Question

On the occasion of New year's day a sweet stall prepared sweet packets. Number of sweet packets and cost of each packet are given as follows.

Cost of packet (in Rs)	$Rs.25$	$Rs.50$	$Rs.75$	$Rs.100$	$Rs.125$	$Rs.150$
No of packets	$20$	$26$	$32$	$29$	$22$	$11$

Find the mean, median and mode of the data

• A. $\overline{x}=82.1428;Median=75;Mode=75$
• B. $\overline{x}=70;Median=75;Mode=50$
• C. $\overline{x}=80;Median=5;Mode=50$
• D. None of these

Answer 1

The correct option is A $\overline{x}=82.1428;Median=75;Mode=75$

Cost of packet	Number of packets	Cumulative frequency
25	20	20
50	26	46
75	32	78
100	29	107
125	22	129
150	11	140

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

So, Mean $=\frac{(25\times 20+50\times 26+75\times 32+100\times 29+125\times 22+150\times 11)}{(20+26+32+29+22+11)}$

Mean $=\frac{(500+1300+2400+2900+2750+1650)}{140}=\frac{11500}{140}=82.1428$

As we can see that the cost of greatest number of packets is Rs. 75 each, so, the mode will be 75.

As we can see there are 140 packets, arranged in increasing order of their cost. So, the middle one will be $\frac{\left(140\right)}{2}=\frac{140}{2}={70}^{th}$ packet in the list.

So, the median will be the cost of packet corresponding to cumulative frequency just greater or equal to $70i.e.75$.

Therefore, median is $75$.