This MCQ module is based on: Mean of Grouped Data
TOPIC 31 OF 35
Mean of Grouped Data
🎓 Class 10
Mathematics
CBSE
Theory
Ch 13 — Statistics
⏱ ~30 min
🌐 Language: [gtranslate]
🧠 AI-Powered MCQ Assessment
▲
📐 Maths Assessment
▲
This mathematics assessment will be based on: Mean 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.
13.1 Introduction
In Class IX, you studied the classification of ungrouped and grouped frequency distributions?, and you learned to find measures of central tendency — mean, median and mode — from ungrouped data. In real life however, the data we collect is usually large. Handling every observation one-by-one becomes impractical, so it is grouped into class intervals. In this chapter we extend those three measures to grouped frequency distributions and also learn to draw conclusions from them.
13.2 Mean of Grouped Data
Recall that for ungrouped data with observations \(x_1, x_2, \ldots, x_n\) occurring with frequencies \(f_1, f_2, \ldots, f_n\), the arithmetic mean is
\(\bar{x} = \dfrac{f_1 x_1 + f_2 x_2 + \cdots + f_n x_n}{f_1 + f_2 + \cdots + f_n} = \dfrac{\sum f_i x_i}{\sum f_i}\)
The Greek letter \(\Sigma\) (sigma) is short for "summation". When the data is organised into class intervals, the individual values \(x_i\) of every observation inside a class are not known. To get past this, we assume that all values inside a class are concentrated at the mid-point of that class, called the class mark?.
Definition — Class Mark
For a class whose lower limit is \(l\) and upper limit is \(u\),
\[ \text{Class mark} = \dfrac{\text{Upper limit} + \text{Lower limit}}{2} = \dfrac{l+u}{2} \]
Example 1 — Marks of 30 Students (Direct Method)
The table below gives marks obtained (out of 100) by 30 students of Class X in a Mathematics test. We want to find the mean of this data.
| Class interval | 10–25 | 25–40 | 40–55 | 55–70 | 70–85 | 85–100 |
|---|---|---|---|---|---|---|
| Number of students (\(f_i\)) | 2 | 3 | 7 | 6 | 6 | 6 |
Step 1. Find the class mark \(x_i\) of each class: \(\tfrac{10+25}{2}=17.5\), \(\tfrac{25+40}{2}=32.5\), and so on.
Step 2. Multiply each class mark by its frequency to get \(f_i x_i\).
Step 3. Add a "Total" row.
| Class | \(f_i\) | \(x_i\) | \(f_i x_i\) |
|---|---|---|---|
| 10–25 | 2 | 17.5 | 35.0 |
| 25–40 | 3 | 32.5 | 97.5 |
| 40–55 | 7 | 47.5 | 332.5 |
| 55–70 | 6 | 62.5 | 375.0 |
| 70–85 | 6 | 77.5 | 465.0 |
| 85–100 | 6 | 92.5 | 555.0 |
| Total | \(\sum f_i=30\) | — | \(\sum f_i x_i=1860\) |
Therefore, \[ \bar{x} = \dfrac{\sum f_i x_i}{\sum f_i} = \dfrac{1860}{30} = 62 \] So the mean marks obtained by a student is 62.
Bar diagram of the frequency distribution. The modal class (tallest bar) is 40–55.
Direct Method — Summary
When the class marks \(x_i\) and frequencies \(f_i\) are small numbers, use
\(\bar{x}=\dfrac{\sum f_i x_i}{\sum f_i}\). This is called the direct method.
Example 2 — Assumed Mean Method
The same data can be handled more briskly. When the \(x_i\) values are large, multiplying them with \(f_i\) gives very large numbers. A convenient short-cut is to subtract a suitable number \(a\) (called the assumed mean?) from each \(x_i\), work with the smaller numbers \(d_i = x_i - a\), and adjust at the end.
| Class | \(f_i\) | \(x_i\) | \(d_i=x_i-47.5\) | \(f_i d_i\) |
|---|---|---|---|---|
| 10–25 | 2 | 17.5 | –30 | –60 |
| 25–40 | 3 | 32.5 | –15 | –45 |
| 40–55 | 7 | 47.5 | 0 | 0 |
| 55–70 | 6 | 62.5 | 15 | 90 |
| 70–85 | 6 | 77.5 | 30 | 180 |
| 85–100 | 6 | 92.5 | 45 | 270 |
| Total | 30 | — | — | \(\sum f_i d_i=435\) |
The mean of deviations \(\bar{d}=\dfrac{\sum f_i d_i}{\sum f_i}=\dfrac{435}{30}=14.5\).
