One Curved Graph Reveals the Middle Mark of Fifty Students

A table of grouped marks shows numbers, but a graph shows the shape — where most students scored, and where the middle lies.

This lesson uses the marks of $50$ students:

- $0$–$10$: $3$   $10$–$20$: $5$   $20$–$30$: $9$   $30$–$40$: $12$
- $40$–$50$: $8$   $50$–$60$: $6$   $60$–$70$: $4$   $70$–$80$: $3$

A histogram helps find the mode; an ogive helps find the median and quartiles. This chapter covers both.

Mark class boundaries on the horizontal axis and frequencies on the vertical axis, then draw adjacent bars whose widths are the class intervals and whose heights are the frequencies; if classes have gaps, first make them continuous.

Steps:

- Choose scales for both axes
- Draw one bar per class, with no gaps between bars
- Label the axes and give the scale

Making classes continuous. For classes $10$–$19$, $20$–$29$, $30$–$39$, the gap between $19$ and $20$ is $1$, so subtract $0.5$ from each lower limit and add $0.5$ to each upper limit:

$$9.5\text{–}19.5,19.5\text{–}29.5,29.5\text{–}39.5$$

Worked example. For the marks data, the tallest bar stands over $30$–$40$ with height $12$, and the shortest bars over $0$–$10$ and $70$–$80$ with height $3$.

An everyday example. A fruit seller grading mangoes by weight into groups can draw a histogram to see which weight group is most common.

The substance. Histogram bars touch, unlike a bar graph, because the classes are continuous; if the first class does not start at $0$, show a small kink on the axis.

In the tallest bar, join its top right corner to the top right corner of the bar before it, and its top left corner to the top left corner of the bar after it; from where these lines cross, drop a perpendicular to the horizontal axis to read the mode.

Worked example. For the marks data, the modal class is $30$–$40$ with frequency $12$; the bar before has $9$ and the bar after has $8$.

Drawing the two lines and dropping the perpendicular gives a mode of **about $34.3$ marks.** The same value follows from the formula:

$$\text{mode}=30+\frac{12-9}{2\times 12-9-8}\times 10=30+\frac{3}{7}\times 10\approx 34.29$$

An everyday example. A shoe shop noting the foot lengths of customers can find the most common size range from a histogram and stock more of it.

The substance. The mode lies closer to the taller neighbouring bar: here the bar before ($9$) is taller than the bar after ($8$), so the mode sits slightly left of the middle of $30$–$40$.

Add the frequencies one class at a time to get cumulative frequencies, then plot each cumulative frequency against the upper boundary of its class, start from the lower boundary of the first class at zero, and join the points with a smooth curve.

Cumulative frequency table:

- **Less than $10$**: $3$
- **Less than $20$**: $3+5=8$
- **Less than $30$**: $8+9=17$
- **Less than $40$**: $17+12=29$
- **Less than $50$**: $29+8=37$
- **Less than $60$**: $37+6=43$
- **Less than $70$**: $43+4=47$
- **Less than $80$**: $47+3=50$

Points to plot: $(0,0)$, $(10,3)$, $(20,8)$, $(30,17)$, $(40,29)$, $(50,37)$, $(60,43)$, $(70,47)$, $(80,50)$.

Check. The last cumulative frequency must equal the total, $50$.

An everyday example. A school analysing a class test uses the ogive to see how many students scored below any chosen mark.

The trap. Plot against upper class boundaries, not mid-points, since a less-than count is complete only at the end of each class.

For the median, go across from half the total on the cumulative frequency axis to the curve and down to the marks axis; for the lower and upper quartiles use one-quarter and three-quarters of the total; for a count below a value, go up from that value to the curve and across.

With $n=50$:

Median — at $\frac{50}{2}=25$:

$$30+\frac{25-17}{12}\times 10\approx 36.7\text{ marks}$$

Lower quartile — at $\frac{50}{4}=12.5$:

$$20+\frac{12.5-8}{9}\times 10=25\text{ marks}$$

Upper quartile — at $\frac{3\times 50}{4}=37.5$:

$$50+\frac{37.5-37}{6}\times 10\approx 50.8\text{ marks}$$

Interquartile range $\approx 50.8-25=25.8$ marks.

Counts. Below $45$ marks: about $29+0.5\times 8=33$ students. Below $65$: about $43+0.5\times 4=45$, so **about $5$ students scored above $65$.

An everyday example. To award prizes to the top $10$ students**, read the mark at a cumulative frequency of $40$: about $50+\frac{3}{6}\times 10=55$.

The substance. Values read from a hand-drawn ogive are estimates; small differences from calculated values are expected.