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 $σ_{1} = 2, 45$ ; $σ_{2} = 3, 21$ and $σ_{3} = 2, 78$ . Find the typical deviation of the total marks of the exam.

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Development:

Applying the formula $σ = \sqrt{\frac{σ_{1}^{2} + σ_{2}^{2} + \dots + σ_{n}^{2}}{n}} = \sqrt{\frac{2, 45^{2} + 3, 21^{2} + 2, 78^{2}}{3}} = \sqrt{\frac{24, 035}{3}} =$ $= \sqrt{8, 012} = 2, 83$

Solution:

$σ = 2, 83$

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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 $σ_{1} = 2, 73$ .
$4$ pupils choose exercise $1$ , whereby standard deviation of the grades turns out to be $σ_{2} = 1, 22$ .
$9$ pupils choose exercise $1$ , whereby standard deviation of the grades turns out to be $σ_{3} = 3, 12$ .
$3$ pupils choose exercise $1$ , whereby standard deviation of the grades turns out to be $σ_{4} = 2, 87$ .

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

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Development:

Applying the formula $σ^{2} = \frac{σ_{1}^{2} k_{1} + σ_{2}^{2} k_{2} + \dots + σ_{n}^{2} k_{n}}{k_{1} + k_{2} + \dots + k_{n}} = \frac{2, 73^{2} \cdot 15 + 1, 22^{2} \cdot 4 + 3, 12^{2} \cdot 9 + 2, 87^{2} \cdot 3}{15 + 4 + 9 + 3} =$ $\frac{111, 79 + 5, 95 + 87, 61 + 24, 71}{31} = \frac{230, 06}{31} = 7, 42$

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

Solution:

$σ^{2} = 7, 42$ and $σ = 2, 72$ .

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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 $\overset{―}{x} = \frac{1 + 1 + 1 + 3 + 3 + 4 + 4 + 5 + 6 + 6}{10} = \frac{34}{10} = 3, 4$ and the variance is found by applying the formula:

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

$σ = \sqrt{3, 44} = 1, 85$

Solution:

$σ^{2} = 3, 44$ ; $σ = 1, 85$ .

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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 $\overset{―}{x} = \frac{11 + 17 + 18 + 18 + 19 + 20 + 21}{7} = \frac{124}{7} ≃ 17, 7.$

The variance is calculated applying the formula:

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

Solution:

$σ = 3, 01$

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

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Development:

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

Solution:

$σ ≃ 3, 45$

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