What are the three methods for finding the mean of grouped data?

The Direct Method (mean = Σfᵢxᵢ/Σfᵢ), the Assumed-Mean Method (pick a class mark as 'a', use the deviations dᵢ=xᵢ-a), and the Step-Deviation Method (also divide those deviations by the class size h). All three always give exactly the same mean -- the choice is purely about which keeps the arithmetic simplest for the numbers in front of you.

What is the modal class, and how is the mode of grouped data found?

The modal class is the class interval with the highest frequency. The mode itself is found using Mode = l + ((f1-f0)/(2f1-f0-f2)) × h, where l is the modal class's lower limit, f1 its frequency, f0 and f2 the frequencies of the classes just before and just after it, and h the class size.

How do you find the median of grouped data?

Locate the median class -- the first class whose cumulative frequency is at least n/2 -- then apply Median = l + ((n/2 - cf)/f) × h, where l is that class's lower limit, cf the cumulative frequency of the class before it, f its own frequency, and h the class size.

Does this chapter cover drawing an ogive (cumulative frequency curve)?

No. The rationalised 2026-27 syllabus drops the old chapter's graphical-ogive section and its worked examples on drawing less-than and more-than curves entirely -- only the cumulative frequency TABLE is used, purely to locate the median class algebraically. The chapter's closing Note to the Reader mentions ogives in one sentence and nothing more.

How many exercises does Class 10 Maths Chapter 13 Statistics have?

Three. Exercise 13.1 (Mean, 9 questions), Exercise 13.2 (Mode, 6 questions) and Exercise 13.3 (Median, 7 questions) -- 22 questions in total, plus 8 worked Examples split 3+3+2 across the three sections.

Class 10 Maths Ch 13 Statistics NCERT Solutions

8 Examples + 22 Questions Solved

Chapter 13: Statistics — NCERT Solutions (rationalised syllabus 2026-27), with all 8 worked Examples from the chapter plus every question of Exercise 13.1, Exercise 13.2 and Exercise 13.3. This chapter extends the mean, median and mode you studied for ungrouped data in Class IX to grouped data — data already condensed into class intervals, where the exact individual values are no longer known and every observation in a class is assumed to sit at that class's mid-point.

Section 13.2 (Examples 1-3, Exercise 13.1) covers the MEAN, by three methods that always agree: Direct, Assumed-Mean, and Step-Deviation. Section 13.3 (Examples 4-6, Exercise 13.2) covers the MODE — the value inside the class of highest frequency, found with a formula rather than by inspection. Section 13.4 (Examples 7-8, Exercise 13.3) covers the MEDIAN, which needs a running CUMULATIVE FREQUENCY column to locate the correct class before the formula can be applied.

Worth knowing before you start: this book's own Note to the Reader points out that the mode and median formulas both require the class intervals to be CONTINUOUS -- an inclusive-type table like '50-52, 53-55, ...' must first be converted by adjusting each boundary by half the gap (see Exercise 13.1 Q5). Also note that the rationalised syllabus does NOT include drawing an ogive (the cumulative-frequency curve) — only the cumulative-frequency TABLE is used here, to algebraically locate the median class.

Key Formulas Used on This Page

Mean of grouped data — three equivalent methods: Direct method: \(\bar{x}=\dfrac{\Sigma f_ix_i}{\Sigma f_i}\). Assumed-mean method (pick any class mark as \(a\), let \(d_i=x_i-a\)): \(\bar{x}=a+\dfrac{\Sigma f_id_i}{\Sigma f_i}\). Step-deviation method (also divide by the class size \(h\), let \(u_i=\dfrac{x_i-a}{h}\)): \(\bar{x}=a+h\!\left(\dfrac{\Sigma f_iu_i}{\Sigma f_i}\right)\). All three always give the SAME mean, whatever class mark is chosen as \(a\).
Mode of grouped data: \(\text{Mode}=l+\left(\dfrac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h\), where \(l\)=lower limit of the modal class (the class of highest frequency), \(h\)=class size, \(f_1\)=frequency of the modal class, \(f_0\)=frequency of the class just before it, \(f_2\)=frequency of the class just after it.
Median of grouped data: \(\text{Median}=l+\left(\dfrac{\frac{n}{2}-\text{cf}}{f}\right)\times h\), where \(l\)=lower limit of the median class (the first class whose cumulative frequency is \(\ge \frac{n}{2}\)), \(n\)=total frequency, cf=cumulative frequency of the class just before the median class, \(f\)=frequency of the median class, \(h\)=class size.
Empirical relationship (a useful cross-check, not exact): \(3\times\text{Median}=\text{Mode}+2\times\text{Mean}\). This is APPROXIMATE, holding closely for roughly symmetric or mildly skewed data -- never treat it as a substitute for computing all three properly.

Worked Examples — Mean of Grouped Data (Examples 1–3)

Example 1

The marks obtained by 30 students of Class X in a Mathematics paper consisting of 100 marks are presented below. Find the mean of the marks obtained by the students, first as ungrouped data, then by converting it to grouped data (class width 15) and finding the mean by all three methods.

