What formula is to be used when solving this (Standard Deviation)

Question

In a statistics lesson $12$ people were asked to think of a number, $x$, between $1$ and $20$ inclusive. From the results Tom found that $\sum x = 186$ and that the standard deviation of $x$ is $4.5$. Assuming that Tom's calculations are correct, find the values of $\sum(x - 10)$ and $\sum(x - 10)^2$.

Please can someone teach me how to do this.

Please help I have a test tomorrow on this.

Answer 1

$$\sum_{i=1}^{12}(X_i-10)=\sum_{i=1}^{12}X_i-12\times10=186-120=66$$

2.

Remember that, by definition

$$ \bbox[5px,border:2px solid black] { 4.5^2=\frac{1}{12}\sum_{i=1}^{12}X^2_i-\Bigg(\frac{1}{12}\sum_{i=1}^{12}X_i\Bigg)^2 \qquad (1) } $$

Using equation (1) you can easily get the value of

$$\sum_{i=1}^{12}X^2_i=(4.5^2+15.5^2)\cdot12=3126$$

Now simply expand

$$\sum_{i=1}^{12}(X_i-10)^2=\sum_{i=1}^{12}X^2_i-20\sum_{i=1}^{12}X_i+12\times 100=606$$

...just substituting and resolve.

Answer 2

Given the question in the image, you can do it this way : $$\sum_1^{12} (x-10) = \sum_1^{12} x -\sum_1^{12} 10 = 186-120 = 66$$

and for the second one, you know that $$4.5 = \sqrt{\frac{\sum_1^{12} (x-15.5)^2}{12}} \implies $$ $$243 = \sum_1^{12} (x-15.5)^2 = \sum_1^{12} (x-10)^2 +\sum _1^{12} 5.5^2 - \sum _1^{12} 11(x-10) $$

I think you can finish from here.

Answer 3

The first solution is $186-12\times10-66$. For the next, note $x-10$ has mean $186/12-10=5.5$ and standard deviation $4.5$, so $\sum(x-10)^2=12(4.5^2+5.5^2)=606$.