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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.25Rs.50Rs.75Rs.100Rs.125Rs.150
No of packets202632292211
Find the mean, median and mode of the data

A
¯¯¯x=82.1428;Median=75;Mode=75
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B
¯¯¯x=70;Median=75;Mode=50
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C
¯¯¯x=80;Median=5;Mode=50
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D
None of these
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Solution

The correct option is A ¯¯¯x=82.1428;Median=75;Mode=75
Cost of packet Number of packets Cumulative frequency
2520 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 =(25×20+50×26+75×32+100×29+125×22+150×11)(20+26+32+29+22+11)
Mean =(500+1300+2400+2900+2750+1650)140=11500140=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 (140)2=1402=70th packet in the list.
So, the median will be the cost of packet corresponding to cumulative frequency just greater or equal to 70 i.e.75.
Therefore, median is 75.

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