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Find the mean and standard deviation for 30 objects given
∑(x - 40) = 315
∑(x - 40)2 = 4022

Normally, questions like these have to deal with expectation and variance, however, I don't really understand how to deal with the additional number within the brackets.

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    $\begingroup$ How many terms are you summing? $\endgroup$ – D.B. May 7 '18 at 23:27
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    $\begingroup$ 30, I forgot to mention that $\endgroup$ – lessaint May 7 '18 at 23:28
  • $\begingroup$ Are the $x$'s random samples? Should they be indexed by $i$? $\endgroup$ – D.B. May 7 '18 at 23:29
  • $\begingroup$ It wasn't mentioned, I'm sorry $\endgroup$ – lessaint May 7 '18 at 23:31
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Here's my attempt at the problem:

$$\sum_{i=1}^{30}(x_i-40) = \sum_{i=1}^{30}x_i + 30(-40)$$

$$\sum_{i=1}^{30}x_i + 30(-40) = 315$$

$$\sum_{i=1}^{30}x_i -1200 = 315$$

$$\sum_{i=1}^{30}x_i = 1515$$

Using the result above, we can determine the mean quite easily.

The mean is the sum of all the data points divided by the number of data points. Therefore,

$$\frac{\sum_{i=1}^{30}x_i}{30} = \frac{1515}{30} = 50.5$$

So, the mean is 50.5.

From here, let's us the other formula to get some other results:

$$\sum_{i=1}^{30}{(x_i-40)^2} = 4022$$

Expanding out $(x_i - 40)^2$, we get

$$x_i^2 - 80x_i + 1600$$

Do this 30 times, and we get:

$$-80\sum_{i=1}^{30}{x_i} + \sum_{i=1}^{30}{x_i^2} + 1600(30) = 4022$$

By substitution and rearranging, we get:

$$\sum_{i=1}^{30}{x_i^2} = 173222$$

From this we can calculate the standard deviation using the formula:

$$\sqrt\frac{\sum_{i=1}^{30}{(x_i-50.5)^2}}{30}$$

Expanding out $(x_i - 50.5)^2$, we get: $$x_i^2 - 101x_i + 2550.25$$

If we do this 30 times, we can rearrange it to obtain:

$$\sum_{i=1}^{30}{x_i^2} - 101\sum_{i=1}^{30}{x_i} + 2550.25(30)$$

Substituting, we obtain:

$$173222 - 101(1515) + 2550.25 = 96714.5$$

Now, substituting back into the original formula, we obtain:

$$\sqrt\frac{96714.5}{30} \approx 56.7$$

In conclusion, the standard deviation is approximately 56.7 and the mean is 50.5.

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Note that you have $30$ objects. If we assume that $\sum_{i=1}^{30} (x_i - 40) = 315$, then $\sum_{i=1}^{30} x_i = 315 + \sum_{i=1}^{30} 40 = 315 + 30*40$.

Can you write the mean in terms of $\sum_{i=1}^{30} x_i$?

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40 is the mean. You have the sum of the differences from the mean in your first equation. You have the sum of the square of the differences from the mean in your second equation. You want to know the square root of the average squared distance from the mean (the standard deviation). Divide the second quantity by (n-1)..... Bessel's correction = (30-1) and take the square root.

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Given: $$n=30, \qquad \sum (x-40)=315, \qquad \sum(x-40)^2=4022,$$ we need to find: $$\bar{x}=\frac{\sum x}{30} \qquad \text{and} \qquad \text{SD}=\sqrt{\frac{\sum(x-\bar{x})^2}{30}}.$$ The mean: $$\sum(x-40)=315 \Rightarrow \sum x -40\cdot 30=315 \Rightarrow \\ \bar{x}=\frac{\sum x}{30}=\frac{315+1200}{30}=50.5.$$ The standard deviation: $$\text{SD}=\sqrt{\frac{\sum(x-\bar{x})^2}{30}}=\sqrt{\frac{\sum(x-50.5)^2}{30}}=\sqrt{\frac{\sum(x-40-10.5)^2}{30}}=\\ \sqrt{\frac{\sum(x-40)^2-21\sum(x-40)+30\cdot 10.5^2}{30}}=\\ \sqrt{\frac{4022-21\cdot 315+3307.5}{30}}=\sqrt{\frac{714.5}{30}}\approx 4.88.$$

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