Derivation of the relation between \(\bar{x}\) and \(\bar{d}\):
\(\bar{d}=\dfrac{\sum f_i d_i}{\sum f_i}=\dfrac{\sum f_i(x_i-a)}{\sum f_i}=\dfrac{\sum f_i x_i}{\sum f_i}-a\cdot\dfrac{\sum f_i}{\sum f_i}=\bar{x}-a\)
\(\therefore\ \boxed{\bar{x}=a+\bar{d}=a+\dfrac{\sum f_i d_i}{\sum f_i}}\)
\(\therefore\ \boxed{\bar{x}=a+\bar{d}=a+\dfrac{\sum f_i d_i}{\sum f_i}}\)
With \(a=47.5\): \(\bar{x}=47.5+14.5=62\). Same answer as the direct method — but the arithmetic is far lighter.
Example 3 — Step-Deviation Method
Notice that every \(d_i\) in the table above is a multiple of the common class-width \(h=15\). We can reduce the numbers further by dividing each \(d_i\) by \(h\). Define
\(u_i=\dfrac{x_i-a}{h}\)
Then
\(\bar{u}=\dfrac{\sum f_i u_i}{\sum f_i}=\dfrac{1}{h}\cdot\dfrac{\sum f_i(x_i-a)}{\sum f_i}=\dfrac{1}{h}(\bar{x}-a)\)
\(\Rightarrow h\bar{u}=\bar{x}-a\) i.e. \(\boxed{\bar{x}=a+h\bar{u}=a+h\left(\dfrac{\sum f_i u_i}{\sum f_i}\right)}\)
\(\Rightarrow h\bar{u}=\bar{x}-a\) i.e. \(\boxed{\bar{x}=a+h\bar{u}=a+h\left(\dfrac{\sum f_i u_i}{\sum f_i}\right)}\)
| Class | \(f_i\) | \(x_i\) | \(u_i=\tfrac{x_i-47.5}{15}\) | \(f_i u_i\) |
|---|---|---|---|---|
| 10–25 | 2 | 17.5 | –2 | –4 |
| 25–40 | 3 | 32.5 | –1 | –3 |
| 40–55 | 7 | 47.5 | 0 | 0 |
| 55–70 | 6 | 62.5 | 1 | 6 |
| 70–85 | 6 | 77.5 | 2 | 12 |
| 85–100 | 6 | 92.5 | 3 | 18 |
| Total | 30 | — | — | 29 |
Therefore \(\bar{x}=47.5+15\times\dfrac{29}{30}=47.5+14.5=62\). Same mean, lightest arithmetic.
Which method to choose?
- Direct method: convenient when \(x_i\) and \(f_i\) are small.
- Assumed-mean method: useful when \(x_i\) are large but class-widths are uneven.
- Step-deviation method: best when all classes share a common width \(h\).
Example 4 — Wages Problem (Direct Method)
The daily wages of 50 workers of a factory are:
| Daily wages (₹) | 500–520 | 520–540 | 540–560 | 560–580 | 580–600 |
|---|---|---|---|---|---|
| Number of workers | 12 | 14 | 8 | 6 | 10 |
Find the mean daily wages of the workers using the direct method.
Class marks: 510, 530, 550, 570, 590.
\(\sum f_i = 12+14+8+6+10 = 50\).
\(\sum f_i x_i = 12(510)+14(530)+8(550)+6(570)+10(590)\)
\(=6120+7420+4400+3420+5900=27260\).
\(\bar{x}=\dfrac{27260}{50}=\boxed{545.2}\). Mean daily wage = ₹545.20.
\(\sum f_i = 12+14+8+6+10 = 50\).
\(\sum f_i x_i = 12(510)+14(530)+8(550)+6(570)+10(590)\)
\(=6120+7420+4400+3420+5900=27260\).
