Statistics teaches you to answer a question with data. You pose a statistical investigative question, collect and organise the data in a frequency table, show it in a suitable graph, and summarise it with the mean, median or mode. The chapter adds the weighted average, used for report cards, mixtures and combined class means, and the stacked and 100% stacked bar graphs, which show totals and shares. It sits in Unit VI, Statistics and Probability, of the Class 9 syllabus and builds the data skills used again in Class 10.
Key Concepts
1. Statistical Investigative Questions and Collecting Data
Statistics begins with a question. A statistical investigative question is a question whose answer comes from data that vary. “How many hours do Class 9 students of our school sleep on a school night?” is statistical, because different students give different answers. “What time does our school start?” has one fixed answer, so it is not statistical. The spread of different answers is called variability, and it is the reason we need statistics at all.
A quick test: if you must ask many people or take many measurements, and you expect the answers to differ, the question is statistical. “How tall is Riya?” is not statistical. “How tall are the girls of Class 9?” is.
Every statistical investigation follows the same cycle:
- Pose a clear question and decide the group it is about, for example all Class 9 students of your school.
- Collect the data by a survey, an experiment or a measurement.
- Organise the data in a table.
- Visualise the data with a suitable graph.
- Interpret the result and use it to answer the question or to make a decision.
Primary and secondary data. Data that you collect yourself, first-hand, for your own purpose are primary data, for example a survey of your classmates. Data taken from a source where someone else has already collected them, such as school records, a newspaper table or census figures, are secondary data.
Collecting fairly. The data should represent the whole group the question is about. Suppose Priya wants to know how Class 9 students travel to school. If she asks only her five closest friends, they may all live in the same area. If she asks only the students waiting at the bus stop, almost all of them will say “bus”. Both plans give a biased picture. A better plan surveys students from every section of Class 9.
2. Organising Data: Frequency Tables and Range
Data written down in the order they were collected are called raw data. They are hard to read. A frequency table organises them: each different value (or category) is listed once, and next to it we write how many times it occurs. That count is its frequency. Tally marks help while counting: four strokes with a fifth stroke across them stand for 5.
Illustration. Marks (out of 10) of 20 students: 6, 8, 5, 9, 7, 6, 8, 10, 4, 6, 7, 8, 6, 9, 5, 7, 6, 8, 7, 3.
| Marks (x) | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | Total |
|---|---|---|---|---|---|---|---|---|---|
| Frequency (f) | 1 | 1 | 2 | 5 | 4 | 4 | 2 | 1 | 20 |
Every observation is counted exactly once, so the frequencies always add up to the total number of observations. This gives a check on your counting, and it lets you find a missing frequency. If shoe sizes 5, 6, 7 and 8 of 40 students have frequencies 6, 14, x and 8, then 6 + 14 + x + 8 = 40, so x = 12.
Range. The range is the simplest measure of how spread out the data are:
Range = Highest value − Lowest value
For the marks above, range = 10 − 3 = 7. A small range means the values are close together; a large range means they are spread out. The range uses only two values, so one unusual value can make it large.
3. Graphical Representation: Choosing a Suitable Graph
A graph lets the reader see a pattern at a glance. The right graph depends on the question.
| Graph | What it shows well | Example |
|---|---|---|
| Bar graph | Comparing counts of separate categories | Number of students choosing each sport |
| Double bar graph | Comparing two groups category by category | Boys and girls choosing each sport |
| Stacked bar graph | Each group’s total and how it is split into parts | Students of each section, split by mode of travel |
| 100% stacked bar graph | Shares (percentages) of parts, for groups of different sizes | Percentage of each school’s students who cycle |
| Line graph | How one quantity changes over time | Daily temperature over a week |
Reading any bar graph. Read the title, the labels on both axes and the scale before reading any bar. Bars must have equal width and equal gaps; only the length of a bar carries information. Check whether the scale starts at 0. If the vertical axis starts at, say, 40 instead of 0, a bar of 50 looks twice as tall as a bar of 45 even though 50 is only about 11% more than 45. A graph with a cut scale can mislead, so always compare the actual readings.
4. Stacked Bar Graphs
In a stacked bar graph each bar stands for one group, such as a class section or a day. The bar is cut into segments, one for each sub-category, placed one above the other. A legend (key) tells which shade stands for which sub-category, and the segments are stacked in the same order in every bar.
- The height of the whole bar is the total for that group.
- Each segment is the part contributed by one sub-category.
- Only the bottom segment starts at 0. Every other segment starts where the one below it ends, so segment value = top reading − bottom reading.
Illustration. A stacked bar graph shows how the students of three sections travel to school. The scale readings of the segments are given in the table.
| Section | Walk segment | Bus segment | Cycle segment |
|---|---|---|---|
| 9A | 0 to 14 | 14 to 30 | 30 to 38 |
| 9B | 0 to 10 | 10 to 32 | 32 to 42 |
| 9C | 0 to 18 | 18 to 28 | 28 to 40 |
For 9B: Walk = 10 − 0 = 10, Bus = 32 − 10 = 22, Cycle = 42 − 32 = 10, and the total is the top of the bar, 42. Check: 10 + 22 + 10 = 42.
Comparing bars. Totals are compared by bar heights. Bottom segments share the baseline, so they can be compared by eye. Upper segments start at different heights, so work out each value before comparing. In the graph above, the Cycle segment of 9B sits highest on the page, yet 9C has more cyclists (12 against 10).
Finding a missing part. If you know the bar total and all but one segment, subtract. A canteen’s Tuesday bar reaches 95 cups, with tea 45 and coffee 30, so juice = 95 − 45 − 30 = 20 cups.
