Question

The donations given to an orphanage home by the students of different classes of a secondary school are given below.

Class	Donation by each students (in Rs)	No. of students donated
VI	$5$	$15$
VII	$7$	$15$
VIII	$10$	$20$
IX	$15$	$16$
X	$20$	$14$

Find the mean, median and mode of the data

• A. $Rs.11.26,Median=Rs.10;Mode=Rs.10$
• B. $Rs.1.26,Median=Rs.10;Mode=Rs.10$
• C. $Rs.11.26,Median=Rs.20;Mode=Rs.10$
• D. None of these

Answer 1

Answer: (A)

$Rs.11.26,Median=Rs.10;Mode=Rs.10$

Donation by each student	No. of students donated	Cumulative frequency
$5$	$15$	$15$
$7$	$15$	$30$
$10$	$20$	$50$
$15$	$16$	$66$
$20$	$14$	$80$

To find the mean we will add up total cost of packets , then divide by total number of packets.

So,

Mean $=\frac{(5\times 15+7\times 15+10\times 20+15\times 16+20\times 14)}{(15+15+20+16+14)}$

Mean $=(75+105+200+240+280)/80=900/80=11.26$

As we can see that the donation of greatest number of student is Rs. $10$ each, so the mode will be $10$

As we can see there are $80$ students arranged in increasing order of their respective donations. So, the middle one will be $\left(80\right)/2={40}^{th}$ and ${41}^{st}$ student in the list.

So, the median will be the donation by student corresponding to cumulative frequency just greater or equal to $40$ and $41$ , i.e., $10$

Therefore , median is $10$