Variance and Standard deviation

We throw a dice $10$ times consecutively, obtaining the results: $1, 1, 1, 3, 3, 4, 4, 5, 6, 6$. Calculate the variance and the typical deviation of the throws.

The average is $\overline{x}=\dfrac{1+1+1+3+3+4+4+5+6+6}{10}=\dfrac{34}{10}=3,4$ and the variance is found by applying the formula:

$$\sigma^2=\dfrac{(1-3,4)^2+(1-3,4)^2+(1-3,4)^2+(3-3,4)^2+(3-3,4)^2+(4-3,4)^2}{10}+$$ $$+\dfrac{(4-3,4)^2+(5-3,4)^2+(6-3,4)^2+(6-3,4)^2}{10}=$$ $$=\dfrac{2,4^2+2,4^2+2,4^2+0,4^2+0,4^2+0,6^2+0,6^2+1,6^2+2,6^2+2,6^2}{10}=$$ $$=\dfrac{5,76+5,76+5,76+0,16+0,16+0,36+0,36+2,56+6,76+6,76}{10}=$$ $$=\dfrac{34,4}{10}=3,44 \\ $$

$$\sigma=\sqrt{3,44}=1,85$$

$\sigma^2=3,44$; $\sigma=1,85$.

The same exam is given to all the first grade classes in a high school. The teachers from the three classes of $30$ students each agree that, when correcting, they will get the same average.

Standard deviations of the grades of each class are, respectively $\sigma_1=2,45$; $\sigma_2=3,21$ and $\sigma_3=2,78$. Find the typical deviation of the total marks of the exam.

Applying the formula $$\sigma=\sqrt{\dfrac{\sigma_1^2+\sigma_2^2+\ldots+\sigma_n^2}{n}}=\sqrt{\dfrac{2,45^2+3,21^2+2,78^2}{3}}=\sqrt{\dfrac{24,035}{3}}=$$ $$=\sqrt{8,012}=2,83$$

$$\sigma=2,83$$

The teacher of mathematics allows each student to choose the exercise that he or she prefers from the 4 exercises proposed in the examination. The teacher guarantees the student that the average of the grades in each exercise will be the same. The results are the followings:

  • $15$ pupils choose exercise $1$, whereby standard deviation of the grades turns out to be $\sigma_1=2,73$.
  • $4$ pupils choose exercise $1$, whereby standard deviation of the grades turns out to be $\sigma_2=1,22$.
  • $9$ pupils choose exercise $1$, whereby standard deviation of the grades turns out to be $\sigma_3=3,12$.
  • $3$ pupils choose exercise $1$, whereby standard deviation of the grades turns out to be $\sigma_4=2,87$.

Find the variance and the typical deviation of the marks of the whole class.

Applying the formula $$\sigma^2=\dfrac{\sigma_1^2k_1+\sigma_2^2k_2+\ldots+\sigma_n^2k_n }{k_1+k_2+\ldots+k_n}=\dfrac{2,73^2\cdot15+1,22^2\cdot4+3,12^2\cdot9+2,87^2\cdot3}{15+4+9+3}=$$ $$\dfrac{111,79+5,95+87,61+24,71}{31}=\dfrac{230,06}{31}=7,42$$

Extracting the root the typical deviation we isolate $$\sigma=\sqrt{7,42}=2,72$$

$\sigma^2=7,42$ and $\sigma=2,72$.

Next, the points of a basketball team are shown. Find the existing typical deviation. $$73-\mbox{PANATINAIKOS}(21+27+8+17): \\ \mbox{Spanoulis}(13), \ \mbox{Pekovic}(6), \ \mbox{Fotsis}(13), \ \mbox{Nicholas}(7), \ \mbox{Perperoglou}(6), \\ \mbox{Batiste}(6), \ \mbox{Diamantidis}(10), \ \mbox{Jasikevicius}(10), \ \mbox{Tsartaris}(2)$$

The average is $\overline{x}=\dfrac{73}{9}\simeq8,1$. Next, we calculate the variance: $$\sigma^2=\dfrac{(13-8,1)^2+(13-8,1)^2+(10-8,1)^2+(10-8,1)^2+(7-8,1)^2+(6-8,1)^2}{9}+$$ $$+\dfrac{(6-8,1)^2+(6-8,1)^2+(2-8,1)^2}{9}=$$ $$=\dfrac{4,9^2+4,9^2+1,9^2+1,1^2+2,1^2+2,1^2+2,1^2+6,1^2}{9}=$$ $$=\dfrac{24,01+24,01+3,61+3,61+1,21+4,41+4,41+4,41+37,21}{9}=$$ $$=\dfrac{106,89}{9}=11,9$$ And the typical deviation is isolated $$\sigma=\sqrt{11,9}\simeq 3,45$$

$$\sigma\simeq3,45$$

We have the temperatures in several cities of Spain: Aviles $(11^\circ C)$, Barcelona $(17^\circ C)$, Madrid $(21^\circ C)$, Mallorca $(18^\circ C)$, Valencia $(18^\circ C)$, Marbella $(19^\circ C)$, Las Palmas $(20^\circ C)$.

Calculate the typical deviation of these temperatures.

The average is $\overline{x}=\dfrac{11+17+18+18+19+20+21}{7}=\dfrac{124}{7}\simeq 17,7.$

The variance is calculated applying the formula:

$$\sigma^2=\dfrac{(11-17,7)^2+(17-17,7)^2+(18-17,7)^2+(18-17,7)^2+(19-17,7)^2}{7}+$$ $$+\dfrac{(20-17,7)^2+(21-17,7)^2}{7}=\dfrac{6,7^2+0,7^2+0,3^2+0,3^2+1,3^2+2,3^2+3,3^2}{7}=$$ $$=\dfrac{44,89+0,49+0,09+0,09+1,69+5,29+10,89}{7}=$$ $$=\dfrac{63,43}{7}\simeq9,06$$ And the typical deviation is isolated $$\sigma=\sqrt{9,06}\simeq 3,01$$

$$\sigma=3,01$$

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