5. 100% Stacked Bar Graphs
Suppose School P has 250 students and School Q has 600. In an ordinary stacked bar graph, Q’s bar is more than twice as tall, and P’s segments look tiny. To compare proportions fairly, every bar is drawn to the same height, 100%, and each segment shows a percentage share of its own group. This is a 100% stacked bar graph.
Percentage share = (Part / Group total) × 100
Building a bar. Section 9C has 40 students: 18 walk, 10 take the bus and 12 cycle. Walk = 18/40 × 100 = 45%, Bus = 10/40 × 100 = 25%, Cycle = 12/40 × 100 = 30%. Check: 45 + 25 + 30 = 100. On the graph the Walk segment runs from 0% to 45%, Bus from 45% to 70% and Cycle from 70% to 100%.
Going back to counts. A count can be found only when the group total is given separately:
Count = (Percentage / 100) × Group total
| School (students) | Walk | Bus | Cycle |
|---|---|---|---|
| P (250) | 40% = 100 | 36% = 90 | 24% = 60 |
| Q (600) | 25% = 150 | 55% = 330 | 20% = 120 |
The key limitation. P has the larger share of cyclists (24% against 20%), but Q has more cyclists (120 against 60). A 100% stacked bar graph hides group sizes, so it answers “what fraction?” and cannot answer “how many?” unless the totals are supplied. Always multiply a percentage by that group’s own total.
Which stacked graph to use. Use an ordinary stacked bar graph when both the totals and their make-up matter. Use a 100% stacked bar graph when you want to compare the make-up of groups of different sizes.
6. Mean, Median and Mode
A measure of central tendency is a single value that represents a whole data set. There are three common ones.
Mean. The mean (arithmetic mean, average) is the value each observation would have if the total were shared equally.
Mean = Sum of all observations / Number of observations
Turned around, Sum = Mean × Number of observations. This form solves missing-value and corrected-mean problems. If a value was misread, correct the sum first, then divide again.
Median. Arrange the data in ascending (or descending) order. The median is the middle value.
- If the number of observations n is odd, the median is the ((n + 1)/2)th observation. For 25 observations, it is the 13th.
- If n is even, the median is the mean of the (n/2)th and (n/2 + 1)th observations. For 7, 9, 11, 12, 15, 20 (n = 6), the median is (11 + 12)/2 = 11.5.
Mode. The mode is the observation that occurs most often. In 3, 5, 7, 5, 9, 5, 7 the mode is 5. A data set can have two modes (bimodal), or no mode when every value occurs once. The mode is the only one of the three that works for categories such as favourite colour.
Mean of a frequency table. When value x occurs f times, it contributes f × x to the sum. So
Mean = Σfx / Σf
where Σ (sigma) means “sum of” and Σf is the total number of observations. For the 20 students’ marks in Concept 2: Σfx = 3 + 4 + 10 + 30 + 28 + 32 + 18 + 10 = 135, so mean = 135/20 = 6.75. The median is the mean of the 10th and 11th values. Counting frequencies from the lowest mark (1, 2, 4, 9, 13, …), the 10th to 13th values are all 7, so median = 7. The mode is 6 (frequency 5). One data set, three different centres: 6.75, 7 and 6.
7. Weighted Average and Combined Mean
The simple mean treats every value as equally important. Often they are not. If you mix 3 glasses of juice costing ₹40 each with 2 glasses costing ₹50 each, the average price per glass is not ₹45. You have more of the cheaper juice, so the total cost is 3 × 40 + 2 × 50 = ₹220 for 5 glasses, which is ₹44 per glass. The number of glasses acts as a weight.
Weighted average = (w1x1 + w2x2 + … + wnxn) / (w1 + w2 + … + wn) = Σwx / Σw
When every weight is 1, this becomes the ordinary mean. The mean of a frequency table is also a weighted average, with the frequencies as weights.
| Setting | Values (x) | Weights (w) |
|---|---|---|
| Final score from several exams | Marks in each exam | Percentage weight of each exam |
| Grade point average | Grade point in each subject | Credits of each subject |
| Price of a mixture | Price per kg of each variety | Quantity (kg) of each variety |
| Combined mean of groups | Mean of each group | Number of members in each group |
Percentage weights. Tests 20%, half-yearly 30% and annual 50%, with scores 80, 70 and 60: weighted score = 0.2 × 80 + 0.3 × 70 + 0.5 × 60 = 16 + 21 + 30 = 67. The simple mean is 70; the final score is lower because the lowest score carries half the weight. When the weights are percentages that add to 100%, dividing by the sum of the weights (1) changes nothing.
Combined mean. For groups of sizes n1 and n2 with means m1 and m2:
Combined mean = (n1m1 + n2m2) / (n1 + n2)
Section A (30 students, mean 60) and Section B (20 students, mean 70) have combined mean (1800 + 1400)/50 = 64, closer to 60 because Section A is larger.
A built-in check. A weighted average always lies between the smallest and the largest value, and it is pulled towards the value with the larger weight. An answer outside that range means an arithmetic slip.
8. Choosing the Right Average for Decisions
Each measure has a job, and the chapter expects you to pick the right one and say why.
| Measure | Strength | Weakness | Use it for |
|---|---|---|---|
| Mean | Uses every observation | Pulled strongly by extreme values (outliers) | Evenly spread data such as heights, test marks |
| Median | Unaffected by extreme values | Ignores the size of the other values | Salaries, house prices, travel times with a few very large values |
| Mode | Shows the most popular value; works for categories | May not exist or may not be unique | Shoe size to stock most, favourite snack, most common answer |
The outlier effect. Data: 10, 12, 12, 15, 16 give mean 13, median 12 and mode 12. Replace 16 by 60: the sum becomes 109 and the mean jumps to 21.8, while the median and mode stay at 12. An outlier is a value far away from the rest.
