Introduction & Mean Of Grouped Data
Introduction
Scope of the chapter
- Data organisation: Data may be presented as ungrouped observations or as grouped frequency distributions.
- Earlier representations: Bar graphs, histograms and frequency polygons display data visually.
- Central tendency: Mean, median and mode represent the centre of a data set.
- This chapter: Extends these measures to grouped data and later studies cumulative frequency distributions and ogives.
Mean of a Frequency Distribution
Mean — The mean is the sum of all observations divided by the total number of observations.
If the value \(x_i\) occurs \(f_i\) times, then its contribution to the total is \(f_i x_i\). The total number of observations is \(\sum f_i\).
Use when values and their frequencies are given.
Find the mean marks when \(\sum f_i=30\) and \(\sum f_i x_i=1779\).
- For the marks data, the total frequency is \(30\) and the sum of frequency-weighted marks is \(1779\).$$\sum f_i=30,\qquad \sum f_i x_i=1779$$
- Apply the frequency mean formula.$$\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{1779}{30}=59.3$$
Mean of Grouped Data: Direct Method
Class mark — The class mark is the midpoint of a class interval and is used as the representative value for that class.
Grouped data essentials
- Class intervals: Grouping condenses a large set of observations into intervals with frequencies.
- Boundary convention: A value at an upper class limit is counted in the next interval; for example, \(40\) belongs to \(40\)–\(55\), not \(25\)–\(40\).
- Midpoint assumption: Treat observations in each class as concentrated at its class mark; therefore, a grouped-data mean is generally approximate.
- Direct method: Multiply each class mark by its frequency, add the products, then divide by the total frequency.
Here, \(L\) is the lower class limit and \(U\) is the upper class limit.
Use the class marks as the representative values \(x_i\).
Direct method steps
- Find each class mark from the class limits.$$x_i=\frac{L+U}{2}$$
- Calculate each product of frequency and class mark.$$f_i x_i$$
- Add the frequencies and the products, then divide.$$\bar{x}=\frac{\sum f_i x_i}{\sum f_i}$$
Find the grouped mean for class intervals \(10\)–\(25\), \(25\)–\(40\), \(40\)–\(55\), \(55\)–\(70\), \(70\)–\(85\), \(85\)–\(100\) with frequencies \(2,3,7,6,6,6\).
- For the six mark intervals, the frequencies are \(2,3,7,6,6,6\) and the class marks are \(17.5,32.5,47.5,62.5,77.5,92.5\).$$\sum f_i=30$$
- The corresponding products add to \(1860\).$$\sum f_i x_i=2(17.5)+3(32.5)+7(47.5)+6(62.5)+6(77.5)+6(92.5)=1860$$
- Calculate the grouped mean using the direct method.$$\bar{x}=\frac{1860}{30}=62$$
Do not treat a grouped mean as necessarily exact: replacing every observation by its class mark introduces the midpoint assumption. For the same marks data, the exact ungrouped mean is \(59.3\), whereas the grouped estimate is \(62\).
Assumed Mean Method
Assumed mean — An assumed mean is a convenient class mark, usually near the centre, chosen to simplify the grouped-mean calculations.
If \(d_i=x_i-a\), then the mean deviation is \(\bar{d}=\frac{\sum f_i d_i}{\sum f_i}\) and the actual mean is \(\bar{x}=a+\bar{d}\).
Assumed mean steps
- Choose a convenient class mark as the assumed mean \(a\).
- Find each deviation from the assumed mean.$$d_i=x_i-a$$
- Multiply each deviation by its frequency and find the totals.$$\sum f_i d_i,\qquad \sum f_i$$
- Add the mean deviation to the assumed mean.$$\bar{x}=a+\frac{\sum f_i d_i}{\sum f_i}$$
Find the grouped mean for the data in the direct-method example using \(a=47.5\).
