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A set X={a_1, a_2, a_3, a_4, a_5} . . .

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A set X={a_1, a_2, a_3, a_4, a_5} . . .

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A set $$X=\left\{a_1,\ a_2,\ a_3,\ a_4,\ a_5\right\}$$ with non-zero elements have mean M. What is the mean of the set $$Y=\left\{a_1-1,\ a_2-2,\ \ a_3-3,\ a_4+5,\ a_5+1\right\}?$$

A. M-3
B. M-2
C. M-1
D. M
E. M+1

The OA is D.

Why the mean of Y is equal to the mean of X? How can I prove it? Experts, I need your help.

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A set $$X=\left\{a_1,\ a_2,\ a_3,\ a_4,\ a_5\right\}$$ with non-zero elements have mean M. What is the mean of the set $$Y=\left\{a_1-1,\ a_2-2,\ \ a_3-3,\ a_4+5,\ a_5+1\right\}?$$

A. M-3
B. M-2
C. M-1
D. M
E. M+1

The OA is D.

Why the mean of Y is equal to the mean of X? How can I prove it? Experts, I need your help.
Hi VJesus12,
Let's take a look at your question.

The mean of the set $$X=\left\{a_1,\ a_2,\ a_3,\ a_4,\ a_5\right\}$$ is M.
We know that mean is equal to the sum of the values divided by the number of the values, therefore,
$$M\ =\ \frac{a_1+a_2+a_3+a_4+a_5}{5}$$

Now , lets find the mean of the set $$Y=\left\{a_1-1,\ a_2-2,\ \ a_3-3,\ a_4+5,\ a_5+1\right\}.$$
$$Mean\ =\ \frac{\left(a_1-1\right)+\left(a_2-2\right)+\left(a_3-3\right)+\left(a_4+5\right)+\left(a_5+1\right)}{5}$$
$$Mean\ =\ \frac{a_1-1+a_2-2+a_3-3+a_4+5+a_5+1}{5}$$
$$Mean\ =\ \frac{a_1+a_2+a_3+a_4+a_5-1-2-3+5+1}{5}$$
$$Mean\ =\ \frac{a_1+a_2+a_3+a_4+a_5-6+6}{5}$$
$$Mean\ =\ \frac{a_1+a_2+a_3+a_4+a_5+0}{5}$$
$$Mean\ =\ \frac{a_1+a_2+a_3+a_4+a_5}{5}$$
Which is equal to the mean of the set $$X=\left\{a_1,\ a_2,\ a_3,\ a_4,\ a_5\right\}$$, i.e. M, therefore,
$$Mean\ =\ M$$

Therefore, Option D is correct.

Hope it helps.
I am available if you'd like any follow up.

_________________
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