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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.

Find the mean, median and mode of the data.

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\underset{̲}{x} = 82.1428; Median = 75; Mode = 75

\underset{̲}{x} = 70; Median = 75; Mode = 50

\underset{̲}{x} = 80; Median = 5; Mode = 50

None of these

answer is A.

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Detailed Solution

Given data,
Question Image

Then,

For 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}

\Rightarrow Mean = \frac{11500}{140}

\Rightarrow Mean = 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,

\Rightarrow \frac{(140)}{2}

\Rightarrow \frac{140}{2}

\Rightarrow 70^{th}

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

.
Hence, the correct option is 1.

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