0) \\ f_1 - f_0 &= t\big[(f_1 - f_0) + (f_1 - f_2)\big] \\ t &= \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \end{aligned}$$

So the cross­ing point is at $l + t \times h$, which is the for­mula.

Method in steps#

1. Check that the classes are con­tin­u­ous and of equal width. If the classes are writ­ten as 10–19, 20–29, con­vert them to 9.5–19.5, 19.5–29.5 and so on.
2. Find the largest fre­quency. Its class is the modal class; write down $l$ and $h$.
3. Read $f_1$, then $f_0$ from the row above and $f_2$ from the row below. Use $0$ if a neigh­bour does not exist.
4. Work out $f_1 - f_0$ and $2f_1 - f_0 - f_2$ sep­a­rately.
5. Sub­sti­tute, sim­plify, and check that the answer lies between $l$ and $l + h$.

Worked exam­ples#

Exam­ple 1: a stan­dard table#

Find the mode of the marks of 50 stu­dents (our own data).

The largest fre­quency is 14, so the modal class is 40–50. Hence $l = 40$, $h = 10$, $f_1 = 14$, $f_0 = 12$, $f_2 = 7$.

$$\displaystyle \begin{aligned} \text{Mode} &= 40 + \dfrac{14 - 12}{2(14) - 12 - 7} \times 10 \\ &= 40 + \dfrac{2}{9} \times 10 = 40 + \dfrac{20}{9} \\ &\approx 40 + 2.22 = 42.22 \end{aligned}$$

The mode is about 42.22 marks: more stu­dents scored around 42 than around any other mark. It lies in 40–50, as it must, and is pulled slightly left because the class before (12) is much larger than the class after (7).

Exam­ple 2: the modal class is the first class#

A clinic records how long 24 patients waited (our own data). Find the modal wait­ing time.

The largest fre­quency, 9, is in the first class 0 to 5. There is no class before it, so $f_0 = 0$. We have $l = 0$, $h = 5$, $f_1 = 9$, $f_2 = 6$.

$$\displaystyle \begin{aligned} \text{Mode} &= 0 + \dfrac{9 - 0}{2(9) - 0 - 6} \times 5 \\ &= \dfrac{9}{12} \times 5 = \dfrac{45}{12} = 3.75 \end{aligned}$$

The modal wait­ing time is 3.75 min­utes. For com­par­i­son, the mean of this table is $\displaystyle \frac{215}{24} \approx 8.96$ min­utes, much larger, because a few patients waited a long time. When data is bunched at one end like this, the mode describes the typ­i­cal patient bet­ter than the mean.

Exam­ple 3: wages, and com­par­ing mode with mean#

For the daily wages of 50 work­ers below (our own data), the mean is ₹523. Find the mode and com­pare the two.

Modal class 500–550: $l = 500$, $h = 50$, $f_1 = 18$, $f_0 = 12$, $f_2 = 10$.

$$\displaystyle \begin{aligned} \text{Mode} &= 500 + \dfrac{18 - 12}{36 - 12 - 10} \times 50 \\ &= 500 + \dfrac{6}{14} \times 50 = 500 + \dfrac{300}{14} \\ &\approx 500 + 21.43 = 521.43 \end{aligned}$$

The modal wage is about ₹521.43, while the mean is ₹523. They are close because the table is nearly bal­anced on both sides of the modal class. The mode tells us the most com­mon wage; the mean tells us what each worker would get if the total wage bill were shared equally.

Exam­ple 4: miss­ing fre­quency from the mode#

The mode of the dis­tri­b­u­tion below is 25. Find $x$.

The mode 25 lies in 20–30, so this is the modal class: $l = 20$, $h = 10$, $f_1 = 15$, $f_0 = x$, $f_2 = 10$.

$$\displaystyle \begin{aligned} 20 + \dfrac{15 - x}{30 - x - 10} \times 10 &= 25 \\ \dfrac{15 - x}{20 - x} &= \dfrac{1}{2} \\ 30 - 2x &= 20 - x \\ x &= 10 \end{aligned}$$

Check: $x = 10$ is less than 15, so 20–30 really is the modal class, and since $f_0 = f_2 = 10$ the mode is the mid­dle of the class, 25. So $x = 10$.

Inter­pret­ing the mode, and mean ver­sus mode#

• The mode answers "what is most typ­i­cal or most pop­u­lar?". It is the nat­ural aver­age for sizes, choices and busiest times.
• The mean uses every obser­va­tion, so a few very large or very small val­ues pull it towards them. The mode is not affected by such extreme val­ues.
• When the data is roughly sym­met­ric, the mean and mode are close (Exam­ple 3). When the data is bunched at one end, they can be far apart (Exam­ple 2).
• A dis­tri­b­u­tion can have two classes with the same high­est fre­quency. Then it is called bimodal, and the sin­gle-mode for­mula should not be forced onto it.

Com­mon mis­takes#

• Tak­ing the class with the largest class mark, or the mid­dle row of the table, as the modal class instead of the one with the largest fre­quency.
• Swap­ping $f_0$ and $f_2$. $f_0$ is always the class before the modal class.
• Writ­ing the denom­i­na­tor as $2f_1 - f_0 + f_2$. Both neigh­bours are sub­tracted.
• For­get­ting that $f_0 = 0$ when the modal class is the first class (or $f_2 = 0$ when it is the last).
• Using the upper limit of the modal class as $l$.
• Get­ting a mode out­side the modal class and not notic­ing the error.

Try these#

1. Find the mode: classes 10 to 20, 20–30, 30–40, 40–50, 50–60 with fre­quen­cies 3, 7, 12, 8, 5. Answer: about 35.56
2. Find the mode: classes 0 to 4, 4–8, 8–12, 12–16 with fre­quen­cies 12, 8, 5, 3. Answer: 3
3. Find the mode: classes 60 to 70, 70–80, 80–90, 90–100, 100–110 with fre­quen­cies 4, 9, 16, 11, 5. Answer: about 85.83
4. Find the mode: classes 0 to 10, 10–20, 20–30, 30–40 with fre­quen­cies 3, 5, 8, 11. Answer: about 32.14

Key terms#

Mode

The value that occurs most often in a set of data.

Modal class

The class inter­val with the high­est fre­quency in a grouped dis­tri­b­u­tion.

$l$

The lower limit of the modal class.

$f_1$

The fre­quency of the modal class.

$f_0$

The fre­quency of the class just before the modal class, or $0$ if there is none.

$f_2$

The fre­quency of the class just after the modal class, or $0$ if there is none.

Class size

The com­mon width $h$ of the class inter­vals.

Bimodal

Describes data with two val­ues or classes shar­ing the high­est fre­quency.

Com­mon ques­tions#

Can the mode for­mula give a value out­side the modal class?#

No. Since $f_1$ is the largest fre­quency, both $f_1 - f_0$ and $f_1 - f_2$ are non-neg­a­tive, so the frac­tion lies between 0 and 1 and the mode lies between $l$ and $l + h$.

What if two classes have the same high­est fre­quency?#

The data is bimodal and the for­mula does not pick a sin­gle answer. Class 10 ques­tions are set so that there is one clear modal class.

Do the class widths have to be equal?#

The for­mula assumes the modal class and its neigh­bours have the same width $h$. Board ques­tions use equal widths.

Is the mode always less than the mean?#

No. It depends on the shape of the data. In Exam­ple 2 the mode is far below the mean; if most val­ues were near the top end, the mode could be above the mean.

Why does the mode not depend on the other classes?#

The for­mula looks only at the modal class and its two neigh­bours, because the mode is about where val­ues are most crowded, not about the total or the extremes.

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.