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Scrat [10]
3 years ago
15

17. When a single set of values is randomly divided into two equal groups, explain how the means of these two groups may be very

different from each other and may be very different
from the mean of the single set of values.
Mathematics
1 answer:
joja [24]3 years ago
7 0

If you take 2 groups of equal cardinality, it could happen that, for example, many of the higher values go to the first group and therefore, many of the low values go to the second group, making their respective means quite different and different from the original sample mean. This could even go worse due to the possible existence of outliers, that is, values that are far different than the sample mean. An outlier tend to disrup the mean of a sample, but for smaller samples the result is much dramatic.

For example, let X be {1,2,3,4,5,6,7,8,9,100}

The elements of X sum 145, hence the mean of X is 14,5. Let divide X in two groups

Y = {1,2,3,5,9}

Z = {4,6,7,8,100}

The elements of Y sum 20, so its mean is 4

The elements of Z sum 125, so its mean is 25

Both means are quite different from each other and quite different from the mean of X. Note that if we take the mean of the means the result is 4+25/2 = 14,5 which is equal to the mean of X.

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Consider the lengths of consecutive 1-2 blocks.

block 1 - 1, 2 - length 2

block 2 - 1, 2, 2 - length 3

block 3 - 1, 2, 2, 2 - length 4

block 4 - 1, 2, 2, 2, 2 - length 5

and so on.


Recall the formula for the sum of consecutive positive integers,

\displaystyle \sum_{i=1}^j i = 1 + 2 + 3 + \cdots + j = \frac{j(j+1)}2 \implies \sum_{i=2}^j = \frac{j(j+1) - 2}2

Now,

1234 = \dfrac{j(j+1)-2}2 \implies 2470 = j(j+1) \implies j\approx49.2016

which means that the 1234th term in the sequence occurs somewhere about 1/5 of the way through the 49th 1-2 block.

In the first 48 blocks, the sequence contains 48 copies of 1 and 1 + 2 + 3 + ... + 47 copies of 2, hence they make up a total of

\displaystyle \sum_{i=1}^48 1 + \sum_{i=1}^{48} i = 48+\frac{48(48+1)}2 = 1224

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\displaystyle \sum_{i=1}^{48} 1 + \sum_{i=1}^{48} 2i = 48 + 48(48+1) = 48\times50 = 2400

This leaves us with the contribution of the first 10 terms in the 49th block, which consist of one 1 and nine 2s with a sum of 1+9\times2=19.

So, the sum of the first 1234 terms in the sequence is 2419.

8 0
2 years ago
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