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Let X be a (data) set, we partition the data $A, B \subset X $ such that A and B are disjoint $(A \cap B = ∅ \ and \ A \cup B = X)$.

Mean(A) = 5

Mean(B) = 10

what is the Mean(X)?

1) Is the answer $5 \le Mean(X)\le 10?$

2) Also what if $ A \cup B \ne X$

It was on an SAT question where they said: there is a group, where the mean for women was N and mean for men was M, where N and M are numbers (we will assume N< M) but don't remember what they actually were. What is the mean for the whole group? I think I remember the answer key saying that it was $N \le Mean(X)\le M?$

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  • $\begingroup$ Can you show your work in attempting to solve the problem? $\endgroup$ – RayDansh Jul 12 '18 at 20:50
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For the disjoint case, find the sum of all the values in $X$ as $5|A|+10|B|$. Since there are $|X| = |A|+|B|$ values in all, our mean is

$$ \mu = \frac{5|A|+10|B|}{|A|+|B|} $$

and note that

$$ 5|A|+5|B| \leq 5|A|+10|B| \leq 10|A|+10|B| $$


For the non-disjoint case, consider $X = \{-10, 20, 0\}$ with $A = \{-10, 20\}$ and $B = \{20, 0\}$. Then the mean of $A$ is $5$ and the man of $B$ is $10$, but the mean of $X$ is $10/3$, which is outside the interval $[5, 10]$. It shouldn't be hard to construct other examples; observe that $\{10-u, u, 20-u\}$ is a generator with $A$ consisting of the first two, and $B$ consisting of the last two.

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Well, obviously the mean is in between $N$ and $M$ provided the groups are disjoint. If there were $a$ women and $b$ men in the group. Then the mean would be given by $$ \frac{aN + bM}{a+b} $$ where you should note that $a N + b M$ equals the sum of the quantity you're measuring of the whole group.

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