This MCQ module is based on: Mode of Grouped Data
TOPIC 32 OF 35
Mode of Grouped Data
🎓 Class 10
Mathematics
CBSE
Theory
Ch 13 — Statistics
⏱ ~25 min
🌐 Language: [gtranslate]
🧠 AI-Powered MCQ Assessment
▲
📐 Maths Assessment
▲
This mathematics assessment will be based on: Mode of Grouped Data
Targeting Class 10 level in Statistics, with Intermediate difficulty.
Upload images, PDFs, or Word documents to include their content in assessment generation.
13.3 Mode of Grouped Data
Recall from Class IX that for ungrouped data, the mode? is the observation that occurs most frequently. For example, in the list 2, 6, 4, 5, 0, 2, 1, 3, 2, 3 the value 2 occurs 3 times — more than any other — so the mode is 2.
But when data is grouped into class intervals, we cannot point to a single observation. Instead we identify the modal class? — the class with the highest frequency — and estimate the mode from it using an interpolation formula obtained by assuming a linear change of frequency on both sides of the modal class.
Formula — 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,
\(h\) = class size (assumed equal),
\(f_1\) = frequency of the modal class,
\(f_0\) = frequency of the class preceding the modal class,
\(f_2\) = frequency of the class succeeding the modal class.
Example 1 — Ages of Hospital Patients
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:
| Family size | 1–3 | 3–5 | 5–7 | 7–9 | 9–11 |
|---|---|---|---|---|---|
| Number of families | 7 | 8 | 2 | 2 | 1 |
Find the mode of this data.
Solution. The highest frequency is 8, so the modal class is 3–5. Here \(l=3\), \(h=2\), \(f_1=8\), \(f_0=7\), \(f_2=2\).
Mode \(=3+\left(\dfrac{8-7}{2(8)-7-2}\right)\times 2 = 3+\dfrac{1}{7}\times 2 = 3+0.286\approx\boxed{3.286}\)
Example 2 — Marks of 80 Students
The marks distribution of 80 students of Class X are shown below. Compare and interpret the modal class of the two measures (mode and mean).
| Marks | 10–25 | 25–40 | 40–55 | 55–70 | 70–85 | 85–100 |
|---|---|---|---|---|---|---|
| Students | 2 | 3 | 7 | 6 | 6 | 6 |
Modal class = 40–55 (highest \(f_1=7\)). \(l=40\), \(h=15\), \(f_0=3\), \(f_2=6\).
Mode \(=40+\left(\dfrac{7-3}{2(7)-3-6}\right)\times 15 = 40+\dfrac{4}{5}\times 15=40+12=\boxed{52}\)
The two dashed lines across the three tallest bars intersect above the modal-class base at the estimated mode.
Interpretation. The mean (62, from Part 1) gives the average performance; the mode (52) tells us the marks most students scored. Both pieces of information together describe the data better than either alone.
Example 3 — Wickets Taken by Bowlers
The distribution below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mode of the data.
| Wickets | 20–60 | 60–100 | 100–150 | 150–250 | 250–350 | 350–450 |
|---|---|---|---|---|---|---|
| Bowlers | 7 | 5 | 16 | 12 | 2 | 3 |
Modal class = 100–150 (\(f_1=16\)). Here \(l=100\), \(h=50\), \(f_0=5\), \(f_2=12\).
Mode \(=100+\dfrac{16-5}{2(16)-5-12}\times 50=100+\dfrac{11}{15}\times 50=100+36.67\approx\boxed{136.67}\)
Although the class widths are unequal, the formula is still used because the modal class has the highest frequency (not frequency density). Students are not expected to adjust for width at this level.
Derivation of the Mode Formula (Optional)
Assume the frequency of the modal class sits between two linear segments: one rising from \((l-h, f_0)\) to \((l, f_1)\) and another falling from \((l+h, f_1)\) to \((l+2h, f_2)\). The peak of the upper triangle inside the modal class — lying at distance \(x\) from \(l\) — is found by setting slopes equal:
\(\dfrac{f_1-f_0}{x}=\dfrac{f_1-f_2}{h-x}\Rightarrow (f_1-f_0)(h-x)=(f_1-f_2)x\)
\((f_1-f_0)h = x[(f_1-f_2)+(f_1-f_0)] = x(2f_1-f_0-f_2)\)
\(\therefore x=\dfrac{(f_1-f_0)h}{2f_1-f_0-f_2}\)
\((f_1-f_0)h = x[(f_1-f_2)+(f_1-f_0)] = x(2f_1-f_0-f_2)\)
\(\therefore x=\dfrac{(f_1-f_0)h}{2f_1-f_0-f_2}\)
Finally \(\text{Mode}=l+x\), giving the boxed formula.