A mean that no one has. A shoe shop learns that the mean shoe size sold is 6.4. No customer wears size 6.4. The shop needs the mode, the size that sells most often.
Looking beyond the centre. Two batters each average 50 runs over five innings. Batter A scored 50, 48, 52, 49, 51 (range 4). Batter B scored 0, 5, 120, 10, 115 (range 120, median 10). The means are equal, but A is far more consistent. For a decision, compare the spread (range) as well as the centre.
Formula and Theorem Sheet
| Result | Formula or rule | When to use |
|---|---|---|
| Range | Highest value − Lowest value | Spread or consistency of data |
| Sum of frequencies | Σf = total number of observations | Checking a table; finding a missing frequency |
| Mean of raw data | (x1 + x2 + … + xn) / n | Values listed one by one |
| Sum from mean | Sum = Mean × n | Missing, removed, added or corrected values |
| Mean of a frequency table | Σfx / Σf | Values with frequencies |
| Median (n odd) | ((n + 1)/2)th observation of ordered data | Data with outliers |
| Median (n even) | Mean of the (n/2)th and (n/2 + 1)th observations | Data with outliers |
| Mode | Value with the highest frequency | Most popular value; categorical data |
| Weighted average | Σwx / Σw | Values of unequal importance |
| Combined mean | (n1m1 + n2m2) / (n1 + n2) | Joining two groups |
| Stacked bar segment | Top reading − Bottom reading | Reading any segment above the bottom one |
| Percentage share | (Part / Group total) × 100 | Drawing a 100% stacked bar |
| Count from share | (Percentage / 100) × Group total | Reading counts from a 100% stacked bar |
Read the rest of the chapter →Hide the rest ↑
Important Definitions
- Statistics: the study of collecting, organising, representing and interpreting data to answer questions and make decisions.
- Data: facts or figures, numbers or categories, collected for a purpose.
- Statistical investigative question: a question that is answered using data and whose answers are expected to vary.
- Variability: the fact that observations of the same quantity differ from one another.
- Primary data: data collected first-hand by the investigator for the purpose in hand.
- Secondary data: data taken from a source where they were already collected by someone else.
- Raw data: data in the form in which they were collected, before any organising.
- Frequency: the number of times a value or category occurs in the data.
- Frequency table: a table listing each value or category with its frequency.
- Range: the difference between the highest and the lowest observation.
- Stacked bar graph: a bar graph in which each bar is divided into segments for sub-categories, so the bar shows a group total and its parts.
- 100% stacked bar graph: a stacked bar graph in which every bar represents 100% and each segment shows the percentage share of a sub-category in that group.
- Mean: the sum of all observations divided by the number of observations.
- Median: the middle value of the data when arranged in ascending or descending order.
- Mode: the observation that occurs most frequently.
- Weighted average: the sum of each value multiplied by its weight, divided by the sum of the weights.
- Outlier: an observation that lies far away from the rest of the data.
Solved Examples
Example 1: Frequency table, range and mode
The number of books read in a month by 16 students is: 2, 3, 1, 4, 3, 2, 3, 5, 0, 3, 2, 4, 3, 1, 2, 3. Make a frequency table and find the range and the mode.
Solution:
| Books (x) | 0 | 1 | 2 | 3 | 4 | 5 | Total |
|---|---|---|---|---|---|---|---|
| Frequency (f) | 1 | 2 | 4 | 6 | 2 | 1 | 16 |
Check: 1 + 2 + 4 + 6 + 2 + 1 = 16. Range = 5 − 0 = 5. The highest frequency is 6, for 3 books, so the mode = 3 books.
Example 2: Mean, median and mode of raw data
The daily rainfall (in mm) in a town over a week was 8, 12, 15, 10, 20, 7, 12. Find the mean, median and mode.
Solution: Sum = 8 + 12 + 15 + 10 + 20 + 7 + 12 = 84. Mean = 84/7 = 12 mm.
Ordered: 7, 8, 10, 12, 12, 15, 20. n = 7 is odd, so the median is the (7 + 1)/2 = 4th value = 12 mm.
12 occurs twice and every other value once, so the mode = 12 mm.
Example 3: Median of an even number of observations
The heights (in cm) of 8 students are 142, 150, 138, 145, 160, 148, 152, 141. Find the median height.
Solution: Ordered: 138, 141, 142, 145, 148, 150, 152, 160. n = 8 is even, so the median is the mean of the 4th and 5th values.
Median = (145 + 148)/2 = 293/2 = 146.5 cm.
Example 4: Finding a missing observation
The mean of six numbers is 15. Five of them are 12, 18, 10, 20 and 14. Find the sixth number.
Solution: Sum of six numbers = 6 × 15 = 90. Sum of the five known numbers = 12 + 18 + 10 + 20 + 14 = 74. Sixth number = 90 − 74 = 16.
Example 5: Correcting a wrong mean
The mean of 25 observations was found to be 36. Later it was found that one observation, 48, had been wrongly copied as 84. Find the correct mean.
Solution: Wrong sum = 25 × 36 = 900. Correct sum = 900 − 84 + 48 = 864. Correct mean = 864/25 = 34.56.
The wrong value was larger than the correct one, so the correct mean must be smaller than 36. It is.
Example 6: Mean, median and mode from a frequency table
A survey of 30 families recorded the number of children in each family.
| Children (x) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Families (f) | 3 | 8 | 12 | 5 | 2 |
Find the mean, median and mode.
Solution: Σf = 3 + 8 + 12 + 5 + 2 = 30. fx values: 0, 8, 24, 15, 8, so Σfx = 55. Mean = 55/30 ≈ 1.83 children.
n = 30 is even, so the median is the mean of the 15th and 16th values. Running totals of frequency: 3 (up to 0), 11 (up to 1), 23 (up to 2). The 12th to 23rd values are all 2, so the 15th and 16th are both 2. Median = 2.
