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Re: The average (arithmetic mean) weight of 5 crates is 250 pou [#permalink]
IlCreatore wrote:
We need to find an interval, thus we can proceed by first computing the sum of the crates weights as mean times number, i.e. 250*5 = 1250.

Now, we can derive x at the two extremes, i.e. when the lightest and heaviest crates have the higher weight and when they have the smaller weight.

In the first case, x = 1250-2*200-2*300= 250. This is the highest value of x because we have taken the minimum weight for the other crates.
In the second one, x = 1250-2*205-2*310= 220. This is the lowest value of x.

Thus the interval is 220-250. Answer A



Why Multiplied by 2??
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Re: The average (arithmetic mean) weight of 5 crates is 250 pou [#permalink]
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