Let $a$ and $b$ be constants and let $y_j = ax_j + b$ for $j=1,2,\ldots,n$.
What are the relationships between the means of $ya$ and $x$, and the standard deviations of $y$ and $x$?
I'm slightly confused with how to approach a theoretical question such as this and was wondering if anyone could help provide me some advice on how to approach this problem.
At the moment here is what I'm thinking, but I'm currently working without certainty:
We know
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$x_j = (y_j – b)/a$
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The mean of $x$ = mean of $y$
In terms of standard deviation, I'm not sure how they correlate at all right now aside from the fact that you need the mean of $x$ or $y$ in order to calculate the corresponding standard deviation.
If someone could help explain this question and help me understand what I'm being asked and how to solve this I would greatly appreciate it!
EDIT: So looking at the second portion of the question I am doing the following:
SD = sqrt(Sigma(y_i – y)^2/(n-1)))
SD(y) = (Sigma(yi – (ax+b)))/(n-1)
SD(y) = (Sigma (ax+b) – (ax+b))/(n-1)
SD(y) = 1/(n-1)
Is the following correct?
Best Answer
This is not true.
The way you should approach this problem is to use the formulas for mean and standard deviation directly: \begin{align*} \text{Mean}(y_1, y_2, \ldots, y_n) &= \frac{y_1 + y_2 + \cdots + y_n}{n} \\ &= \frac{(ax_1 + b) + (ax_2 + b) + \cdots + (ax_n + b)}{n} \\ &= \frac{a(x_1 + x_2 + \cdots + x_n) + nb}{n} \\ &= a \cdot \text{Mean}(x_1, x_2, \ldots, x_n) + b \\ \end{align*}
See if you can do a similar algebraic manipulation for standard deviation.