The highest frequency is 12, for 2 children. Mode = 2.
Example 7: Weighted final score
A school gives weights of 20% to unit tests, 30% to the half-yearly exam and 50% to the annual exam. Anjali scores 85, 72 and 64 (out of 100) in them. Find her final score and compare it with the simple mean.
Solution: Final score = 0.2 × 85 + 0.3 × 72 + 0.5 × 64 = 17 + 21.6 + 32 = 70.6.
Simple mean = (85 + 72 + 64)/3 = 221/3 ≈ 73.67. The final score is lower because the annual exam, her lowest score, carries the largest weight.
Example 8: Price of a mixture
A shopkeeper mixes 5 kg of rice costing ₹48 per kg with 3 kg of rice costing ₹56 per kg. Find the cost per kg of the mixture.
Solution: Total cost = 5 × 48 + 3 × 56 = 240 + 168 = ₹408. Total quantity = 5 + 3 = 8 kg. Cost per kg = 408/8 = ₹51.
Check: 51 lies between 48 and 56 and is closer to 48, which has the larger weight.
Example 9: Combined mean of two sections
Section A has 35 students with mean marks 62. Section B has 25 students with mean marks 74. Find the mean marks of all 60 students.
Solution: Total marks of A = 35 × 62 = 2170. Total marks of B = 25 × 74 = 1850. Combined mean = (2170 + 1850)/60 = 4020/60 = 67.
The simple mean of 62 and 74 is 68; the correct answer is lower because the section with the lower mean is larger.
Example 10: Finding an unknown group size
Section A has 30 students with mean 50. Section B has some students with mean 60. The combined mean is 54. How many students are in Section B?
Solution: Let Section B have n students. Then (30 × 50 + 60n)/(30 + n) = 54.
1500 + 60n = 54(30 + n) = 1620 + 54n, so 6n = 120 and n = 20.
Check: (1500 + 1200)/50 = 2700/50 = 54.
Example 11: Reading a stacked bar graph
Use the stacked bar graph of Concept 4 (9A: Walk 0 to 14, Bus 14 to 30, Cycle 30 to 38; 9B: Walk 0 to 10, Bus 10 to 32, Cycle 32 to 42; 9C: Walk 0 to 18, Bus 18 to 28, Cycle 28 to 40). (i) Find the number of students in each section who travel by each mode. (ii) Which section has the most bus riders? (iii) How many students in all three sections cycle?
Solution: (i)
| Section | Walk | Bus | Cycle | Total |
|---|---|---|---|---|
| 9A | 14 | 30 − 14 = 16 | 38 − 30 = 8 | 38 |
| 9B | 10 | 32 − 10 = 22 | 42 − 32 = 10 | 42 |
| 9C | 18 | 28 − 18 = 10 | 40 − 28 = 12 | 40 |
(ii) 9B, with 22 bus riders. (iii) 8 + 10 + 12 = 30 students cycle.
Example 12: A 100% stacked bar graph and the share trap
A 100% stacked bar graph shows that in School P (250 students) 24% cycle, and in School Q (600 students) 20% cycle. Riya says, “More students cycle in School P.” Is she right?
Solution: Cyclists in P = (24/100) × 250 = 60. Cyclists in Q = (20/100) × 600 = 120.
Riya is wrong. P has the larger share of cyclists, but Q has twice as many cyclists. A 100% stacked bar graph compares shares; counts need the group totals.
Example 13: Choosing the right average
The times (in minutes) that 15 students take to reach school are: 12, 15, 10, 18, 15, 20, 14, 15, 11, 16, 13, 15, 17, 12, 55. Find the mean, median and mode, and say which best describes a typical student.
Solution: Sum = 258, so mean = 258/15 = 17.2 min.
Ordered: 10, 11, 12, 12, 13, 14, 15, 15, 15, 15, 16, 17, 18, 20, 55. n = 15, so the median is the 8th value = 15 min. 15 occurs four times, so the mode = 15 min.
The value 55 is an outlier. It pulls the mean up to 17.2, although 12 of the 15 students take less than 17.2 minutes. The median, 15 minutes, is the better typical value. Without the 55, the mean would be 203/14 = 14.5, while the median stays at 15.
Competency-Based Questions (with answers)
1. Case-based: The school canteen
A canteen records sales of three items on three days in a stacked bar graph. Monday: Samosa 0 to 40, Sandwich 40 to 65, Fruit cup 65 to 80. Tuesday: Samosa 0 to 35, Sandwich 35 to 70, Fruit cup 70 to 90. Wednesday: Samosa 0 to 50, Sandwich 50 to 70, Fruit cup 70 to 95.
(a) How many sandwiches were sold on Tuesday? 70 − 35 = 35.
(b) On which day were the most fruit cups sold? Monday 80 − 65 = 15, Tuesday 90 − 70 = 20, Wednesday 95 − 70 = 25. Wednesday.
(c) Find the mean number of items sold per day, and the item the canteen should prepare most. Totals are 80, 90 and 95, so mean = 265/3 ≈ 88.3 items per day. Over three days: samosa 40 + 35 + 50 = 125, sandwich 25 + 35 + 20 = 80, fruit cup 15 + 20 + 25 = 60. Samosa is the modal item, so the canteen should prepare the most samosas.
2. Case-based: Screen time in two schools
A 100% stacked bar graph shows daily screen time. School X (300 students): less than 1 hour 30%, 1 to 3 hours 50%, more than 3 hours 20%. School Y (500 students): less than 1 hour 20%, 1 to 3 hours 45%, more than 3 hours 35%.
(a) Which school has the larger share of students using screens for more than 3 hours? School Y, 35% against 20%.
