Problems from Variance and Standard deviation

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.

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Development:

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

Solution:

$$\sigma=2,83$$

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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.

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Development:

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 $$$\sigma=\sqrt{7,42}=2,72$$$

Solution:

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

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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.

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Development:

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:

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

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

Solution:

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

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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.

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Development:

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 $$\sigma=\sqrt{9,06}\simeq 3,01$$

Solution:

$$\sigma=3,01$$

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

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Development:

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 $$\sigma=\sqrt{11,9}\simeq 3,45$$

Solution:

$$\sigma\simeq3,45$$

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