\(\bar{x}=\dfrac{27260}{50}=\boxed{545.2}\). Mean daily wage = ₹545.20.
Example 5 — Heartbeats of Women (Step-deviation)
The following table gives the number of heartbeats per minute of 30 women. Find the mean.
| Beats/min | 65–68 | 68–71 | 71–74 | 74–77 | 77–80 | 80–83 | 83–86 |
|---|---|---|---|---|---|---|---|
| No. of women | 2 | 4 | 3 | 8 | 7 | 4 | 2 |
Take \(a=75.5\), \(h=3\).
Class marks: 66.5, 69.5, 72.5, 75.5, 78.5, 81.5, 84.5. Corresponding \(u_i\): –3, –2, –1, 0, 1, 2, 3.
\(\sum f_i u_i = 2(-3)+4(-2)+3(-1)+8(0)+7(1)+4(2)+2(3)=-6-8-3+0+7+8+6=4\).
\(\sum f_i=30\). \(\bar{x}=75.5+3\times\dfrac{4}{30}=75.5+0.4=\boxed{75.9}\) beats/min.
Class marks: 66.5, 69.5, 72.5, 75.5, 78.5, 81.5, 84.5. Corresponding \(u_i\): –3, –2, –1, 0, 1, 2, 3.
\(\sum f_i u_i = 2(-3)+4(-2)+3(-1)+8(0)+7(1)+4(2)+2(3)=-6-8-3+0+7+8+6=4\).
\(\sum f_i=30\). \(\bar{x}=75.5+3\times\dfrac{4}{30}=75.5+0.4=\boxed{75.9}\) beats/min.
In-text: Why is the mean obtained by all three methods always equal? Because the three formulas are algebraic identities: multiplying, subtracting, or scaling each \(x_i\) by a constant only rearranges the summation — the final weighted average is unchanged.
Activity — Hands-on: Mean by Three Methods
L3 Apply
Materials: Classmates' heights (in cm), paper, calculator.
Predict: If you compute the mean of the grouped heights using direct, assumed-mean and step-deviation methods, which will give the smallest arithmetic and which the largest?
- Measure the height of 30 students of your class. Group the data into classes of width 5 cm (e.g. 140–145, 145–150, …).
- Write the frequency distribution and compute the class marks \(x_i\).
- Compute \(\bar{x}\) using the direct method.
- Choose \(a\) = middle class mark. Compute \(d_i = x_i-a\) and then \(\bar{x}\) using the assumed-mean method.
- With \(h=5\), compute \(u_i=\tfrac{x_i-a}{h}\) and then \(\bar{x}\) using the step-deviation method.
Observe: The three answers agree to the last decimal. Step-deviation involves the smallest numbers; direct the largest. Your prediction was right!
Competency-Based Questions
Scenario: In a district survey, the electricity consumption (in units) per household of 68 consumers is recorded as the grouped frequency distribution: 65–85 (4), 85–105 (5), 105–125 (13), 125–145 (20), 145–165 (14), 165–185 (8), 185–205 (4).
Q1. Using the step-deviation method with \(a=135\) and \(h=20\), compute the mean monthly consumption.
L3 Apply
Class marks: 75, 95, 115, 135, 155, 175, 195. \(u_i\): –3, –2, –1, 0, 1, 2, 3.
\(\sum f_i=68\). \(\sum f_i u_i = 4(-3)+5(-2)+13(-1)+20(0)+14(1)+8(2)+4(3)=-12-10-13+0+14+16+12=7\).
\(\bar{x}=135+20\cdot\tfrac{7}{68}=135+\tfrac{140}{68}\approx 135+2.06=\mathbf{137.06}\) units.
\(\sum f_i=68\). \(\sum f_i u_i = 4(-3)+5(-2)+13(-1)+20(0)+14(1)+8(2)+4(3)=-12-10-13+0+14+16+12=7\).
\(\bar{x}=135+20\cdot\tfrac{7}{68}=135+\tfrac{140}{68}\approx 135+2.06=\mathbf{137.06}\) units.
Q2. A student claims the mean can be read directly off the frequency distribution as "the class mark of the class with the highest frequency." Analyse whether this claim is true, using the data above.