(b) How many students in each school use screens for more than 3 hours? X: 20% of 300 = 60. Y: 35% of 500 = 175.
(c) A student looks only at the graph and says School X has more students using screens for less than 1 hour. Comment. The graph alone cannot show this, because it hides group sizes. Using the totals: X has 30% of 300 = 90 and Y has 20% of 500 = 100. Y has more such students, so the claim is wrong.
3. Case-based: The report card
A school’s final score gives 20% weight to periodic tests, 30% to the half-yearly exam and 50% to the annual exam. Kabir has scored 70 in periodic tests and 60 in the half-yearly exam. He wants a final score of at least 65.
(a) Write the final score in terms of his annual mark a. 0.2 × 70 + 0.3 × 60 + 0.5a = 14 + 18 + 0.5a = 32 + 0.5a.
(b) Find the least annual mark he needs. 32 + 0.5a ≥ 65 gives 0.5a ≥ 33, so a ≥ 66.
(c) If the three exams had equal weight, what annual mark would give 65? (70 + 60 + a)/3 = 65 gives 130 + a = 195, so a = 65. With the weights he needs one mark more. His weaker score, 60, carries 30% weight against 20% for the 70, so his two earlier scores have a weighted average of 32/0.5 = 64, which is below 65, and the annual exam must make up the gap.
4. Source-based: Picking a batter
A selector compares two batters over five matches. Batter A: 50, 48, 52, 49, 51. Batter B: 0, 5, 120, 10, 115.
(a) Find the mean score of each. A: 250/5 = 50. B: 250/5 = 50.
(b) Find the median and range of each. A: ordered 48, 49, 50, 51, 52, median 50, range 4. B: ordered 0, 5, 10, 115, 120, median 10, range 120.
(c) Whom should the selector pick for a steady performer, and why? Batter A. Both means are 50, but A’s range is 4 and B’s is 120, so A is far more consistent. B’s median of 10 shows that in most matches B scored little.
5. Assertion-Reason
Assertion (A): For the data 3, 5, 7, 9, 100, the median describes a typical value better than the mean. Reason (R): The median depends only on the middle value of the ordered data, so making an extreme value larger or smaller does not change it.
(a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true.
Answer: (a). Mean = 124/5 = 24.8, larger than four of the five values. Median = 7. The median resists the outlier 100 for exactly the reason given in R.
6. Assertion-Reason
Assertion (A): The weighted average of 60 with weight 3 and 80 with weight 1 is 70. Reason (R): A weighted average is pulled towards the value with the larger weight.
(a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true.
Answer: (d). Weighted average = (3 × 60 + 1 × 80)/4 = 260/4 = 65. The value 70 is the simple mean. R is true: 65 lies closer to 60, the value with the larger weight.
7. Assertion-Reason
Assertion (A): The mode is the suitable average for finding the favourite sport of a class. Reason (R): The mean of a data set need not be one of its observations.
(a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true.
Answer: (b). Sports are categories, so a mean or median cannot be found and the most frequent choice (the mode) is the answer. R is true (the mean of 1, 2, 4 is 7/3), but it is a different fact and does not explain A.
8. Investigation design: Asking the right question
Three students plan a class project. Aman asks, “How many periods are there on Monday?” Neha asks, “How much pocket money do Class 9 students of our school get in a week?” Rohit asks, “How much pocket money do I get?” (a) Whose question is a statistical investigative question? (b) Suggest a fair way to collect the data.
Answer: (a) Neha’s. Its answers vary from student to student. Aman’s and Rohit’s each have a single fixed answer. (b) Neha should survey students from every section of Class 9 (for example, a few students chosen at random from each section), so that the data represent the whole of Class 9. A survey of only her friends would be biased. These would be primary data.
9. Decision-making: The shoe shop
A shop sold these sizes of school shoes in a week: size 4: 6 pairs, size 5: 15 pairs, size 6: 22 pairs, size 7: 12 pairs, size 8: 5 pairs. The owner can reorder only one size in bulk. Which size should she choose, and which measure did you use?
Answer: Size 6, the mode, because it sold the most pairs (22). The mean size is Σfx/Σf = (24 + 75 + 132 + 84 + 40)/60 = 355/60 ≈ 5.92, which is not a size anyone can buy. For “which is most popular” decisions, the mode is the right measure.
Important Questions for Exams
1-Mark Questions
- Define the range of a data set. The difference between the highest and the lowest observation.
- Find the mode of 2, 4, 4, 5, 6, 6, 6, 7. 6, since it occurs three times.
- Find the mean of the first five natural numbers. (1 + 2 + 3 + 4 + 5)/5 = 15/5 = 3.
- Is “What is the height of our school flagpole?” a statistical investigative question? No. It has one fixed answer, so there is no variability in the data.
- In a 100% stacked bar, two segments are 35% and 40%. What is the third segment? 100 − 35 − 40 = 25%.
2-Mark Questions
- Find the median of 23, 19, 31, 27, 15, 22. Ordered: 15, 19, 22, 23, 27, 31. n = 6, so median = (22 + 23)/2 = 22.5.
- The mean of 5 observations is 24. If the observation 30 is removed, find the new mean. Sum = 5 × 24 = 120. New sum = 120 − 30 = 90. New mean = 90/4 = 22.5.
- If the mean of x, x + 2, x + 4, x + 6 and x + 8 is 11, find x. Sum = 5x + 20, so (5x + 20)/5 = x + 4 = 11, giving x = 7.
- Distinguish between primary and secondary data with one example of each. Primary data are collected first-hand by the investigator, for example a survey of classmates’ favourite fruit. Secondary data are taken from an existing source, for example rainfall figures from a newspaper.