Marks obtained (x₀)	10	20	36	40	50	56	60	70	72	80	88	92	95
Number of students (f₀)	1	1	3	4	3	2	4	4	1	1	2	3	1

Solution

Step 1 — as ungrouped data. Multiply each mark by its frequency and add:

\(\Sigma f_ix_i = 1(10)+1(20)+3(36)+4(40)+3(50)+2(56)+4(60)\)

\(\phantom{\Sigma f_ix_i =}\ {}+4(70)+1(72)+1(80)+2(88)+3(92)+1(95)=1779\)

\(\bar{x}=\frac{\Sigma f_ix_i}{\Sigma f_i}=\frac{1779}{30}=59.3\)

So the exact mean, treating every mark individually, is 59.3.

Step 2 — convert to grouped data. Real classroom data is often condensed into class intervals. Splitting the marks into 6 classes of width 15 (10–25, 25–40, ..., 85–100) gives:

Class interval	10-25	25-40	40-55	55-70	70-85	85-100	Total
Number of students	2	3	7	6	6	6	Σf₀=30
Class mark (x₀)	17.5	32.5	47.5	62.5	77.5	92.5
f₀x₀	35.0	97.5	332.5	375.0	465.0	555.0	Σf₀x₀=1860.0

Step 3 — Direct Method. Each class is represented by its class mark (the midpoint), assumed to concentrate the whole class there:

\(\bar{x}=\frac{\Sigma f_ix_i}{\Sigma f_i}=\frac{1860}{30}=62\)

This is 62, not the exact 59.3 — the difference comes entirely from the mid-point assumption. Grouping trades a little accuracy for far less arithmetic on large data sets.

Step 4 — Assumed-Mean Method. Pick any class mark as the "assumed mean" \(a\) — a class near the centre keeps the numbers small. Take \(a=47.5\), and let \(d_i=x_i-a\):

Class interval	10-25	25-40	40-55	55-70	70-85	85-100	Total
f₀	2	3	7	6	6	6	Σf₀=30
x₀	17.5	32.5	47.5	62.5	77.5	92.5
d₀=x₀−47.5	−30	−15	0	15	30	45
f₀d₀	−60	−45	0	90	180	270	Σf₀d₀=435

\(\bar{x}=a+\frac{\Sigma f_id_i}{\Sigma f_i}=47.5+\frac{435}{30}=47.5+14.5=62\)

Step 5 — Step-Deviation Method. Every d₀ above happens to be a multiple of 15 (the class size h), so dividing by h first keeps the numbers even smaller. Let \(u_i=\dfrac{x_i-a}{h}=\dfrac{d_i}{15}\):

Class interval	10-25	25-40	40-55	55-70	70-85	85-100	Total
f₀	2	3	7	6	6	6	Σf₀=30
u₀	−2	−1	0	1	2	3
f₀u₀	−4	−3	0	6	12	18	Σf₀u₀=29

\(\bar{x}=a+h\left(\frac{\Sigma f_iu_i}{\Sigma f_i}\right)=47.5+15\left(\frac{29}{30}\right)=47.5+14.5=62\)

All three methods on the grouped data agree exactly: the mean is 62. (Choosing any OTHER class mark as \(a\) in Step 4/5 — try \(a=62.5\) — still gives 62; the mean never depends on which class was chosen as the assumed mean.)

Remark. The step-deviation method is worth using whenever every \(d_i\) shares a common factor. All three methods are algebraically forced to agree — the assumed-mean and step-deviation formulas are just the direct-method formula rearranged, never a different calculation.

Example 2

The table below gives the percentage distribution of female teachers in the primary schools of rural areas of various states and union territories (U.T.) of India. Find the mean percentage of female teachers by all three methods discussed above.

Percentage of female teachers	15-25	25-35	35-45	45-55	55-65	65-75	75-85
Number of States/U.T.	6	11	7	4	4	2	1

Solution

Step 1 — class marks. The class marks (midpoints) are 20, 30, 40, 50, 60, 70, 80. Take \(a=50\) (a central class) and \(h=10\) (the common class width):

Percentage	15-25	25-35	35-45	45-55	55-65	65-75	75-85	Total
f₀	6	11	7	4	4	2	1	Σf₀=35
x₀	20	30	40	50	60	70	80
d₀=x₀−50	−30	−20	−10	0	10	20	30
u₀=d₀/10	−3	−2	−1	0	1	2	3
f₀x₀	120	330	280	200	240	140	80	Σf₀x₀=1390
f₀d₀	−180	−220	−70	0	40	40	30	Σf₀d₀=−360
f₀u₀	−18	−22	−7	0	4	4	3	Σf₀u₀=−36

Step 2 — Direct Method.

\(\bar{x}=\frac{\Sigma f_ix_i}{\Sigma f_i}=\frac{1390}{35}=39.71\)

Step 3 — Assumed-Mean Method.

\(\bar{x}=a+\frac{\Sigma f_id_i}{\Sigma f_i}=50+\frac{-360}{35}=50-10.29=39.71\)

Step 4 — Step-Deviation Method.

\(\bar{x}=a+h\left(\frac{\Sigma f_iu_i}{\Sigma f_i}\right)=50+10\left(\frac{-36}{35}\right)=50-10.29=39.71\)

All three methods agree: the mean percentage of female teachers in the primary schools of rural areas is 39.71.