- Choose \(a=47.5\). For the class marks \(17.5,32.5,47.5,62.5,77.5,92.5\), the deviations are \(-30,-15,0,15,30,45\).$$d_i=x_i-47.5$$
- Using frequencies \(2,3,7,6,6,6\), the weighted deviations total \(435\), and the total frequency is \(30\).$$\sum f_i d_i=2(-30)+3(-15)+7(0)+6(15)+6(30)+6(45)=435,\qquad \sum f_i=30$$
- Add the mean deviation to \(a\) to obtain the grouped mean.$$\bar{x}=47.5+\frac{435}{30}=47.5+14.5=62$$
Keep the sign of every deviation: class marks below \(a\) have negative deviations and those above \(a\) have positive deviations. The assumed mean is only a calculation aid, not the final mean.
Find the mean for values \(x_i=2,4,6\) with frequencies \(f_i=3,2,5\).
Find the grouped mean for intervals \(0\)–\(10\), \(10\)–\(20\), \(20\)–\(30\) with frequencies \(2,5,3\).
Quick recap
- For values with frequencies, the mean is \(\bar{x}=\frac{\sum f_i x_i}{\sum f_i}\).
- A class mark is the midpoint \(x_i=\frac{L+U}{2}\) of its class interval.
- The direct method uses class marks and their frequency-weighted products.
- Grouped means are usually approximate because class marks represent all observations in each interval.
- The assumed mean method uses deviations \(d_i=x_i-a\) to reduce calculation size.
- Recover the mean with \(\bar{x}=a+\frac{\sum f_i d_i}{\sum f_i}\).
Mean Of Grouped Data (Activity )
Step-Deviation Method and Choosing a Method
| Wickets | Number of bowlers (\(f_i\)) | Class mark (\(x_i\)) |
|---|---|---|
| \(20\)–\(60\) | \(7\) | \(40\) |
| \(60\)–\(100\) | \(5\) | \(80\) |
| \(100\)–\(150\) | \(16\) | \(125\) |
| \(150\)–\(250\) | \(12\) | \(200\) |
| \(250\)–\(350\) | \(2\) | \(300\) |
| \(350\)–\(450\) | \(3\) | \(400\) |
The direct, assumed-mean and step-deviation methods give the same mean. Since \(u_i = \frac{x_i-a}{h}\), their relationship is \(\bar{x}=a+h\bar{u}\).
Find the mean number of wickets taken by the bowlers.
- For the wickets data, take \(a=200\) and \(h=20\). Calculate \(u_i=(x_i-a)/h\) for each class mark.$$u_i = -8,\ -6,\ -3.75,\ 0,\ 5,\ 10$$
- Multiply each coded deviation by its frequency and add the products; the frequency total is \(45\).$$\sum f_i u_i = -56-30-60+0+10+30=-106, \qquad \sum f_i=45$$
- Apply the step-deviation formula.$$\bar{x}=200+20\left(\frac{-106}{45}\right)=152.888\ldots$$
For female-teacher percentages, the class marks are \(20,30,40,50,60,70,80\) and frequencies are \(6,11,7,4,4,2,1\). Using \(a=50\) and \(h=10\), find the mean.
A grouped distribution has \(a=47.5\), \(h=15\), \(\sum f_i u_i=29\), and \(\sum f_i=30\). Find its mean.
Do not use \(\bar{x}=a+h\frac{\sum f_i u_i}{\sum f_i}\) without multiplying the coded-deviation average by \(h\). Also, use class marks—not class limits—in the mean calculation.
Choose the direct method when the values are small. For larger values, use the assumed-mean or step-deviation method; step-deviation is especially convenient when deviations share a common factor.
Quick recap
- Find each class mark using the midpoint of its class interval.
- The direct mean is \(\bar{x}=\frac{\sum f_i x_i}{\sum f_i}\).
- For assumed mean, use \(d_i=x_i-a\) and \(\bar{x}=a+\frac{\sum f_i d_i}{\sum f_i}\).
- For step-deviation, use \(u_i=\frac{x_i-a}{h}\) and \(\bar{x}=a+h\frac{\sum f_i u_i}{\sum f_i}\).
- All three methods produce the same mean; choose the one that simplifies the arithmetic.
- Unequal class sizes do not prevent step-deviation when a suitable non-zero \(h\) is chosen.
Mode Of Grouped Data
Mode of Ungrouped Data
Mode — The mode is the observation that occurs most often—the value with the greatest frequency.
Key points
- Multimodal data: If more than one value shares the maximum frequency, the data has more than one mode.
