Draw a histogram to represent the following information:

Marks 0 − 10 10 − 30 30 − 45 45 − 50 50 − 60

Number of students 8 32 18 10 6

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Solution

Here the class intervals are of unequal size. So, we calculate the adjusted frequency using the formula,
$Adjusted freq = \frac{\min class size}{class size of this class} \times freq$
Minimum class size = 5

Marks Frequency Width of class Height of rectangle

0-10 8 10 $\frac{5}{10} \times 8 = 4$

10-30 32 20 $\frac{5}{20} \times 32 = 8$

30-45 18 15 $\frac{5}{15} \times 18 = 6$

45-50 10 5 $\frac{5}{5} \times 10 = 10$

50-60 6 10 $\frac{5}{10} \times 6 = 3$

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Similar questions

\begin{matrix} Marks & 0 - 10 & 10 - 30 & 30 - 45 & 45 - 50 & 50 - 60 Number of students & 8 & 32 & 18 & 10 & 6 \end{matrix}

Draw a histogram for the marks of students given below:

$\begin{matrix} M a r k s : & 0 - 10 & 10 - 30 & 30 - 45 & 45 - 50 & 50 - 60 N o . o f s t u d e n t s & 8 & 32 & 18 & 10 & 6 \end{matrix}$

Class interval	5− 10	10 − 15	15 − 25	25 − 45	45 − 75
Frequency	6	12	10	8	18

Marks	0−10	10−20	20−30	30−40	40−50	50−60	60−70	70−80
Number of Students	2	18	30	45	35	20	6	3

Marks	0−10	10−20	20−30	30−40	40−50	50−60	60−70	70−80
No. of Students	2	18	30	45	35	20	6	3

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Marks	0 − 10	10 − 30	30 − 45	45 − 50	50 − 60
Number of students	8	32	18	10	6