Remark. The choice of method depends on how large the numbers are. If \(x_i\) and \(f_i\) are small, the direct method is simplest. If they are large, the assumed-mean or step-deviation method saves arithmetic — and if the class sizes are unequal, the step-deviation method can still be used by picking \(h\) as any convenient common divisor of the \(d_i\)'s, as Example 3 shows next.

Example 3

The distribution below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method. What does the mean signify?

Number of wickets	20-60	60-100	100-150	150-250	250-350	350-450
Number of bowlers	7	5	16	12	2	3

Solution

Step 1 — notice the class sizes are UNEQUAL. The classes are 20–60 (width 40), 60–100 (width 40), 100–150 (width 50), 150–250 (width 100), 250–350 (width 100), 350–450 (width 100). The class marks are still just (lower+upper)/2: 40, 80, 125, 200, 300, 400.

Step 2 — choose the step-deviation method anyway. Even though the class sizes vary, the \(x_i\) themselves are large, so step-deviation is still the easiest route — \(h\) is simply taken as a CONVENIENT common divisor of the deviations, not the (varying) real class width. Take \(a=200\) and \(h=20\):

Class	20-60	60-100	100-150	150-250	250-350	350-450	Total
f₀	7	5	16	12	2	3	Σf₀=45
x₀	40	80	125	200	300	400
d₀=x₀−200	−160	−120	−75	0	100	200
u₀=d₀/20	−8	−6	−3.75	0	5	10
f₀u₀	−56	−30	−60	0	10	30	Σf₀u₀=−106

\(\bar{u}=\frac{-106}{45}\ \Rightarrow\ \bar{x}=a+h\bar{u}=200+20\left(\frac{-106}{45}\right)=200-47.11=152.89\)

So the mean number of wickets taken by these 45 bowlers is 152.89.

Step 3 — what it signifies. This tells us that, on average, a bowler in this one-day list has taken about 153 wickets in his career — a single representative number standing in for the whole spread of 20 to 450 wickets across 45 different bowlers.

Exercise 13.1

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?

Number of plants	0-2	2-4	4-6	6-8	8-10	10-12	12-14
Number of houses	1	2	1	5	6	2	3

Answer

Step 1 — class marks. 1, 3, 5, 7, 9, 11, 13. The numbers here are small, so the Direct Method is the natural choice — there is no real arithmetic to save by shifting to an assumed mean.

Class	0-2	2-4	4-6	6-8	8-10	10-12	12-14	Total
f₀	1	2	1	5	6	2	3	Σf₀=20
x₀	1	3	5	7	9	11	13
f₀x₀	1	6	5	35	54	22	39	Σf₀x₀=162

\(\bar{x}=\frac{\Sigma f_ix_i}{\Sigma f_i}=\frac{162}{20}=8.1\)

The mean number of plants per house is 8.1, found by the Direct Method since the class marks are already small, single-digit numbers.

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.

Daily wages (in ₹)	500-520	520-540	540-560	560-580	580-600
Number of workers	12	14	8	6	10

Answer

Step 1 — class marks are large. 510, 530, 550, 570, 590. Numbers this large make the Direct Method tedious, so the Assumed-Mean Method is more appropriate. Take \(a=550\):

Class	500-520	520-540	540-560	560-580	580-600	Total
f₀	12	14	8	6	10	Σf₀=50
x₀	510	530	550	570	590
d₀=x₀−550	−40	−20	0	20	40
f₀d₀	−480	−280	0	120	400	Σf₀d₀=−240

\(\bar{x}=a+\frac{\Sigma f_id_i}{\Sigma f_i}=550+\frac{-240}{50}=550-4.8=545.2\)

The mean daily wage of the workers is ₹545.20.

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

Daily pocket allowance (in ₹)	11-13	13-15	15-17	17-19	19-21	21-23	23-25
Number of children	7	6	9	13	f	5	4

Answer

Step 1 — set up the direct-method equation. Class marks: 12, 14, 16, 18, 20, 22, 24.

Class	11-13	13-15	15-17	17-19	19-21	21-23	23-25	Total
f₀	7	6	9	13	f	5	4	44+f
x₀	12	14	16	18	20	22	24
f₀x₀	84	84	144	234	20f	110	96	752+20f

Step 2 — solve for f. The mean is given as 18, so:

\(18=\frac{752+20f}{44+f}\)

\(18(44+f)=752+20f \ \Rightarrow\ 792+18f=752+20f\)

\(792-752=20f-18f \ \Rightarrow\ 40=2f \ \Rightarrow\ f=20\)

The missing frequency is f = 20.

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.

Number of heartbeats per minute	65-68	68-71	71-74	74-77	77-80	80-83	83-86
Number of women	2	4	3	8	7	4	2

Answer

Step 1 — class marks. These classes have width 3, so the class marks are 66.5, 69.5, 72.5, 75.5, 78.5, 81.5, 84.5 — not round numbers, but there is no common factor to exploit for step-deviation, so the Direct Method is the most straightforward here.