- Ungrouped data: Count the frequency of each distinct value; the value with the highest count is the mode.
| Wickets | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| Number of matches | 1 | 1 | 3 | 2 | 1 | 1 | 1 |
| Conclusion | — | — | \(2\) occurs most often | — | — | — | — |
Mode of Grouped Data
Modal class — The class interval with the highest frequency is called the modal class; the mode is an estimated value within this class.
Key points
- Grouped frequencies: They do not reveal the exact most frequent observation; they identify the modal class.
- Scope: This method is used here for grouped distributions with a single mode.
- Class widths: The formula below assumes equal class sizes. Unequal class sizes require a different treatment.
For equal class sizes, estimate the mode from the modal class and the frequencies of the classes immediately before and after it.
Use for a single-mode grouped distribution with equal class sizes.
Symbols
- \(l\): Lower limit of the modal class.
- \(h\): Class size, assuming equal class sizes.
- \(f_1\): Frequency of the modal class.
- \(f_0\) and \(f_2\): Frequencies of the classes immediately before and after the modal class.
Finding the grouped-data mode
- Identify the class with the greatest frequency; this is the modal class.
- Read off \(l\), \(h\), \(f_1\), \(f_0\), and \(f_2\) from the table.
- Substitute these values into the formula and calculate the estimated mode.
Find the mode of the grouped family-size data.
- The family-size intervals and frequencies are \(1\)–\(3:7\), \(3\)–\(5:8\), \(5\)–\(7:2\), \(7\)–\(9:2\), and \(9\)–\(11:1\). The modal class is \(3\)–\(5\).
- Use \(l=3\), \(h=2\), \(f_1=8\), \(f_0=7\), and \(f_2=2\).
- Substitute into the grouped-data mode formula.$$3+\frac{8-7}{2(8)-7-2}\times2=3+\frac{2}{7}\approx3.286$$
Interpreting mode and mean
- Mode: Describes the value or level occurring most often; in the marks example, the mode was \(52\) marks.
- Mean: Describes the average across all observations; in that example, the mean was \(62\) marks.
- Comparison: Mode may be less than, equal to, or greater than the mean. Choose the measure that matches the question.
- Use: Mean answers questions about the overall average; mode answers questions about what occurs most often.
Do not use only the modal-class frequency: \(f_0\) and \(f_2\) are the frequencies of the immediately preceding and succeeding classes. The formula assumes equal class sizes and a single mode.
For class intervals \(0\)–\(10\), \(10\)–\(20\), \(20\)–\(30\), and \(30\)–\(40\), the frequencies are \(4\), \(9\), \(12\), and \(5\). Find the mode.
Marks are grouped into \(45\)–\(55\), \(55\)–\(65\), and \(65\)–\(75\) with frequencies \(4\), \(7\), and \(6\). If the mean is \(62\), find the mode and compare it with the mean.
Quick recap
- Mode is the observation with the greatest frequency; tied maximum frequencies make ungrouped data multimodal.
- For grouped data, the class with the highest frequency is the modal class; this lecture's method considers a single mode.
- For equal class sizes, use \(\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h\).
- \(f_0\) and \(f_2\) are the frequencies of the classes immediately before and after the modal class.
- Mode describes what occurs most often, while mean describes the overall average; mode may be below, equal to, or above mean.
Median Of Grouped Data
Median of Ungrouped Data
Use the middle observation for odd \(n\) and average the two middle observations for even \(n\).
Find the median of the marks-frequency data in the table.
- There are \(100\) students, so the middle positions are the \(50\)th and \(51\)st observations.$$n=100,\qquad \frac{n}{2}=50$$
- From the cumulative frequencies, the \(50\)th observation is \(28\) and the \(51\)st is \(29\).$$x_{50}=28,\qquad x_{51}=29$$
- Average the two middle observations.$$\text{Median}=\frac{28+29}{2}=28.5$$
Cumulative Frequency
Find the less-than cumulative frequency below \(20\) and the more-than cumulative frequency at \(20\).
