Exercise 7.3 — Graphs

This problem gives the distribution of IQ scores for 45 students across seven classes, each of width 10. This is the simplest case for drawing a histogram, because the classes are already continuous and of equal width — the upper boundary of one class (say 70) is exactly the lower boundary of the next class. There is no extra calculation needed before plotting.

Steps of Construction

1. Draw two perpendicular lines — the horizontal x-axis for IQ scores and the vertical y-axis for the number of students.
2. Choose a scale: since the data starts at 60 (not 0), let 1 unit on the x-axis = 10 IQ points, and draw a small zig-zag (kink) near the origin to show that the scale does not start from zero.
3. On the y-axis, take 1 unit = 1 student.
4. For each class interval, draw a rectangle starting at its lower boundary and ending at its upper boundary, with height equal to the frequency (number of students). Because the data is continuous, every bar touches the next one — there should be no gaps.

Scale used: x-axis — 1 unit = 10 IQ points; y-axis — 1 unit = 1 student.

Notice the overall shape of this histogram — the bars rise gradually, peak at the 90–100 class (10 students), and then fall again. This bell-like shape means most students' IQ scores cluster around the middle of the range, which is a first hint at the idea of central tendency (mean, median, mode) that you will study in later chapters.

This problem has a twist. The table below gives the marks obtained by 600 students, but the numbers 360, 400, 440, 480, 520 and 560 are not class boundaries — they are the class marks (mid-points) of the actual classes. Before a histogram can be drawn, these class marks must be converted into proper class intervals.

Step 1: Find the Class Width (h)

Look at the difference between two consecutive class marks: 400 − 360 = 40. This difference is called h, the class width. Half of this value, h/2, is the amount that must be added and subtracted from each class mark to get its lower and upper boundary.

Step 2: Apply the Formula to Each Class Mark

1. For class mark 360: interval = (360 − 20) to (360 + 20) = 340 – 380
2. For class mark 400: interval = (400 − 20) to (400 + 20) = 380 – 420
3. For class mark 440: interval = (440 − 20) to (440 + 20) = 420 – 460
4. Continuing the same pattern, the remaining intervals are 460 – 500, 500 – 540 and 540 – 580.

Notice that the upper boundary of each interval is exactly the lower boundary of the next one (380, 420, 460 …), which makes the data continuous and ready for a histogram — exactly like in Problem 1.

Scale used: x-axis — 1 unit = 40 marks; y-axis — 1 unit = 10 students.

This problem asks for both a histogram and a frequency polygon on the same graph. The class intervals (₹500 – 550 up to ₹750 – 800) are equal and continuous, so the histogram is drawn exactly as in Problem 1. The frequency polygon is then added on top of it.

Finding the Class Marks

The class mark of an interval is its mid-point — found by adding the lower and upper boundaries and dividing by 2. For example, the class mark of 500 – 550 is (500 + 550) ÷ 2 = 525. The frequency polygon is drawn by plotting the point (class mark, frequency) for each class and joining these points with straight lines.

Steps of Construction

1. Draw the histogram for the given data exactly as in Problem 1, with bars touching each other.
2. Mark the class mark of each bar at its top centre and plot the point (class mark, frequency).
3. Join these six points in order using straight lines — this gives the frequency polygon.
4. To "close" the polygon so it touches the x-axis, imagine one extra class before the first (class mark 475, frequency 0) and one extra class after the last (class mark 825, frequency 0), and extend the lines to these two points.

Scale used: x-axis — 1 unit = ₹50; y-axis — 1 unit = 10 workers.

This problem shows that a frequency polygon does not need a histogram at all — it can be plotted directly from the class marks and frequencies. The same data is then used to draw a frequency curve, which is a smoothed version of the polygon.

Steps of Construction — Frequency Polygon

1. Take ages on the x-axis (1 unit = 4 years) and number of teachers on the y-axis (1 unit = 2 teachers).
2. Plot the six points (class mark, frequency) listed in the table above.
3. Add an extra point at class mark 22 with frequency 0 (one class width before the first class), and another at class mark 50 with frequency 0 (one class width after the last class).
4. Join all the points, including the two zero-frequency points, using straight lines to get a closed polygon that touches the x-axis at both ends.

Scale used in both graphs: x-axis — 1 unit = 4 years; y-axis — 1 unit = 2 teachers.

The data here is already given in "less than" cumulative form — for example, "Less than 10" tells us that 8 students scored below 10 marks in total, not that exactly 8 students scored between 5 and 10. To draw the graphs, we first need to find the actual class intervals and the individual (non-cumulative) frequency of each class.

Step 1: Find the Individual Frequencies

Since each value is a running total, the frequency of a class is found by subtracting the previous cumulative frequency from the current one. For example, the frequency of the class 5 – 10 is 8 − 2 = 6, since 8 students scored below 10 and 2 of them already scored below 5.

Step 2: The "Less Than" Ogive

For the "less than" ogive, the cumulative frequency is plotted against the upper boundary of each class. The point (0, 0) is also included, since "0 students scored less than 0".

Scale used: x-axis — 1 unit = 5 marks; y-axis — 1 unit = 5 students.

Step 3: The "Greater Than" Ogive

The "greater than" ogive uses the same individual frequencies, but the cumulative total is built up from the last class backwards — each value tells us how many students scored that mark or more. These cumulative totals are plotted against the lower boundary of each class.

Scale used: x-axis — 1 unit = 5 marks; y-axis — 1 unit = 5 students.