3-Mark Questions
- Find the mean, median and mode of the marks: x = 10, 20, 30, 40, 50 with frequencies 4, 6, 9, 7, 4. Σf = 30; Σfx = 40 + 120 + 270 + 280 + 200 = 910; mean = 910/30 ≈ 30.33. Running totals 4, 10, 19, so the 15th and 16th values are both 30; median = 30. Highest frequency 9, so mode = 30.
- In the table x = 2, 4, 6, 8, 10 with frequencies 3, 2, 3, 1, p, the mean is 5. Find p. Σf = 9 + p; Σfx = 6 + 8 + 18 + 8 + 10p = 40 + 10p. (40 + 10p)/(9 + p) = 5 gives 40 + 10p = 45 + 5p, so 5p = 5 and p = 1.
- A student takes three subjects with credits 4, 3 and 1 and grade points 9, 8 and 5. Find the weighted grade point average and explain why it differs from the simple mean. Weighted average = (4 × 9 + 3 × 8 + 1 × 5)/(4 + 3 + 1) = (36 + 24 + 5)/8 = 65/8 = 8.125. The simple mean 22/3 ≈ 7.33 gives the 1-credit subject, where the grade point is lowest, the same importance as the 4-credit subject.
- A stacked bar for a family’s monthly spending has Books from 0 to 150, Snacks from 150 to 230 and Travel from 230 to 400 (in rupees). Find each amount, the total, and the percentage spent on travel. Books ₹150, Snacks ₹80, Travel ₹170, total ₹400. Travel share = 170/400 × 100 = 42.5%.
5-Mark Questions
- The times (in minutes) that 15 students take to reach school are 12, 15, 10, 18, 15, 20, 14, 15, 11, 16, 13, 15, 17, 12, 55. (i) Find the range. (ii) Find the mean, median and mode. (iii) Which measure best represents a typical student, and why? (i) Range = 55 − 10 = 45. (ii) Mean = 258/15 = 17.2; median = 8th ordered value = 15; mode = 15. (iii) The median. The value 55 is an outlier that pulls the mean above the times of 12 of the 15 students; the median ignores extreme values and stays at 15.
- Section A has 30 students with mean marks 50. Section B has 20 students with mean marks 60. (i) Find the combined mean. (ii) In Section A, one mark of 72 was wrongly entered as 27. Find the correct mean of Section A and the new combined mean. (iii) State why the combined mean is closer to Section A’s mean. (i) (30 × 50 + 20 × 60)/50 = (1500 + 1200)/50 = 54. (ii) Correct sum of A = 1500 − 27 + 72 = 1545, so correct mean = 1545/30 = 51.5. New combined mean = (1545 + 1200)/50 = 2745/50 = 54.9. (iii) The combined mean is a weighted average with group sizes as weights; Section A has more students (30 against 20), so it pulls the combined mean towards its own mean.
Common Mistakes and Examiner Tips
- Finding the median of unordered data. Always rewrite the data in ascending order first and show that line in your answer.
- Taking one middle value when n is even. For an even count there are two middle values; the median is their mean.
- Using the position as the median. (n + 1)/2 tells you which observation to pick. The 8th observation is the median, the number 8 is not.
- Calling the largest value the mode. The mode is the most frequent value. In a frequency table, look for the largest f, then write its x.
- Dividing by the old count. After removing or adding an observation, divide the new sum by the new number of observations.
- Changing the mean directly for a misread value. Correct the sum (subtract the wrong value, add the right one), then divide.
- Dividing Σfx by the number of rows. In a frequency table, divide by Σf, the total number of observations, never by how many different x-values there are.
- Taking the simple mean of weighted values. Group means, mixture prices and exam scores with weights all need Σwx/Σw. Check that the answer lies between the smallest and largest value.
- Reading the top of a stacked segment as its value. Only the bottom segment starts at 0. Subtract the bottom reading from the top reading.
- Treating a larger share as a larger number. In a 100% stacked bar graph, convert shares to counts with each group’s own total before saying “more”.
- Judging with the mean alone. When the question asks about consistency or reliability, compare ranges too. When there is an outlier, prefer the median and say why in one line.
Quick Revision Points
- A statistical investigative question expects answers that vary.
- Investigation cycle: pose, collect, organise, visualise, interpret.
- Primary data are collected first-hand; secondary data come from an existing source.
- Collect from all parts of the group to avoid bias.
- Frequencies in a table add up to the total number of observations.
- Range = highest − lowest; it measures spread and consistency.
- Bar graphs compare categories; line graphs show change over time.
- Check the scale: a vertical axis that does not start at 0 exaggerates differences.
- Stacked bar: whole bar = group total; segment = top reading − bottom reading.
- Compare upper segments only after calculating their values.
- 100% stacked bar: every bar is 100%; segments are percentage shares.
- Share = part/total × 100; count = percentage/100 × that group’s total.
- A 100% stacked bar graph hides group sizes.
- Mean = sum/n; sum = mean × n.
- Median: order first; n odd, ((n + 1)/2)th value; n even, mean of the (n/2)th and (n/2 + 1)th values.
- Mode = most frequent value; a set may have more than one mode or none.
- Mean of a frequency table = Σfx/Σf.
- Weighted average = Σwx/Σw; it lies between the smallest and largest values.
- Combined mean = (n1m1 + n2m2)/(n1 + n2).
- Outliers pull the mean; the median resists them.
- Use the mode for the most popular choice and for categorical data.
- Equal means can hide very different spreads; compare ranges for consistency.