- For the less-than cumulative frequency below \(20\), add frequencies for the intervals \(0\)–\(10\) and \(10\)–\(20\).$$5+3=8$$
- For the more-than cumulative frequency at \(20\), subtract the frequencies below \(20\) from the total.$$53-5-3=45$$
Median of Grouped Data
Use this formula after locating the median class; \(h\) is its class width.
Finding the Grouped-Data Median
- Add the frequencies to find the total number of observations.$$n=\sum f$$
- Calculate half the total and use cumulative frequencies to find the median class.$$\frac{n}{2}$$
- Identify the lower limit \(l\), preceding cumulative frequency \(cf\), median-class frequency \(f\), and class width \(h\).
- Substitute into the grouped-data median formula and simplify.$$\text{Median}=l+\left(\frac{\frac{n}{2}-cf}{f}\right)h$$
Find the median of the grouped marks distribution of \(53\) students.
- There are \(53\) students, so half the observations is \(26.5\). The first cumulative frequency at least this large is \(29\), for class \(60\)–\(70\).$$n=53,\qquad \frac{n}{2}=26.5$$
- Read the required values from the median class and the class before it.$$l=60,\qquad cf=22,\qquad f=7,\qquad h=10$$
- Substitute into the median formula.$$\text{Median}=60+\left(\frac{26.5-22}{7}\right)10$$
- Simplify and round to one decimal place.$$\text{Median}=60+\frac{45}{7}\approx66.4$$
Do not use the cumulative frequency of the median class for \(cf\); use the cumulative frequency of the class immediately before it.
Find the median for the grouped data: \(0\)–\(10\): \(4\), \(10\)–\(20\): \(6\), \(20\)–\(30\): \(10\), \(30\)–\(40\): \(8\), \(40\)–\(50\): \(2\).
For the grouped distribution \(0\)–\(10\): \(5\), \(10\)–\(20\): \(3\), \(20\)–\(30\): \(4\), find the less-than cumulative frequency below \(30\) and the more-than cumulative frequency at \(10\).
Quick recap
- Arrange ungrouped observations in order; use the middle value or average the two middle values.
- Less-than cumulative frequency is built by successive addition; more-than cumulative frequency is built by successive subtraction.
- For grouped data, the median class is the first class whose cumulative frequency is at least \(\frac{n}{2}\).
- Use \(\text{Median}=l+\left(\frac{\frac{n}{2}-cf}{f}\right)h\) for equal-sized classes.
- In the grouped-data formula, \(cf\) is the cumulative frequency before the median class.
Median Of Grouped Data ( Examples Continue)
Median from a Less-Than Cumulative Frequency Table
Cumulative frequency — The cumulative frequency of a class is the total frequency up to and including that class.
- Convert to class frequencies: For a less-than table, take successive differences between cumulative frequencies; the first cumulative frequency is the first class frequency.
- Median class: It is the class whose cumulative frequency first reaches or exceeds \(n/2\).
For continuous grouped data, use the median class and the formula below; \(l\) is its lower limit, \(cf\) is the cumulative frequency before it, \(f\) is its frequency, and \(h\) is its class width.
Use after identifying the median class.
Find a median from less-than data
- Convert cumulative frequencies into class frequencies using successive differences.
- Find half the total frequency and locate the first cumulative frequency that reaches or exceeds it.$$n/2$$
- Read the median-class values and substitute them into the grouped-data median formula.$$\text{Median}=l+\left(\frac{n/2-cf}{f}\right)h$$
Find the median height from these less-than cumulative frequencies: below 140 cm: 4; below 145 cm: 11; below 150 cm: 29; below 155 cm: 40; below 160 cm: 46; below 165 cm: 51.
- Find the class frequencies by taking successive differences.$$4,\quad 11-4=7,\quad 29-11=18,\quad 40-29=11,\quad 46-40=6,\quad 51-46=5$$
- The first cumulative frequency at least as large as half the total is 29, so the median class is 145–150.$$n=51,\qquad n/2=25.5$$
- Identify the median-class values.$$l=145,\quad cf=11,\quad f=18,\quad h=5$$
- Substitute and evaluate.$$\text{Median}=145+\left(\frac{25.5-11}{18}\right)5\approx149.03\text{ cm}$$
Finding Unknown Frequencies
Unknown frequency in a grouped distribution — Find missing class frequencies by combining the total-frequency condition with the grouped-data median formula.
