$$N=30 \\ \sum_{i=1}^{30}x_i=120 \\ \\ \sum_{i=1}^{30}x_i^{2}=750\\ $$

Find a standard deviation of x:

$$ \bar{x}=\frac{120}{30}=4 \\ sd=\sqrt[]{\frac{\sum_{i=1}^{n}(x_i-\bar{x})^{2}}{N}}=\sqrt[]{\frac{\sum_{i=1}^{n}(x_i^{2}-2\bar{x}x_i+\bar{x}^{2})}{N}}=\sqrt[]{\frac{\sum_{i=1}^{n}x_i^{2}-2n\bar{x}\sum_{i=1}^{n}x_i+n\bar{x}^{2}}{N}}=\sqrt[]{\frac{750-2*30*4*120+30*{4}^{2}}{30}}=30.1$$

But in the textbook the answer is 4


There is another formula for the variance of a random variable derived from the first and second moments.


As it turns out, you have all the information you need to estimate this.

  • $\begingroup$ Oh my god, I forgot about that formula. Thanks. But it gives a different answer (27) $\endgroup$ – PHP Useless Jan 19 at 11:50
  • $\begingroup$ @PHPUseless Are you sure about that? Is that the result of the variance of $X$ or the standard deviation of $X$? $\endgroup$ – user2974951 Jan 19 at 11:59
  • $\begingroup$ yeah, sd=sqrt(27), it is about 5, but the answer is 4. And what's wrong with my calculation? $\endgroup$ – PHP Useless Jan 19 at 12:13
  • $\begingroup$ @PHPUseless $SD(X)=\sqrt{750/30-(120/30)^2}=3$ for me. $\endgroup$ – user2974951 Jan 20 at 9:50
  • $\begingroup$ why should we divide 750 by N=30? $\endgroup$ – PHP Useless Jan 24 at 18:55

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