Class	65-68	68-71	71-74	74-77	77-80	80-83	83-86	Total
f₀	2	4	3	8	7	4	2	Σf₀=30
x₀	66.5	69.5	72.5	75.5	78.5	81.5	84.5
f₀x₀	133	278	217.5	604	549.5	326	169	Σf₀x₀=2277

\(\bar{x}=\frac{\Sigma f_ix_i}{\Sigma f_i}=\frac{2277}{30}=75.9\)

The mean number of heartbeats per minute for these women is 75.9.

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?

Number of mangoes	50-52	53-55	56-58	59-61	62-64
Number of boxes	15	110	135	115	25

Answer

Step 1 — these are INCLUSIVE-type classes. "50-52" and "53-55" have a gap between them (52 and 53 are both included, nothing lies strictly between). To apply the grouped-data formulas, the classes must first be made continuous by subtracting/adding half the gap (0.5) at each boundary:

\(50\text{-}52\to 49.5\text{-}52.5,\ 53\text{-}55\to 52.5\text{-}55.5,\ \ldots,\ 62\text{-}64\to 61.5\text{-}64.5\)

Each continuous class still has width 3, and its midpoint is unchanged (51, 54, 57, 60, 63) — the adjustment only matters for correctly locating boundaries in the mode/median formulas later, not for this mean calculation. The frequencies are large, so use the Step-Deviation Method with \(a=57\), \(h=3\):

Class (continuous)	49.5-52.5	52.5-55.5	55.5-58.5	58.5-61.5	61.5-64.5	Total
f₀	15	110	135	115	25	Σf₀=400
x₀	51	54	57	60	63
u₀=(x₀−57)/3	−2	−1	0	1	2
f₀u₀	−30	−110	0	115	50	Σf₀u₀=25

\(\bar{x}=a+h\left(\frac{\Sigma f_iu_i}{\Sigma f_i}\right)=57+3\left(\frac{25}{400}\right)=57+0.1875=57.19\)

The mean number of mangoes kept in a packing box is 57.19, found by the Step-Deviation Method (the frequencies are large and the deviations share a common factor of 3).

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.

Daily expenditure (in ₹)	100-150	150-200	200-250	250-300	300-350
Number of households	4	5	12	2	2

Answer

Step 1 — class marks are large, and evenly spaced by 50. Use the Step-Deviation Method with \(a=225\), \(h=50\):

Class	100-150	150-200	200-250	250-300	300-350	Total
f₀	4	5	12	2	2	Σf₀=25
x₀	125	175	225	275	325
u₀=(x₀−225)/50	−2	−1	0	1	2
f₀u₀	−8	−5	0	2	4	Σf₀u₀=−7

\(\bar{x}=225+50\left(\frac{-7}{25}\right)=225-14=211\)

The mean daily expenditure on food is ₹211.

To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city, given below. Find the mean concentration of SO₂ in the air.

Concentration of SO₂ (in ppm)	0.00-0.04	0.04-0.08	0.08-0.12	0.12-0.16	0.16-0.20	0.20-0.24
Frequency	4	9	9	2	4	2

Answer

Step 1 — class marks. 0.02, 0.06, 0.10, 0.14, 0.18, 0.22. These decimals are small enough that the Direct Method works cleanly:

Class	0.00-0.04	0.04-0.08	0.08-0.12	0.12-0.16	0.16-0.20	0.20-0.24	Total
f₀	4	9	9	2	4	2	Σf₀=30
x₀	0.02	0.06	0.10	0.14	0.18	0.22
f₀x₀	0.08	0.54	0.90	0.28	0.72	0.44	Σf₀x₀=2.96

\(\bar{x}=\frac{\Sigma f_ix_i}{\Sigma f_i}=\frac{2.96}{30}=0.099\)

The mean concentration of SO₂ in the air is 0.099 ppm.

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.

Number of days	0-6	6-10	10-14	14-20	20-28	28-38	38-40
Number of students	11	10	7	4	4	3	1

Answer

Step 1 — notice the class widths are UNEQUAL (6, 4, 4, 6, 8, 10, 2) — there is no common step-deviation factor worth using, so apply the Direct Method with the true class marks (still just the midpoint of each interval, regardless of width):

Class	0-6	6-10	10-14	14-20	20-28	28-38	38-40	Total
f₀	11	10	7	4	4	3	1	Σf₀=40
x₀	3	8	12	17	24	33	39
f₀x₀	33	80	84	68	96	99	39	Σf₀x₀=499

\(\bar{x}=\frac{\Sigma f_ix_i}{\Sigma f_i}=\frac{499}{40}=12.475\)

The mean number of days a student was absent is 12.475.

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

Literacy rate (in %)	45-55	55-65	65-75	75-85	85-95
Number of cities	3	10	11	8	3

Answer

Step 1 — class marks, evenly spaced by 10. Use the Step-Deviation Method with \(a=70\), \(h=10\):

Class	45-55	55-65	65-75	75-85	85-95	Total
f₀	3	10	11	8	3	Σf₀=35
x₀	50	60	70	80	90
u₀=(x₀−70)/10	−2	−1	0	1	2
f₀u₀	−6	−10	0	8	6	Σf₀u₀=−2

\(\bar{x}=70+10\left(\frac{-2}{35}\right)=70-0.571=69.43\)

The mean literacy rate is 69.43%.

