Question

Draw the histogram of the following frequency distribution:

Class-Interval	Frequency
$0-9$	$5$
$10-19$	$8$
$20-29$	$12$
$30-39$	$18$
$40-49$	$22$
$50-59$	$10$

Answer 1

The given distribution is in inclusive form. It should be converted into exclusive form. This can be done by applying a correction factor $\frac{d}{2}$, where
$d=$ (lower limit of class) $-$ (upper limit of a class before it)
Here, we have
actual upper limit $=statedlimit+\frac{d}{2}$;
actual lower limit $=statedlimit-\frac{d}{2}$.
For example, consider the class limit $10-19$. We get
$d=$ lower limit of the class interval $-$ upper limit of class before it $=10-9=1$.
Hence, $d=1$ or $\frac{d}{2}=0.5$. Now,
actual upper limit $=$ (state upper limit) $+\frac{d}{2}=19+0.5=19.5$.
actual lower limit $=$ (state lower limit) $-\frac{d}{2}=10-0.5=9.5$
Converting into exclusive form, we get the table as below:

Stated Class interval	Actual Class interval	Frequency
$0-9$	$-0.5-9.5$	$5$
$10-19$	$9.5-19.5$	$8$
$20-29$	$19.5-29.5$	$12$
$30-39$	$29.5-39.5$	$18$
$40-49$	$39.5-49.5$	$22$
$50-59$	$49.5-59.5$	$10$