Statistics
CBSE Class 9 · Mathematics · Notes, formulas and practice questions
This chapter explains how to collect and present data using frequency distribution tables and graphs. It covers ungrouped and grouped distributions, class intervals, and drawing bar graphs, histograms (including unequal class widths) and frequency polygons.
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This chapter introduces statistics as the branch dealing with collection, organisation and presentation of numerical data. At the outset we gather raw data and usually its volume is large. To make it meaningful, we first organise it into a frequency distribution table, grouping observations into classes when necessary. The way we present data determines which graphical representation, such as a bar graph, histogram or frequency polygon, is most suitable.
The frequency distribution can be ungrouped if the data are discrete with few distinct values; we just list each value and its count. For continuous or numerous values, we form a grouped frequency distribution. In grouped distribution, we choose class intervals of a fixed size, write the lower and upper class limits, and count the number of observations falling in each interval. All class intervals should be continuous and non-overlapping; if an inclusive form is given, we first convert it to an exclusive (continuous) form by applying a suitable correction.
The visual representation of data is an important skill. A bar graph is used for discrete categories, where bars are separated by equal gaps. A histogram is used for grouped continuous data, where bars are adjacent without gaps. In a histogram, the area of each rectangle is proportional to the frequency of the corresponding class. When all class widths are equal, one can compare frequencies directly by heights.
A histogram becomes slightly more complex when class intervals are of unequal widths. In such a case, using the heights alone would misrepresent frequencies. We must adjust the heights according to the class width, usually by taking the smallest class width as the base and calculating adjusted frequencies for other classes. The frequency polygon is an alternative that uses class marks instead of rectangles; joining the midpoints creates a simple line graph that is especially useful for comparing distributions.
Key terms
- Frequency
- The number of times a particular observation or class of observations occurs in a data set. For example, if 5 students scored 14 out of 20, the frequency of the observation 14 is 5.
- Ungrouped frequency distribution
- A table that records each distinct value of a variable and its frequency. It is usually prepared when the data are discrete and the number of different values is small, such as marks scored by 20 students.
- Grouped frequency distribution
- A table in which data are divided into class intervals and the total number of observations falling in each interval is recorded. It is used when data are continuous or contain many different values.
- Class interval
- A continuous range of values into which data are grouped, written as lower limit–upper limit, e.g., 10–20. The lower limit and upper limit are the two ends of the interval, and every observation lies within these limits. Intervals must be continuous and non-overlapping.
- Class size
- The difference between the upper limit and lower limit of a class interval, e.g., 10–20 has class size 10. It is also the difference between the lower limits of two successive classes when the intervals are continuous.
- Class mark
- The midpoint of a class interval, obtained by adding the lower and upper limits and dividing by 2. For the interval 10–20, the class mark is (10+20)/2 = 15. It is used as the representative value of the interval when drawing a frequency polygon.
- Histogram
- A graphical representation of a grouped continuous frequency distribution in which adjacent rectangles are drawn for each class. The width of each rectangle equals the class width, and its height is chosen so that the area of the rectangle is proportional to the class frequency. For classes of equal width, the heights are simply proportional to frequencies.
- Frequency polygon
- A line graph obtained by plotting class marks against the corresponding frequencies and joining the points in order with straight line segments. To close the polygon, the line is extended to meet the x-axis at the class marks of an imaginary class before the first and after the last, each having zero frequency.
- Bar graph
- A graph using equally spaced rectangular bars to represent frequencies of distinct or discrete categories. The length or height of a bar is proportional to the corresponding frequency, and gaps are left between consecutive bars. Bar graphs are not used for continuous grouped data; histograms serve that purpose.
Formula sheet
| What | Formula | Notes |
|---|---|---|
| Class mark | (L + U) / 2 | L is the lower limit of the class interval and U is the upper limit. It gives the midpoint of the class. |
| Class size | U − L | U and L are the upper and lower limits of the interval. It gives the width of the class. For continuous classes, it also equals the difference between the lower limits of two successive classes. |
| Adjusted frequency for histogram with unequal class widths | f′ = f × B / w | f is the actual frequency of the class, B is the smallest class width (taken as the base), and w is the width of the class under consideration. The result gives the height of the rectangle when the graph is drawn with the smallest class width as standard. |
Practice questions with answers
1. Define class mark. Find the class mark of the class interval 45−55.
Class mark is the average of the lower and upper limits of a class interval, i.e., class mark = (lower limit + upper limit)/2. For 45−55, it is (45 + 55)/2 = 100/2 = 50.
2. Convert the following inclusive class intervals into continuous exclusive intervals: 10−19, 20−29, 30−39.
The gap between the upper limit of one interval and the lower limit of the next is 1 (19 to 20, 29 to 30). Subtract half the gap (0.5) from each lower limit and add 0.5 to each upper limit. The exclusive intervals are 9.5−19.5, 19.5−29.5, 29.5−39.5.
3. Consider the class intervals 5−10, 10−15, 15−20. Write the lower limit, upper limit and class mark of the middle class, and state its class size.
The middle class is 10−15. Its lower limit is 10 and upper limit is 15. Class mark = (10 + 15)/2 = 12.5. Class size = difference between upper and lower limit = 15 − 10 = 5.
4. Why is a histogram used instead of a bar graph to represent grouped continuous data?
A bar graph shows discrete or separate categories and therefore bars are drawn with gaps, suggesting discontinuity. A histogram is meant for continuous data grouped into class intervals, so the rectangles are adjacent and the boundaries merge. This matches the idea that one class interval ends exactly where the next begins, and the area of each rectangle represents frequency.
5. When class widths in a grouped distribution are unequal, what adjustment is made to the heights of rectangles in a histogram? Give the relation.
The height is adjusted so that the area of a rectangle is proportional to its frequency. Choose the smallest class width as the base. For a class of width w, the adjusted frequency (or height) = actual frequency × (smallest width)/w.
6. A class 20−35 has frequency 15, and the smallest class width in the table is 5. Find the adjusted frequency that should be taken as the height of its rectangle.
Width of the class 20−35 is 35 − 20 = 15. Using the formula, adjusted frequency = 15 × (5/15) = 5. So the height of the rectangle is 5 units when the standard base width is 5.
7. The range of a continuous data set is 40. If the desired class size is 8, what is the minimum number of classes required to represent the data in a grouped frequency distribution?
Number of classes = range ÷ class size = 40 ÷ 8 = 5. Therefore at least 5 class intervals are required, each of width 8, to cover the whole range.
8. In a histogram, the heights of three rectangles are in the ratio 3 : 5 : 7. If the total frequency of these three classes is 300, find the frequency of the class whose rectangle height is 5.
Since the class widths are equal, frequencies are proportional to the heights. Let the frequencies be 3k, 5k and 7k. Their sum is 15k = 300, so k = 20. Hence the required frequency = 5 × 20 = 100.
9. Describe the steps to draw a frequency polygon without first drawing a histogram.
Steps: (1) Find the class mark of each class interval. (2) Plot points whose x-coordinate is the class mark and y-coordinate is the frequency of that class. (3) Join these points successively with straight line segments. (4) To close the polygon, assume an imaginary class with zero frequency before the first class and one after the last; take their class marks on the x-axis and join the endpoints to these two points.
10. Two successive classes have equal width. Their frequencies are 20 and 15. What is the ratio of the heights of their rectangles in a histogram?
For equal class widths, the area and height of a rectangle are both proportional to the frequency of the class. Therefore the ratio of heights is the same as the ratio of frequencies, 20 : 15, which simplifies to 4 : 3.
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