Worked Examples — Mode of Grouped Data (Examples 4–6)

Example 4

The wickets taken by a bowler in 10 cricket matches are as follows: 2, 6, 4, 5, 0, 2, 1, 3, 2, 3. Find the mode of the data.

Solution

Step 1 — recall the mode for ungrouped data. The mode is simply the value that occurs most often. List how many times each value appears:

Number of wickets	0	1	2	3	4	5	6
Number of matches	1	1	3	2	1	1	1

Step 2 — read off the maximum. The value 2 occurs 3 times, more than any other value.

The mode of this data is 2.

Example 5

A survey conducted on 20 households in a locality by a group of students resulted in the following frequency table for the number of family members in a household. Find the mode of this data.

Family size	1-3	3-5	5-7	7-9	9-11
Number of families	7	8	2	2	1

Family size of 20 households (Example 5) — modal class 3–5 highlighted

Solution

Step 1 — locate the modal class. In a grouped distribution we cannot pick out one exact repeated value, only the CLASS with the highest frequency. Here the maximum frequency is 8, in the class 3-5. So the modal class is 3-5.

\(l=3,\ h=2,\ f_1=8\ (\text{modal class}),\ f_0=7\ (\text{class before}),\ f_2=2\ (\text{class after})\)

Step 2 — apply the mode formula.

\(\text{Mode}=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h=3+\left(\frac{8-7}{2(8)-7-2}\right)\times 2\)

\(=3+\left(\frac{1}{7}\right)\times 2=3+\frac{2}{7}=3.286\)

The mode of this data (the most common family size) is 3.286.

Example 6

The marks distribution of 30 students in a mathematics examination are given in Table 13.3 of Example 1. Find the mode of this data. Also compare and interpret the mode and the mean.

Marks distribution of 30 students (Example 1/6 data, Table 13.3) — modal class 40–55 highlighted, with its neighbouring classes f0 and f2

Solution

Step 1 — locate the modal class. Referring back to Table 13.3 (Example 1): the maximum frequency is 7, in the class 40-55. So the modal class is 40-55.

\(l=40,\ h=15,\ f_1=7,\ f_0=3\ (\text{class 25-40, before}),\ f_2=6\ (\text{class 55-70, after})\)

\(\text{Mode}=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h=40+\left(\frac{7-3}{2(7)-3-6}\right)\times 15\)

\(=40+\left(\frac{4}{5}\right)\times 15=40+12=52\)

The mode marks is 52.

Step 2 — compare with the mean. Example 1 already found the mean of this same data (via the direct/assumed-mean/step-deviation methods) to be 62.

\(\text{Mode}=52 \ <\ \text{Mean}=62\)

Interpretation. The maximum number of students (7 of them) obtained marks in the interval containing 52, while the average performance across ALL 30 students was 62. The mode is NOT always less than the mean — for some other data it can equal or exceed the mean — which measure is "better" depends on what question is being asked: the mean if the overall average performance matters, the mode if the most frequently obtained score matters.

Exercise 13.2

The following table shows the ages of the patients admitted in a hospital during a year. Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Age (in years)	5-15	15-25	25-35	35-45	45-55	55-65
Number of patients	6	11	21	23	14	5

Answer

Step 1 — the mode. The maximum frequency is 23, in the class 35-45. So the modal class is 35-45, with \(l=35,\ h=10,\ f_1=23,\ f_0=21,\ f_2=14\):

\(\text{Mode}=35+\left(\frac{23-21}{2(23)-21-14}\right)\times 10=35+\left(\frac{2}{11}\right)\times 10=35+\frac{20}{11}=36.8\)

Step 2 — the mean (Step-Deviation Method, \(a=40,\ h=10\)):

Class	5-15	15-25	25-35	35-45	45-55	55-65	Total
f₀	6	11	21	23	14	5	Σf₀=80
x₀	10	20	30	40	50	60
u₀=(x₀−40)/10	−3	−2	−1	0	1	2
f₀u₀	−18	−22	−21	0	14	10	Σf₀u₀=−37

\(\bar{x}=40+10\left(\frac{-37}{80}\right)=40-4.625=35.375\)

Interpretation. Mode ≈ 36.8 years, Mean = 35.375 years — the two are very close. This tells us that the age group most commonly admitted (the mode) is very near the average age of ALL patients admitted (the mean): most hospital admissions in this data are of patients around 35–37 years old.

The following data gives the information on the observed lifetimes (in hours) of 225 electrical components. Determine the modal lifetimes of the components.

Lifetimes (in hours)	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	10	35	52	61	38	29

Answer

Step 1 — locate the modal class. The maximum frequency is 61, in the class 60-80. So \(l=60,\ h=20,\ f_1=61,\ f_0=52,\ f_2=38\):

\(\text{Mode}=60+\left(\frac{61-52}{2(61)-52-38}\right)\times 20=60+\left(\frac{9}{32}\right)\times 20=60+5.625=65.625\)

The modal lifetime of the components is 65.625 hours.

The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.