In-text: Why must the denominator \(2f_1-f_0-f_2\) be positive for the formula to make sense? Because \(f_1\) is the largest frequency, both \(f_1-f_0>0\) and \(f_1-f_2>0\); their sum is positive.
Example 4 — States-wise Teacher–Student Ratio
The following distribution gives the state-wise teacher–student ratio in higher secondary schools of India. Find the mode and mean.
| Students / teacher | 15–20 | 20–25 | 25–30 | 30–35 | 35–40 | 40–45 | 45–50 | 50–55 |
|---|---|---|---|---|---|---|---|---|
| States / UT | 3 | 8 | 9 | 10 | 3 | 0 | 0 | 2 |
Mode: modal class = 30–35 (\(f_1=10\)). \(l=30, h=5, f_0=9, f_2=3\).
Mode \(=30+\dfrac{10-9}{2(10)-9-3}\times 5=30+\dfrac{1}{8}\times 5=30+0.625=\mathbf{30.625}\).
Mean (step-deviation, \(a=32.5, h=5\)): class marks 17.5, 22.5, 27.5, 32.5, 37.5, 42.5, 47.5, 52.5; \(u_i\): –3,–2,–1,0,1,2,3,4. \(\sum f_i u_i=3(-3)+8(-2)+9(-1)+10(0)+3(1)+0+0+2(4)=-9-16-9+0+3+8=-23\). \(\sum f_i=35\). \(\bar{x}=32.5+5\cdot(-23/35)\approx 32.5-3.29=\mathbf{29.21}\).
Interpretation: Most states have around 30.6 students per teacher (mode), while the average is 29.2 — the two values being close means the distribution is fairly symmetric around this range.
Mode \(=30+\dfrac{10-9}{2(10)-9-3}\times 5=30+\dfrac{1}{8}\times 5=30+0.625=\mathbf{30.625}\).
Mean (step-deviation, \(a=32.5, h=5\)): class marks 17.5, 22.5, 27.5, 32.5, 37.5, 42.5, 47.5, 52.5; \(u_i\): –3,–2,–1,0,1,2,3,4. \(\sum f_i u_i=3(-3)+8(-2)+9(-1)+10(0)+3(1)+0+0+2(4)=-9-16-9+0+3+8=-23\). \(\sum f_i=35\). \(\bar{x}=32.5+5\cdot(-23/35)\approx 32.5-3.29=\mathbf{29.21}\).
Interpretation: Most states have around 30.6 students per teacher (mode), while the average is 29.2 — the two values being close means the distribution is fairly symmetric around this range.
Example 5 — Lifetimes of Neon Lamps
The following data gives the lifetime (in hours) of 400 neon lamps. Find the modal lifetime.
| Lifetime (h) | 1500–2000 | 2000–2500 | 2500–3000 | 3000–3500 | 3500–4000 | 4000–4500 | 4500–5000 |
|---|---|---|---|---|---|---|---|
| Lamps | 14 | 56 | 60 | 86 | 74 | 62 | 48 |
Modal class = 3000–3500 (\(f_1=86\)). \(l=3000, h=500, f_0=60, f_2=74\).
Mode \(=3000+\dfrac{86-60}{2(86)-60-74}\times 500=3000+\dfrac{26}{38}\times 500\approx 3000+342.10=\mathbf{3342.10}\) h.
Mode \(=3000+\dfrac{86-60}{2(86)-60-74}\times 500=3000+\dfrac{26}{38}\times 500\approx 3000+342.10=\mathbf{3342.10}\) h.
Activity — Finding Modes in Real Class Data
L3 Apply
Materials: Tape, class roll.
Predict: Will the modal class for shoe sizes in your class lie exactly at the class mark, or will the interpolation nudge it away?
- Record the shoe size of every student in your class.
- Group the sizes into classes of width 1 (for example 36–37, 37–38, …).
- Identify the modal class (highest frequency).
- Apply the mode formula and compare the answer with the class mark of the modal class.
The mode lies inside the modal class, usually leaning toward whichever neighbour is taller: if \(f_0>f_2\) the mode sits before the class-mid, if \(f_2>f_0\) the mode sits after.
Competency-Based Questions
Scenario: A student noted the number of cars passing through a spot on a road for 100 periods (each of 3 minutes). The summary is: 0–10 (7), 10–20 (14), 20–30 (13), 30–40 (12), 40–50 (20), 50–60 (11), 60–70 (15), 70–80 (8).
Q1. Compute the mode of the number of cars per 3-minute period.