L4 Analyse
The highest-frequency class is 125–145 with class mark 135. The actual mean computed above is 137.06, not 135. The claim is false: the class-mark of the modal class equals the mode (approximately), not the mean. Mean is a weighted average of all class marks.
Q3. Evaluate: if every observation were increased by 10 units (because of new meters being installed), what would happen to the mean? Justify algebraically.
L5 Evaluate
Each \(x_i\) becomes \(x_i+10\). Then new \(\bar{x}'=\dfrac{\sum f_i(x_i+10)}{\sum f_i}=\dfrac{\sum f_i x_i}{\sum f_i}+10\cdot\dfrac{\sum f_i}{\sum f_i}=\bar{x}+10\). So new mean = 137.06+10 = 147.06 units. The mean shifts by the same constant.
Q4. Create a grouped frequency distribution of 50 observations over 5 classes of equal width, whose mean computed by the step-deviation method equals exactly 50. Share your table.
L6 Create
Sample answer: classes 30–40, 40–50, 50–60, 60–70, 70–80 (width \(h=10\)), frequencies 10, 10, 10, 10, 10 (\(\sum f_i=50\)). Class marks 35, 45, 55, 65, 75. With \(a=55\), \(u_i\): –2, –1, 0, 1, 2; \(\sum f_i u_i = 10(-2-1+0+1+2)=0\). So \(\bar{x}=55+10\cdot 0/50=55\). To hit 50, shift all classes down by 5, e.g. 25–35,…,65–75 with the same frequencies — then mean = 50. Many valid constructions exist.
Assertion–Reason Questions
Assertion (A): The step-deviation method always gives the same mean as the direct method.
Reason (R): The formula \(\bar{x}=a+h\bar{u}\) is obtained from \(\bar{x}=\sum f_i x_i/\sum f_i\) by an algebraic substitution.
Reason (R): The formula \(\bar{x}=a+h\bar{u}\) is obtained from \(\bar{x}=\sum f_i x_i/\sum f_i\) by an algebraic substitution.
(a) — both true; the identity \(u_i=(x_i-a)/h\) is purely algebraic, so substituting back gives the same mean.
Assertion (A): The class mark of the class 40–55 is 47.5.
Reason (R): The class mark equals the sum of the lower and upper limits.
Reason (R): The class mark equals the sum of the lower and upper limits.
(c) — A is true (class mark = 47.5); R is false (class mark = average of lower and upper limits, not the sum).
Assertion (A): For grouped data, the mean computed by assuming all observations are concentrated at class marks is only an approximation of the true mean.
Reason (R): Inside each class we lose information about the exact values of observations.
Reason (R): Inside each class we lose information about the exact values of observations.
(a) — both true; the approximation arises precisely because individual values are replaced by the class mid-point.
Historical Note
The idea of taking an "assumed mean" to ease arithmetic dates back to Bhāskara-II (1114 CE) and later European computers of the 19th century who tabulated astronomical observations by hand. Modern spreadsheets make the direct method trivial — yet the step-deviation method is still taught because it reveals how linear transformations act on averages.
Frequently Asked Questions
What is the mean of grouped data?
The mean of grouped data is the arithmetic average computed by assuming every value in a class interval is concentrated at the class mark (midpoint), then applying x-bar = (sum of f_i * x_i) / (sum of f_i).
What is a class mark?
The class mark of an interval is its midpoint: (lower limit + upper limit) / 2. It represents all observations inside that class when computing the mean.
What is the direct method for finding mean?
Multiply each class mark x_i by its frequency f_i, sum the products, and divide by the total frequency. Best when class marks and frequencies are small numbers.
What is the assumed-mean method?
Choose an assumed mean 'a' (usually a central class mark), compute deviations d_i = x_i - a, and then mean = a + (sum f_i d_i) / (sum f_i). It reduces arithmetic when class marks are large.
What is the step-deviation method?
Let u_i = (x_i - a) / h where h is the class width. Then mean = a + h * (sum f_i u_i) / (sum f_i). Simplifies work when all class widths are equal.
When should I use each method?
Direct method for small data sets, assumed-mean when class marks are large, and step-deviation when class widths are uniform. All three give the same answer.
AI Tutor
Mathematics Class 10
Ready