Expenditure (in ₹)	1000-1500	1500-2000	2000-2500	2500-3000	3000-3500	3500-4000	4000-4500	4500-5000
Number of families	24	40	33	28	30	22	16	7

Answer

Step 1 — the mode. The maximum frequency is 40, in the class 1500-2000. So \(l=1500,\ h=500,\ f_1=40,\ f_0=24,\ f_2=33\):

\(\text{Mode}=1500+\left(\frac{40-24}{2(40)-24-33}\right)\times 500=1500+\left(\frac{16}{23}\right)\times 500=1500+347.83=1847.83\)

The modal monthly expenditure is ₹1847.83.

Step 2 — the mean (Step-Deviation Method, \(a=2750,\ h=500\)):

Class	1000-1500	1500-2000	2000-2500	2500-3000	3000-3500	3500-4000	4000-4500	4500-5000	Total
f₀	24	40	33	28	30	22	16	7	Σf₀=200
u₀	−3	−2	−1	0	1	2	3	4
f₀u₀	−72	−80	−33	0	30	44	48	28	Σf₀u₀=−35

\(\bar{x}=2750+500\left(\frac{-35}{200}\right)=2750-87.5=2662.5\)

The mean monthly expenditure is ₹2662.50.

The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Students per teacher	15-20	20-25	25-30	30-35	35-40	40-45	45-50	50-55
Number of States/U.T.	3	8	9	10	3	0	0	2

Answer

Step 1 — the mode. The maximum frequency is 10, in the class 30-35. So \(l=30,\ h=5,\ f_1=10,\ f_0=9,\ f_2=3\):

\(\text{Mode}=30+\left(\frac{10-9}{2(10)-9-3}\right)\times 5=30+\left(\frac{1}{8}\right)\times 5=30+0.625=30.625\)

Step 2 — the mean (Step-Deviation Method, \(a=32.5,\ h=5\)):

Class	15-20	20-25	25-30	30-35	35-40	40-45	45-50	50-55	Total
f₀	3	8	9	10	3	0	0	2	Σf₀=35
u₀	−3	−2	−1	0	1	2	3	4
f₀u₀	−9	−16	−9	0	3	0	0	8	Σf₀u₀=−23

\(\bar{x}=32.5+5\left(\frac{-23}{35}\right)=32.5-3.286=29.214\)

Interpretation. Mode ≈ 30.6 students/teacher, Mean ≈ 29.2 students/teacher — close together. Most states/U.T.s have a teacher-student ratio of around 30–31 students per teacher (the mode), which is also close to the national average across all states (the mean).

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Find the mode of the data.

Runs scored	3000-4000	4000-5000	5000-6000	6000-7000	7000-8000	8000-9000	9000-10000	10000-11000
Number of batsmen	4	18	9	7	6	3	1	1

Answer

Step 1 — locate the modal class. The maximum frequency is 18, in the class 4000-5000. So \(l=4000,\ h=1000,\ f_1=18,\ f_0=4,\ f_2=9\):

\(\text{Mode}=4000+\left(\frac{18-4}{2(18)-4-9}\right)\times 1000=4000+\left(\frac{14}{23}\right)\times 1000\)

\(=4000+608.7=4608.7\)

The modal number of runs scored is 4608.7.

A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data.

Number of cars	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	7	14	13	12	20	11	15	8

Answer

Step 1 — locate the modal class. The maximum frequency is 20, in the class 40-50. So \(l=40,\ h=10,\ f_1=20,\ f_0=12,\ f_2=11\):

\(\text{Mode}=40+\left(\frac{20-12}{2(20)-12-11}\right)\times 10=40+\left(\frac{8}{17}\right)\times 10=40+4.706=44.7\)

The modal number of cars is 44.7.

Worked Examples — Median of Grouped Data (Examples 7–8)

Example 7

A survey regarding the heights (in cm) of 51 girls of Class X of a school was conducted and the following data was obtained. Find the median height.

Height (in cm)	Less than 140	Less than 145	Less than 150	Less than 155	Less than 160	Less than 165
Number of girls	4	11	29	40	46	51

Less-than ogive plotted from the printed cumulative-frequency table (Example 7, heights of 51 girls) — the book gives this table but never draws the curve itself; median read off the crosshair matches the algebraic answer below.

Solution

Step 1 — this is CUMULATIVE-frequency data of the "less than" type, not plain class frequencies. "Less than 140" gives the upper limits directly, so the actual classes are below 140, 140-145, 145-150, ..., 160-165. Recover the ordinary frequency of each class by successive subtraction:

Class	Below 140	140-145	145-150	150-155	155-160	160-165	Total
Frequency	4	11−4=7	29−11=18	40−29=11	46−40=6	51−46=5	n=51
Cumulative frequency	4	11	29	40	46	51

Step 2 — locate the median class.

\(n=51 \ \Rightarrow\ \frac{n}{2}=25.5\)

Find the first class whose cumulative frequency is greater than (and nearest to) 25.5: that is 145-150, with cumulative frequency 29. So the median class is 145-150. Its lower boundary \(l=145\), class size \(h=5\), and frequency \(f=18\). The cumulative frequency of the class before the median class (i.e. of 140-145) is \(\text{cf}=11\).

