Name: Mean of Grouped Data: Example
Uploaded: 2019-07-19
Duration: 5 min 16 s
Description: Want to find out how to calculate the mean of grouped data when the class intervals are different? Watch this interesting video and learn more!

Question

The following frequency distribution gives the monthly consumption of electricity of $68$ consumers of a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption (in units)	Number of consumers
$65 - 85$	$4$
$85 - 105$	$5$
$105 - 125$	$13$
$125 - 145$	$20$
$145 - 165$	$14$
$165 - 185$	$8$
$185 - 205$	$4$

Accepted Answer

First of all, we have to find class mark for each interval.

We know that,

$C l a s s m a r k = \frac{Upper class limit + Lower class limit}{2}$

Let us take $135$ as assumed mean ( $a$ ), according to step deviation method we may find $d_{i}, u_{i}, f_{i} \times u_{i},$ as following

Monthly consumption (in units) Number of consumers $(f_{i})$ $X_{i}$ class mark $d_{i} = x_{i} - 135$ $4$ $f_{i} u_{i}$

$65 - 85$ $4$ $75$ $- 60$ $- 3$ $- 12$

$85 - 105$ $5$ $95$ $- 40$ $- 2$ $- 10$

$105 - 125$ $13$ $115$ $- 20$ $- 1$ $- 13$

$125 - 145$ $20$ $135$ $0$ $0$ $0$

$145 - 165$ $14$ $155$ $20$ $1$ $14$

$165 - 185$ $8$ $175$ $40$ $2$ $16$

$185 - 205$ $4$ $195$ $60$ $3$ $12$

Total $68$ $7$

We can observe from the table

$\sum f_{i} u_{i} = 7$

$\sum f_{i} = 68$

Class size $(h) = 20$

Mean $\bar{x} = a + (\frac{\sum f_{i} u_{i}}{\sum f_{i}}) \times h$
$= 135 + \frac{7}{68} \times 20$
$= 135 + \frac{140}{68}$
$= 137.058$

It is clear from the table that maximum class frequency is $20$ which belongs to the class interval $125 - 145$ .

Modal class $= 125 - 145$

Lower limit ( $l$ ) of modal class $= 125$

Class size ( $h$ ) $= 20$

Frequency ( $f_{1}$ ) of modal class $= 20$

Frequency ( $f_{0}$ ) of class preceding modal class $= 13$

Frequency ( $f_{2}$ ) of class succeeding the modal class $= 14$

$M o d e = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h$
$= 125 + [\frac{20 - 13}{2 (20) - 13 - 14}] \times 20$
$= 125 + \frac{7}{13} \times 20$
$= 125 + \frac{140}{13} = 135.76$

We know that,

$3$ median = mode + $2$ mean

$= 135.76 + 2 (137.058)$

$= 135.76 + 274.116$

$= 409.876$

⇒ Median $=$ $\frac{409.876}{3}$

Median $= 136.625$

Therefore, the median, mode, mean of given data is $136.625, 135.76, 137.05$ respectively.

Class interval	Frequency
$16$	$5$
$4$	$x$
$20 - 30$	$20$
$30 - 40$	$15$
$40 - 50$	$y$
$50 - 60$	$5$
Total	$60$

Age (in years)	Number of policy holders
Below $20$	$2$
Below $25$	$125 - 145$
Below $30$	$24$
Below $l$	$45$
Below $40$	$78$
Below $45$	$= 20$
Below $50$	$92$
Below $55$	$98$
Below $60$	$100$

Length $(in mm)$	Number or leaves $(f_{i})$
$118 - 126$	$3$
$127 - 135$	$5$
$136 - 144$	$9$
$145 - 153$	$= 135.76 + 2 (137.058)$
$154 - 162$	$5$
$163 - 171$	$4$
$172 - 180$	$2$

Lifetime (in hours)	Number of lamps
$1500 - 2000$	$14$
$2000 - 2500$	$56$
$2500 - 3000$	$60$
$3000 - 3500$	$86$
$3500 - 4000$	$74$
$4000 - 4500$	$62$
$4500 - 5000$	$48$

Number of letters	$1 - 4$	$4 - 7$	$7 - 10$	$10 - 13$	$13 - 16$	$16 - 19$
Number of surnames	$6$	$30$	$40$	$16$	$4$	$4$

Exercise 14.3

Jharkhand Board Mathematics Solutions for Exercise 14.3

Questions from Exercise 14.3 with Hints & Solutions

Learn and Prepare for any exam you want !

Weight (in kg)	$40 - 45$	$45 - 50$	$50 - 55$	$55 - 60$	$60 - 65$	$65 - 70$	$70 - 75$
Number of students	$2$	$3$	$8$	$6$	$6$	$3$	$2$