Weightage and Exam Pattern
Class 9 has no board exam. This chapter is assessed in school exams: periodic tests, the half-yearly exam and the annual exam. In the CBSE Class IX Mathematics course structure for 2026-27, Unit VI, Statistics and Probability, carries 10 marks out of the 80-mark theory paper. Those 10 marks are shared between this chapter and Introduction to Probability; the course structure does not split them further. The unit is allotted 24 teaching periods. The remaining 20 marks of the year are internal assessment: pen-paper tests and multiple assessment (10), portfolio (5) and lab practical (5).
| Unit | Marks (out of 80) |
|---|---|
| I. Number System | 07 |
| II. Algebra | 20 |
| III. Coordinate Geometry | 04 |
| IV. Geometry | 25 |
| V. Mensuration | 14 |
| VI. Statistics and Probability | 10 |
The course structure lists the competencies this chapter builds: collecting, organising, visualising and interpreting data to answer a statistical investigative question; computing and applying a weighted average in different settings; reading and interpreting stacked and 100% stacked bar graphs; and interpreting mean, median and mode for decision-making. Expect questions in all of these forms: direct calculations of mean, median and mode, weighted average problems set in real contexts, graph-reading items described in a case, and short justifications of which average to use. The question paper design also gives weight to applying and analysing, so a one-line reason after each calculation is good practice. The same averages and graph-reading skills are used again in Class 10.
Class 9 Maths – swipe through all 9 cards to understand the whole chapter.
Statistical Questions
A statistical investigative question expects answers that vary, so it needs data.
Cycle: pose, collect, organise, visualise, interpret.
- How long do Class 9 students sleep? is statistical.
- What time does school start? has one fixed answer.
- Survey students from every section to avoid bias.
Frequency Table and Range
List each value with how often it occurs, then measure the spread.
Frequencies add up to the total number of observations.
- Sizes 5, 6, 7, 8 with f = 6, 14, x, 8 and total 40 give x = 12.
- Marks 3 to 10 give range 10 − 3 = 7.
- Primary data you collect; secondary data come from a source.
Stacked Bar Graph
Each bar is a group total, cut into segments for sub-categories.
Only the bottom segment starts at 0.
- Bus from 14 to 30 means 16 students.
- Missing part = bar total − other segments.
- Work out upper segments before comparing them.
100% Stacked Bar Graph
Every bar is 100% tall; segments show percentage shares of each group.
It hides group sizes: a bigger share can be fewer people.
- 18 of 40 walk gives 18/40 × 100 = 45%.
- 24% of 250 = 60, but 20% of 600 = 120.
- Segments in each bar add up to 100%.
Mean
The mean is the sum of all observations divided by their number.
For a frequency table, Mean = Σfx / Σf.
- Mean of 6 is 15 and five sum to 74, so sixth = 90 − 74 = 16.
- Misread value: fix the sum, then divide again.
- Removing a value: divide by the new count.
Median and Mode
Median is the middle of ordered data; mode is the most frequent value.
Always arrange the data in order before finding the median.
- 7, 9, 11, 12, 15, 20 has median (11 + 12)/2 = 11.5.
- 3, 5, 7, 5, 9, 5, 7 has mode 5.
- A set can have two modes or no mode.
Weighted Average
Multiply each value by its weight, add, and divide by the sum of the weights.
The answer lies between the smallest and largest value, nearer the heavier weight.
- Weights 20%, 30%, 50% on 80, 70, 60 give 16 + 21 + 30 = 67.
- 5 kg at ₹48 and 3 kg at ₹56 give ₹408 / 8 = ₹51 per kg.
- Credits 4, 3, 1 with grade points 9, 8, 5 give 65/8 = 8.125.
Combined Mean
Group sizes act as weights when two groups are joined.
Use the simple mean of the two group means only when the groups are equal in size.
- 35 students at 62 and 25 at 74 give 4020 / 60 = 67.
- Unknown size: 1500 + 60n = 54(30 + n) gives n = 20.
- The result is pulled towards the larger group.
Choosing the Right Average
Pick the measure that suits the data and the decision.
Equal means can hide very different spreads; compare ranges.
- Outliers such as salaries: use the median.
- Most popular size or favourite sport: use the mode.
- Evenly spread data with no outliers: use the mean.
📝 Practice Statistics - 10 questions
Concept-check questions · with answers & explanations
Start →Close ✕
Why C: The attendance of Class 9A today is one number found by a single count; there is no variability in the answer, so A is true. But R reverses the definition: a statistical question expects many varying answers, not a single fixed value, so R is false.
Why not A: R is false, so it cannot explain A.
Why not B: Option B needs both statements true, but R is false.
Why not D: A is true, since today’s attendance of one class is a single fixed count.
Remember: Statistical questions expect variability; one fixed answer means not statistical.
Why C: Class 9A total = 12 + 18 + 10 = 40 and Class 9B total = 15 + 15 + 5 = 35, so 9A’s bar is taller. Cyclists: 10 in 9A against 5 in 9B, so 9A also has more cyclists.
Why not A: 9B has 15 bus riders and 9A has 18, so 9A has more.
Why not B: The Walk segments are 12 and 15, which are different.
Why not D: 9B’s total is 35, less than 9A’s 40.
Remember: Compare totals by bar height, and compare parts only after working out each segment’s value.
Why D: Step 1 (count = % × total, School P): 30 / 100 × 200 = 60 cyclists. Step 2 (same rule, School Q): 20 / 100 × 500 = 100 cyclists. Q has more cyclists (100 > 60), while its share (20%) is smaller than P’s (30%).
Why not A: P’s share is larger but its count is smaller, 60 against 100.
Why not B: The counts are 60 and 100, not equal.
Why not C: Q’s share is 20%, which is smaller than P’s 30%.
Remember: A 100% stacked bar compares shares; to compare numbers you need each group’s total.
Why C: Step 1 (sum = mean × n): wrong sum = 20 × 10 = 200. Step 2 (replace the wrong value): correct sum = 200 – 43 + 34 = 191. Step 3 (divide by n): 191 / 10 = 19.1.
