Mean and Median Class 9 ICSE — Arun Chapter 19
Looking for mean and median Class 9 ICSE explained section by section? Chapter 19 of the Arun Deep Class 9 ICSE Maths book covers both for ungrouped data: the mean and the median. You will learn the mean of raw data, three methods for frequency tables, and the median for raw data and discrete series. The chapter closes with two exercise blocks, Exercise 19 (A) and Exercise 19 (B).
ICSE Class 9 Mathematics Arun Deep Chapter 19 Mean and Median of Ungrouped Data PDF
What is inside Chapter 19
12 pages · 1 sections · 25 exercise questions
| Section | Title |
|---|---|
| 19 | Mean and Median of ungrouped data |
- EXERCISE 19 (A): 16 questions
- EXERCISE 19 (B): 9 questions
19 Mean and Median of ungrouped data
The chapter opens with a “Points to Remember” summary. It lists the definition of the mean, the three mean methods, and the two median rules. The exercises at the end test each formula. You can see the full set of chapters on the Class 9 ICSE Maths page.
The chapter first defines the mean of raw data. The mean \bar{x} is the sum of observations divided by the number of observations:
\bar{x} = \frac{\text{Sum of observations}}{\text{Number of observations}} = \frac{\Sigma x_i}{n}
In full, \bar{x} = \frac{x_1 + x_2 + \cdots + x_n}{n}. Here x_1, x_2, \ldots, x_n are the n observations, and \Sigma means “sum of”. Rearranged, the same relation reads x_1 + x_2 + \cdots + x_n = n \times \bar{x}. This form lets you find a missing observation when the mean is known.
Worked pattern. The heights of 6 students are 152, 148, 155, 150, 149 and 146 cm. Their sum is 900 cm, and n = 6. So \bar{x} = \frac{900}{6} = 150 cm.
Mean for ungrouped data: the direct method
Use the direct method when the data is given as variates with their frequencies. The mean is the total of all observations divided by the total frequency.
\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}
The symbol \Sigma f_i x_i means “multiply each variate by its frequency, then add.” The symbol \Sigma f_i is the total number of observations. Why does the formula work? Each variate x_i repeats f_i times, so the sum of all observations is \Sigma f_i x_i. Dividing by the total frequency gives one average observation.
Set up three columns: x, f, and f \times x. Total the f column and the f \times x column, then divide.
Worked pattern. Find the mean of these marks.
| Marks x | Frequency f | f \times x |
|---|---|---|
| 40 | 3 | 120 |
| 50 | 5 | 250 |
| 60 | 4 | 240 |
| 70 | 2 | 140 |
| Total | 14 | 750 |
\bar{x} = \frac{750}{14} = 53.6 \text{ marks (approx.)}
The total frequency 14 is the number of observations. The product column total 750 is the sum of all marks. So the mean is about 53.6 marks.
The same table handles missing-value questions. If one frequency is unknown, write the mean as the total divided by n, set it equal to the given mean, and solve for the unknown.
Assumed mean method
The assumed mean method cuts the arithmetic when variates are large or awkward. You choose a convenient value A near the middle of the data. Then you measure how far each variate sits from it.
\bar{x} = A + \frac{\Sigma f_i d_i}{\Sigma f_i}
Here d_i = x_i - A is the deviation, and f_i is the frequency of x_i. The fraction \frac{\Sigma f_i d_i}{\Sigma f_i} is the weighted average deviation. Adding it to A moves the assumed value to the true mean. Keep every deviation signed, because negative deviations cancel positive ones.
Worked pattern. Find the mean of these weekly wages.
- Wages x: 100, 120, 140, 160; frequencies f: 2, 3, 3, 2.
- Choose A = 140. Deviations d = x - A: −40, −20, 0, 20.
- Products f \times d: −80, −60, 0, 40; total −100.
- Total frequency \Sigma f = 10.
\bar{x} = 140 + \frac{-100}{10} = 140 - 10 = 130
Check with the direct method. The total of all wages is 1300, and \frac{1300}{10} = 130. The two methods agree.
Step deviation method
The step deviation method shrinks the numbers even further. You divide each deviation by a common step h, work with the smaller values, then convert back. It requires equally spaced variates, so h = x_2 - x_1 is one constant.
As printed in this chapter, the formula is \bar{x} = A + h + \frac{\Sigma f_i u_i}{\Sigma f_i}, where u_i = \frac{x_i - A}{h}. Treat the printed “A + h +” as a misprint. The worked example in the chapter applies the multiplication form, and that gives the correct result.
\bar{x} = A + h \cdot \frac{\Sigma f_i u_i}{\Sigma f_i}
Why multiply by h at the end? Dividing the deviations by h changes their units. Multiplying the average of u by h restores the original units.
Worked pattern. Take variates 20, 30, 40, 50 with frequencies 2, 3, 4, 1. Choose A = 40 and h = 10.
- u values: \frac{20-40}{10} = -2, \frac{30-40}{10} = -1, 0, 1.
- Products f \times u: −4, −3, 0, 1; total −6.
- Total frequency \Sigma f = 10.
\bar{x} = 40 + 10 \cdot \frac{-6}{10} = 40 - 6 = 34
Direct check: the total of the data is 340, and \frac{340}{10} = 34. Same mean. Note that the printed form of this formula varies by edition, so match the pattern your school book shows.