Step 3 — apply the median formula.

\(\text{Median}=l+\left(\frac{\frac{n}{2}-\text{cf}}{f}\right)\times h=145+\left(\frac{25.5-11}{18}\right)\times 5\)

\(=145+\frac{72.5}{18}=145+4.03=149.03\)

The median height of the girls is 149.03 cm. This means about half the girls are shorter than 149.03 cm and the other half are taller. (The ogive figure above plots this same printed cumulative-frequency table and reads the same median off the crosshair — two independent routes to the same 149.03.)

Example 8

The median of the following data is 525. Find the values of x and y, if the total frequency is 100.

Class intervals	0-100	100-200	200-300	300-400	400-500	500-600	600-700	700-800	800-900	900-1000
Frequency	2	5	x	12	17	20	y	9	7	4

Solution

Step 1 — build the cumulative-frequency column in terms of x and y.

Class	0-100	100-200	200-300	300-400	400-500	500-600	600-700	700-800	800-900	900-1000
f	2	5	x	12	17	20	y	9	7	4
cf	2	7	7+x	19+x	36+x	56+x	56+x+y	65+x+y	72+x+y	76+x+y

Step 2 — use the given total.

\(n=100 \ \Rightarrow\ 76+x+y=100 \ \Rightarrow\ x+y=24 \ (1)\)

Step 3 — the median (525) lies in the class 500-600.

\(l=500,\ f=20,\ \text{cf}=36+x,\ h=100,\ \frac{n}{2}=50\)

\(525=500+\left(\frac{50-(36+x)}{20}\right)\times 100\)

\(25=(14-x)\times 5 \ \Rightarrow\ 5=14-x \ \Rightarrow\ x=9\)

Step 4 — substitute back into (1).

\(9+y=24 \ \Rightarrow\ y=15\)

So x = 9 and y = 15.

Exercise 13.3

The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption (in units)	65-85	85-105	105-125	125-145	145-165	165-185	185-205
Number of consumers	4	5	13	20	14	8	4

Answer

Step 1 — the median. Cumulative frequencies: 4, 9, 22, 42, 56, 64, 68. Here \(n=68\), so \(\frac{n}{2}=34\). The first cumulative frequency exceeding 34 is 42, in the class 125-145. So \(l=125,\ \text{cf}=22,\ f=20,\ h=20\):

\(\text{Median}=125+\left(\frac{34-22}{20}\right)\times 20=125+12=137\)

Step 2 — the mean (Step-Deviation Method, \(a=135,\ h=20\)):

Class	65-85	85-105	105-125	125-145	145-165	165-185	185-205	Total
f₀	4	5	13	20	14	8	4	Σf₀=68
x₀	75	95	115	135	155	175	195
u₀	−3	−2	−1	0	1	2	3
f₀u₀	−12	−10	−13	0	14	16	12	Σf₀u₀=7

\(\bar{x}=135+20\left(\frac{7}{68}\right)=135+2.06=137.06\)

Step 3 — the mode. Maximum frequency is 20, in the class 125-145. So \(l=125,\ f_1=20,\ f_0=13,\ f_2=14,\ h=20\):

\(\text{Mode}=125+\left(\frac{20-13}{2(20)-13-14}\right)\times 20=125+\left(\frac{7}{13}\right)\times 20=125+10.77=135.77\)

Comparison. Median ≈ 137, Mean ≈ 137.06, Mode ≈ 135.77 — all three measures are very close, which tells us this consumption data is fairly evenly spread with no extreme values pulling any one measure away from the others.

If the median of the distribution given below is 28.5, find the values of x and y.

Answer

(Table: class intervals 0-10, 10-20, 20-30, 30-40, 40-50, 50-60 with frequencies 5, x, 20, 15, y, 5; total frequency 60.)

Step 1 — use the given total.

\(5+x+20+15+y+5=60 \ \Rightarrow\ x+y=15 \ (1)\)

Step 2 — the median 28.5 lies in the class 20-30.

\(l=20,\ f=20,\ \text{cf}=5+x,\ h=10,\ \frac{n}{2}=30\)

\(28.5=20+\left(\frac{30-(5+x)}{20}\right)\times 10\)

\(8.5=(25-x)\times \frac{1}{2} \ \Rightarrow\ 17=25-x \ \Rightarrow\ x=8\)

Step 3 — substitute back into (1).

\(8+y=15 \ \Rightarrow\ y=7\)

So x = 8 and y = 7.

A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

Age (in years)	Below 20	Below 25	Below 30	Below 35	Below 40	Below 45	Below 50	Below 55	Below 60
Number of policy holders	2	6	24	45	78	89	92	98	100

Answer

Step 1 — recover the class frequencies. Since policies start from age 18, the first class is 18-20, not 15-20 or 0-20:

Class	18-20	20-25	25-30	30-35	35-40	40-45	45-50	50-55	55-60	Total
Frequency	2	6−2=4	24−6=18	45−24=21	78−45=33	89−78=11	92−89=3	98−92=6	100−98=2	n=100
Cumulative frequency	2	6	24	45	78	89	92	98	100

Step 2 — locate the median class.