- Total frequency: The sum of all class frequencies equals \(n\).
- Median class: Use its frequency and the cumulative frequency of the preceding classes in the median formula.
Use this equation when some class frequencies are unknown.
Solve for missing frequencies
- Write cumulative frequencies in terms of the unknown frequencies.
- Use the total-frequency condition to form one equation.$$\sum f_i=n$$
- Use the given median in the grouped-data formula to form another equation, then solve the equations together.$$\text{Median}=l+\left(\frac{n/2-cf}{f}\right)h$$
The frequencies for successive classes 0–100 through 900–1000 are \(2,5,x,12,17,20,y,9,7,4\). If the total frequency is 100 and the median is 525, find \(x\) and \(y\).
- Add the known frequencies and use the total frequency of 100.$$76+x+y=100\quad\Rightarrow\quad x+y=24$$
- The median 525 lies in the class 500–600; identify the median-class values.$$l=500,\quad f=20,\quad cf=36+x,\quad h=100,\quad n/2=50$$
- Substitute the median and solve for x.$$525=500+\left(\frac{50-(36+x)}{20}\right)100\quad\Rightarrow\quad25=5(14-x)\quad\Rightarrow\quad x=9$$
- Use the total-frequency equation to find y.$$9+y=24\quad\Rightarrow\quad y=15$$
Choosing a Measure of Central Tendency
Central tendency — Mean, median, and mode each describe a central or typical value, but they are suited to different purposes.
| Measure | Best suited for | Key point or example |
|---|---|---|
| Mean | Summarising and comparing distributions | Most frequently used; uses all observations and lies between the smallest and largest values, but extremes can distort it. |
| Median | Finding a typical value when extreme values may occur | Useful for typical worker productivity or wages in a country. |
| Mode | Finding the most frequent or popular value | Useful for the most-watched programme, most-demanded item, or most common vehicle colour. |
Interpreting the measures
- Mean: It is often a good representative when class frequencies are fairly similar; extreme values can make it unrepresentative.
- Median: Prefer it when individual observations are less important and a typical value is needed despite extreme values.
The empirical relation among the three measures is commonly written as the median equal to one-third of the mode plus twice the mean.
Grouped-Data Formula Recap
Key grouped-data points
- Continuous classes: Ensure class intervals are continuous before applying grouped-data median or mode formulas, or constructing an ogive.
- Unequal class sizes: A grouped median can also be calculated for unequal class widths, though that method is not covered here.
- Ogives: The scales on the two axes do not have to be the same.
- Class mark: In grouped-mean calculations, each class is represented by its midpoint.
Here, \(x_i\) is a class mark, \(a\) is an assumed mean, \(d_i=x_i-a\), and \(u_i=d_i/h\) for step deviation.
Here, \(l\) is the modal-class lower limit, \(h\) is its width, \(f_1\) is its frequency, and \(f_0\) and \(f_2\) are the frequencies of the preceding and succeeding classes.
Here, \(l\) is the median-class lower limit, \(cf\) is the cumulative frequency before it, \(f\) is its frequency, and \(h\) is its class width.
Find the median for the continuous classes 0–10, 10–20, 20–30, and 30–40 with frequencies 3, 5, 8, and 4.
A less-than table has cumulative frequencies below 10: 4, below 20: 13, below 30: 25, and below 40: 34. Find the median.
Quick recap
- Convert less-than cumulative frequencies into class frequencies by taking successive differences.
- Locate the median class using the first cumulative frequency at least as large as \(n/2\).
- Use \(\text{Median}=l+\left(\frac{n/2-cf}{f}\right)h\) for continuous grouped data.
- Find unknown frequencies by combining the total-frequency equation with the median formula.
- The mean is frequently used, uses all observations, and lies between the extremes, but extreme values can distort it.
- Use the median for typical values when extremes matter; use the mode to identify the most frequent value.
- Ensure class intervals are continuous before applying grouped median or mode formulas.
Formula sheet
Mathematics — ALL CHAPTERS
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Real Numbers
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Pair Of Linear Equations In Two Variables
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Introduction To Trigonometry
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