Why not A: 20.9 adds the error of 9 instead of subtracting it.
Why not B: 11 subtracts the whole error of 9 from the mean, 20 – 9, without sharing it over the 10 observations.
Why not D: 18.2 subtracts the error of 9 twice, (200 – 18) / 10.
Remember: Fix the sum, not the mean: remove the wrong value, add the right one, divide again.
Why A: Step 1 (total = mean × size): Group 1 total = 40 × 50 = 2000; Group 2 total = 56n. Step 2 (combined mean equation): 2000 + 56n = 52(40 + n) = 2080 + 52n. Step 3 (solve): 4n = 80, so n = 20. Check: (2000 + 1120) / 60 = 52.
Why not B: With 40 people, the combined mean would be exactly halfway, 53, not 52.
Why not C: 30 people give (2000 + 1680) / 70 ≈ 52.6, not 52.
Why not D: 10 people give (2000 + 560) / 50 = 51.2, not 52.
Remember: Write total = mean × size for each group, set up one equation, then check your answer.
Why B: Before: mean = 65 / 5 = 13, median = 12, mode = 12. After: data 10, 12, 12, 15, 60, mean = 109 / 5 = 21.8, median still 12 (3rd value), mode still 12. Only the mean changes.
Why not A: The median stays 12, because the middle (3rd) value is still 12.
Why not C: Neither the median nor the mode changes.
Why not D: The median is the one that does not change; the mean does.
Remember: Changing only the largest value moves the mean but leaves the median and mode alone.
Why B: The question is about all Class 9 students, so the data should come from students across every section. This avoids favouring any one area or mode of travel.
Why not A: Close friends often live near each other and travel the same way, so the sample is biased and too small.
Why not C: Everyone at the bus stop uses the bus, so walkers and cyclists are left out.
Why not D: A guess is not data at all; statistics needs collected observations.
Remember: Ask the whole group, or a fair spread of it, not just the people who are easy to reach.
Why B: Books = 150 – 0 = ₹150, Snacks = 230 – 150 = ₹80, Travel = 400 – 230 = ₹170. The largest is Travel, ₹170.
Why not A: Books at ₹150 looks large because it starts at zero, but it is less than Travel’s ₹170.
Why not C: Snacks is the smallest at ₹80, even though it sits above Books on the graph.
Why not D: ₹150 and ₹170 are not equal.
Remember: A segment that sits higher on the graph is not automatically bigger; subtract first.
Why A: Because every bar is drawn to the same height (100%) regardless of how many members the group has, the graph shows only shares. Group sizes are therefore invisible, which is exactly why A is true. So R is true and explains A.
Why not B: R is not just a true side fact; it is the very cause of A.
Why not C: R is true; all bars in a 100% stacked bar graph are the same height.
Why not D: A is true; the graph gives no information about group totals.
Remember: Same height for every bar means shares are visible and sizes are hidden.
Why B: Step 1 (sum = mean × n): 18 × 5 = 90. Step 2 (remove the value): 90 – 26 = 64. Step 3 (new mean = new sum / new n): 64 / 4 = 16.
Why not A: 12.8 is 64 / 5, dividing by the old count after one number was removed.
Why not C: The mean changes because 26 is not equal to the old mean 18.
Why not D: 20 would follow only if the removed number were smaller than 18; removing 26, which is above the mean, lowers the mean.
Remember: Work with the sum, then divide by the new count.
Chapter Navigation
Previous: Sequences and Progressions Class 9 Notes
Related Chapters in Class 9 Maths
- Coordinate Geometry Class 9 Notes
- Introduction to Polynomials Class 9 Notes
- Number System Class 9 Notes
- Exploring Algebraic Identities Class 9 Notes
- Circles Class 9 Notes
- Area and Perimeter Class 9 Notes
Explore More
Frequently Asked Questions
It is a question that is answered with data and whose answers are expected to vary. How many hours do Class 9 students of our school sleep on a school night is statistical, because each student gives a different answer. What time does our school start has one fixed answer, so it is not statistical.
Arrange the data in ascending order. With n observations and n even, there are two middle values, the (n/2)th and the (n/2 + 1)th. The median is their mean. For 7, 9, 11, 12, 15, 20 the middle values are 11 and 12, so the median is 11.5.
A weighted average multiplies each value by its weight, adds the products and divides by the sum of the weights: Σwx / Σw. It is used when values are not equally important, such as exams with different percentage weights, subjects with different credits, mixtures of items at different prices, or combining the means of groups of different sizes.
Only the bottom segment starts at zero. Every other segment starts where the one below it ends, so its value is its top reading minus its bottom reading. If a Bus segment runs from 14 to 30, then 16 students travel by bus. The height of the whole bar is the total for that group.
In a stacked bar graph the bar height is the actual group total, so it shows counts. In a 100% stacked bar graph every bar is the same height, 100%, and each segment is a percentage share of its group. It compares proportions across groups of different sizes but hides the group sizes, so counts need the totals.
The mean uses every value, so one very large or very small value pulls it away from most of the data. The median depends only on the middle position of the ordered data, so making an extreme value larger or smaller does not change it. For 10, 12, 12, 15, 60 the mean is 21.8 while the median is 12, which is closer to most values.
Multiply each group’s mean by its size to get each group’s total, add the totals, and divide by the total number of members: (n1m1 + n2m2) / (n1 + n2). For 30 students with mean 60 and 20 with mean 70, the combined mean is 3200 / 50 = 64, closer to the larger group’s mean.
In the CBSE Class IX course structure for 2026-27, Unit VI, Statistics and Probability, carries 10 marks out of 80 in the theory paper. The 10 marks are shared between Statistics and Introduction to Probability, and the course structure does not split them further. Class 9 has no board exam; this chapter is assessed in school exams.