\(\frac{n}{2}=50\)

The first cumulative frequency exceeding 50 is 78, in the class 35-40. So \(l=35,\ \text{cf}=45,\ f=33,\ h=5\):

\(\text{Median}=35+\left(\frac{50-45}{33}\right)\times 5=35+\frac{25}{33}=35+0.7576=35.76\)

The median age of the policy holders is 35.76 years.

The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table. Find the median length of the leaves. (Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5-126.5, 126.5-135.5, ..., 171.5-180.5.)

Length (in mm)	118-126	127-135	136-144	145-153	154-162	163-171	172-180
Number of leaves	3	5	9	12	5	4	2

Answer

Step 1 — convert to continuous classes. As the hint says, subtract 0.5 from every lower limit and add 0.5 to every upper limit: 117.5-126.5, 126.5-135.5, 135.5-144.5, 144.5-153.5, 153.5-162.5, 162.5-171.5, 171.5-180.5 (each still 9 units wide).

Class (continuous)	117.5-126.5	126.5-135.5	135.5-144.5	144.5-153.5	153.5-162.5	162.5-171.5	171.5-180.5	Total
Frequency	3	5	9	12	5	4	2	n=40
Cumulative frequency	3	8	17	29	34	38	40

Step 2 — locate the median class.

\(\frac{n}{2}=20\)

The first cumulative frequency exceeding 20 is 29, in the class 144.5-153.5. So \(l=144.5,\ \text{cf}=17,\ f=12,\ h=9\):

\(\text{Median}=144.5+\left(\frac{20-17}{12}\right)\times 9=144.5+2.25=146.75\)

The median length of the leaves is 146.75 mm.

The following table gives the distribution of the life-time of 400 neon lamps. Find the median life time of a lamp.

Life time (in hours)	1500-2000	2000-2500	2500-3000	3000-3500	3500-4000	4000-4500	4500-5000
Number of lamps	14	56	60	86	74	62	48

Answer

Step 1 — locate the median class. Cumulative frequencies: 14, 70, 130, 216, 290, 352, 400. Here \(n=400\), so \(\frac{n}{2}=200\). The first cumulative frequency exceeding 200 is 216, in the class 3000-3500. So \(l=3000,\ \text{cf}=130,\ f=86,\ h=500\):

\(\text{Median}=3000+\left(\frac{200-130}{86}\right)\times 500=3000+\frac{70}{86}\times 500=3000+406.98=3406.98\)

The median life time of a lamp is 3406.98 hours.

100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows. Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.

Number of letters	1-4	4-7	7-10	10-13	13-16	16-19
Number of surnames	6	30	40	16	4	4

Answer

Step 1 — the median. Cumulative frequencies: 6, 36, 76, 92, 96, 100. Here \(n=100\), so \(\frac{n}{2}=50\). The first cumulative frequency exceeding 50 is 76, in the class 7-10. So \(l=7,\ \text{cf}=36,\ f=40,\ h=3\):

\(\text{Median}=7+\left(\frac{50-36}{40}\right)\times 3=7+\frac{14}{40}\times 3=7+1.05=8.05\)

Step 2 — the mean (Step-Deviation Method, \(a=8.5,\ h=3\)):

Class	1-4	4-7	7-10	10-13	13-16	16-19	Total
f₀	6	30	40	16	4	4	Σf₀=100
x₀	2.5	5.5	8.5	11.5	14.5	17.5
u₀	−2	−1	0	1	2	3
f₀u₀	−12	−30	0	16	8	12	Σf₀u₀=−6

\(\bar{x}=8.5+3\left(\frac{-6}{100}\right)=8.5-0.18=8.32\)

Step 3 — the mode. Maximum frequency is 40, in the class 7-10. So \(l=7,\ f_1=40,\ f_0=30,\ f_2=16,\ h=3\):

\(\text{Mode}=7+\left(\frac{40-30}{2(40)-30-16}\right)\times 3=7+\left(\frac{10}{34}\right)\times 3=7+0.882=7.88\)

Median ≈ 8.05, Mean = 8.32, Mode ≈ 7.88 letters — a typical surname in this data has about 7–8 letters.

The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Weight (in kg)	40-45	45-50	50-55	55-60	60-65	65-70	70-75
Number of students	2	3	8	6	6	3	2

Answer

Step 1 — locate the median class. Cumulative frequencies: 2, 5, 13, 19, 25, 28, 30. Here \(n=30\), so \(\frac{n}{2}=15\). The first cumulative frequency exceeding 15 is 19, in the class 55-60. So \(l=55,\ \text{cf}=13,\ f=6,\ h=5\):

\(\text{Median}=55+\left(\frac{15-13}{6}\right)\times 5=55+\frac{10}{6}=55+1.67=56.67\)

The median weight of the students is 56.67 kg.

Chapter 12: Surface Areas and Volumes

All Chapters

Chapter 14: Probability

Statistics

Key Formulas Used on This Page

Worked Examples — Mean of Grouped Data (Examples 1–3)

Exercise 13.1

Worked Examples — Mode of Grouped Data (Examples 4–6)

Exercise 13.2

Worked Examples — Median of Grouped Data (Examples 7–8)

Exercise 13.3

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