L3 ApplyModal class = 40–50 (\(f_1=20\)). \(l=40, h=10, f_0=12, f_2=11\). Mode \(= 40 + \dfrac{20-12}{2(20)-12-11}\times 10 = 40 + \dfrac{8}{17}\times 10 \approx 40 + 4.71 = \mathbf{44.71}\) cars.
Q2. Analyse: Which bar(s) of the histogram uniquely decide the numerator \(f_1-f_0\) in the mode formula, and why is it incorrect to use the tallest-but-one bar instead of the immediate neighbour?
L4 AnalyseOnly the modal bar (\(f_1\)) and the bar immediately before it (\(f_0\)) decide the numerator. The formula is built by interpolating the ascending edge directly adjacent to the modal class. Using the tallest-but-one (60–70, f=15) would imply frequencies jump continuously over non-adjacent classes, breaking the linearity assumption of the derivation.
Q3. Evaluate: Two distributions P and Q have the same mean but very different modes. Is this possible? Illustrate or refute using simple 3-class examples.
L5 EvaluateYes. P: freq (1,10,1) over classes 0–10, 10–20, 20–30 — mean ≈ 15, mode ≈ 15. Q: freq (5,2,5) over the same classes — mean = 15 but mode lies in 0–10 (or 20–30, bimodal). Hence equal means can hide very different modal shapes — both measures are needed.
Q4. Create a frequency distribution of 50 observations over 4 classes of width 5, whose mode is exactly the lower boundary of the second class. Explain the structural property of such a distribution.
L6 CreateFor mode = \(l\), the offset \(\dfrac{f_1-f_0}{2f_1-f_0-f_2}\times h\) must equal 0, i.e., \(f_1=f_0\). Example classes 0–5, 5–10, 10–15, 15–20 with frequencies 10, 10, 20, 10: modal class = 10–15, \(f_1=20, f_0=10, f_2=10\); mode = \(10+\dfrac{20-10}{40-10-10}\times 5=10+\dfrac{10}{20}\times 5=12.5\) — not lower boundary. Adjust to 10, 20, 10, 10 making modal class 5–10, \(f_0=f_1=20\) not allowed (then modal class is unique only if \(f_1>f_0\)). A valid construction: 10, 12, 20, 8 — modal class 10–15, \(f_0=12=f_1-8\), etc. Many valid answers: the key property is \(f_0\) close to \(f_1\) while \(f_2\) is much smaller pushes the mode toward the left edge.
Assertion–Reason Questions
Assertion (A): For any grouped data, \(l \le \text{Mode} \le l+h\) where \(l,l+h\) are the limits of the modal class.
Reason (R): The quantity \(\dfrac{f_1-f_0}{2f_1-f_0-f_2}\) always lies between 0 and 1.
Reason (R): The quantity \(\dfrac{f_1-f_0}{2f_1-f_0-f_2}\) always lies between 0 and 1.
(a) — \(f_1-f_0\ge 0\) and \(f_1-f_2\ge 0\), so the fraction \((f_1-f_0)/[(f_1-f_0)+(f_1-f_2)]\in[0,1]\); multiplying by \(h\) keeps the offset inside \([0,h]\).
Assertion (A): If \(f_0=f_2\) then Mode = \(l+h/2\), the class mark of the modal class.
Reason (R): When the two neighbouring classes have equal frequency, the modal peak is symmetric, hence centred.
Reason (R): When the two neighbouring classes have equal frequency, the modal peak is symmetric, hence centred.
(a) — Substituting \(f_0=f_2\) gives \((f_1-f_0)/(2f_1-2f_0)=1/2\), so Mode = \(l+h/2\). R explains the algebra geometrically.
Frequently Asked Questions
What is the mode of grouped data?
The mode is the value that occurs most frequently. For grouped data, we find it using the modal class (class with highest frequency) and the mode formula rather than reading a single observation.
What is the modal class?
The modal class is the class interval with the highest frequency in a grouped frequency distribution. All mode calculations begin with identifying this class.
What is the formula for mode of grouped data?
Mode = l + ((f1 - f0) / (2f1 - f0 - f2)) * h, where l = lower limit of modal class, h = class width, f1 = modal class frequency, f0 = previous class frequency, f2 = next class frequency.
Can a distribution have more than one mode?
Yes. If two classes share the highest frequency the distribution is bimodal; for three, trimodal. In Class 10 problems we usually deal with unimodal distributions.
How is mode different from mean and median?
Mean is the arithmetic average, median is the middle value, and mode is the most frequent value. Each measures central tendency differently and may give different answers for the same data.
When is mode the most useful measure?
Mode is most useful for categorical data or when the 'most common' value matters more than an average - such as shoe sizes sold, most popular exam score, or peak demand.
AI Tutor
Mathematics Class 10
Ready