Choosing between the three mean methods
All three methods give the same mean, so pick the one with the least arithmetic. The choice depends on the shape of the data.
| Method | When to use it | What the table needs | Formula |
|---|---|---|---|
| Direct method | Small frequency tables with few variates | Columns x, f, f \times x | \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} |
| Assumed mean method | Larger data with awkward values | Columns x, f, d = x - A, f \times d | \bar{x} = A + \frac{\Sigma f_i d_i}{\Sigma f_i} |
| Step deviation method | Large, equally spaced variates | Columns x, f, u = \frac{x - A}{h}, f \times u | \bar{x} = A + h \cdot \frac{\Sigma f_i u_i}{\Sigma f_i} |
Each method is the same division rewritten. The direct method divides the total by the count. The assumed mean method adds the average deviation to a guess. The step deviation method scales the deviations first, then un-scales the result. Because every step is algebraically the same, the answer never changes.
Follow this practical rule:
- Direct method: use for three or four small variates.
- Assumed mean: use when the variates end in awkward digits, such as 7 or 9.
- Step deviation: use when the variates are large but equally spaced.
Median of ungrouped data
The median is the middle value of a data set. It splits the ordered data into two equal halves. The first step is always to arrange the observations. Either ascending or descending order works.
If n is odd, the median is the \left( \frac{n+1}{2} \right)^{\text{th}} term. If n is even, the median is the average of the \frac{n}{2}^{\text{th}} and \left( \frac{n}{2} + 1 \right)^{\text{th}} terms.
\text{Median} = \frac{1}{2} \left[katex display="true"]\frac{n}{2}^{\text{th}} \text{ term} + \left( \frac{n}{2} + 1 \right)^{\text{th}} \text{ term} \right[/katex]
Why two rules? With an odd count, one value sits exactly in the middle. With an even count, two values share the middle, so you take their average.
Worked pattern (odd). Scores 5, 9, 3, 8, 6 sort to 3, 5, 6, 8, 9. Here n = 5. The median is the \frac{5+1}{2} = 3^{\text{rd}} term, which is 6.
Worked pattern (even). Values 4, 11, 7, 9 sort to 4, 7, 9, 11. Here n = 4. The median is \frac{7 + 9}{2} = 8.
Skipping the sort is the most common error. The median is a positional value, so jumbled data gives the wrong term. A related set of problems from the same book appears on the Chapter 18 statistics page.
Median of a discrete series
In a discrete series, each variate comes with a frequency. You still look for the middle term, but now the terms are repeated. A cumulative frequency column shows where each term position sits.
Method: arrange the variates in order, add a cumulative frequency column, and write the total frequency as n. If n is odd, take the \left( \frac{n+1}{2} \right)^{\text{th}} term. If n is even, average the \frac{n}{2}^{\text{th}} and \left( \frac{n}{2} + 1 \right)^{\text{th}} terms.
Worked pattern. Variates 2, 5, 8, 11 have frequencies 4, 6, 8, 2.
| Variate | Frequency | Cumulative frequency |
|---|---|---|
| 2 | 4 | 4 |
| 5 | 6 | 10 |
| 8 | 8 | 18 |
| 11 | 2 | 20 |
Here n = 20, which is even. The 10th term is the last 5, because the cumulative frequency reaches 10 at that row. The 11th term is the first 8. So the median is \frac{5 + 8}{2} = 6.5.
The cumulative frequency column is how you locate the middle term. Read down the column until it reaches the position you need. That row’s variate is the term.
What to practise from this chapter
Two exercise blocks close the chapter. Each tests a different skill.
- Exercise 19 (A) — 16 questions on the mean. They start with raw-data means, then move to the frequency-table direct method, missing-value problems, and the two shortcut methods.
- Exercise 19 (B) — 9 questions on the median. They cover raw data with odd and even counts, then discrete series with cumulative frequency tables.
Watch for these common mistakes while you practise.
| Mistake | Correct rule | How to check |
|---|---|---|
| Skipping the sort before the median | Arrange the data in ascending or descending order first | Read the ordered list before picking the term |
| Using the odd rule when n is even | For even n, average the two middle terms | Count n first, then apply the even formula |
| Forgetting the f \times x product in the direct method | Multiply each frequency by its variate before adding | Recheck the f \times x column against the data |
| Dropping the sign of d in the assumed mean | Keep deviations signed; negatives cancel positives | Re-add the f \times d column with signs included |
| Dividing by h in step deviation and not multiplying back | Use \bar{x} = A + h \cdot \frac{\Sigma f_i u_i}{\Sigma f_i} | Compare the answer with the direct method |
Look at the question mix inside the blocks. In Exercise 19 (A), raw-data mean sums and the direct-method table appear in most questions; the assumed mean and step deviation methods appear once each. In Exercise 19 (B), the even-n average rule appears in most questions. Practise those two patterns first. The chapter sits inside the ICSE syllabus published by the official CISCE website. The other chapters in this class are listed on the Class 9 ICSE books list.
Frequently asked questions
How do I find the mean of ungrouped data in class 9 ICSE?
Add all the observations and divide by the number of observations: \bar{x} = \frac{\Sigma x_i}{n}. If the data is a frequency table, use the direct method \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}.
When should I use the assumed mean method instead of the direct method?
Use it when the variates are large or the products f \times x become unwieldy. Choose A near the middle, find each deviation d_i = x_i - A, then apply \bar{x} = A + \frac{\Sigma f_i d_i}{\Sigma f_i}.
How do I find the median when the number of observations is even?
Sort the data first. Then average the \frac{n}{2}^{\text{th}} and \left( \frac{n}{2} + 1 \right)^{\text{th}} terms. For example, with n = 4, average the 2nd and 3rd terms.
Do I have to sort the data before finding the median?
Yes. The median is the middle value, and a jumbled list hides the middle position. Sort in ascending or descending order, then apply the odd or even rule.
What is the difference between mean and median?
The mean uses every value, so one extreme value pulls it. The median is the middle value after sorting, so it resists extremes. Both measure the centre of the data.
Reference: ICSE Class 9 Mathematics textbook material, chapter 19: Mean and Median of Ungrouped Data — CISCE-aligned.