Variability

In the table below, the number of wrong produced goods in two shifts: morning shift: 2; 0; 6; 10; 2; 2; 4; 2; 5; 2; afternoon shift: 4; 4; 0; 2; 10; 2; 6; 2; 3; 10; Compare the variability in both shifts, compare the average number of wrong goods on both shifts as well as other variability and positioning rates.

Correct answer:

a₁ = 3.5
a₂ = 4.3
s₁ = 2.7295
s₂ = 3.2265

Step-by-step explanation:

a_{1} = (2 + 0 + 6 + 10 + 2 + 2 + 4 + 2 + 5 + 2) / 10 = 3.5

a_{2} = (4 + 4 + 0 + 2 + 10 + 2 + 6 + 2 + 3 + 10) / 10 = \frac{4 3}{1 0} = 4 \frac{3}{1 0} = 4.3 a_{2} > a_{1}

b_{1} = (2 - a_{1})^{2} + (0 - a_{1})^{2} + (6 - a_{1})^{2} + (10 - a_{1})^{2} + (2 - a_{1})^{2} + (2 - a_{1})^{2} + (4 - a_{1})^{2} + (2 - a_{1})^{2} + (5 - a_{1})^{2} + (2 - a_{1})^{2} = (2 - 3.5)^{2} + (0 - 3.5)^{2} + (6 - 3.5)^{2} + (10 - 3.5)^{2} + (2 - 3.5)^{2} + (2 - 3.5)^{2} + (4 - 3.5)^{2} + (2 - 3.5)^{2} + (5 - 3.5)^{2} + (2 - 3.5)^{2} = \frac{1 4 9}{2} = 74 \frac{1}{2} = 74.5 r_{1} = 1 / 10 \cdot b_{1} = 1 / 10 \cdot 74.5 = \frac{1 4 9}{2 0} = 7 \frac{9}{2 0} = 7.45 s_{1} = r_{1} = 7.45 = 2.7295

b_{2} = (4 - a_{2})^{2} + (4 - a_{2})^{2} + (0 - a_{2})^{2} + (2 - a_{2})^{2} + (10 - a_{2})^{2} + (2 - a_{2})^{2} + (6 - a_{2})^{2} + (2 - a_{2})^{2} + (3 - a_{2})^{2} + (10 - a_{2})^{2} = (4 - 4.3)^{2} + (4 - 4.3)^{2} + (0 - 4.3)^{2} + (2 - 4.3)^{2} + (10 - 4.3)^{2} + (2 - 4.3)^{2} + (6 - 4.3)^{2} + (2 - 4.3)^{2} + (3 - 4.3)^{2} + (10 - 4.3)^{2} = \frac{1 0 4 1}{1 0} = 104 \frac{1}{1 0} = 104.1 r_{2} = 1 / 10 \cdot b_{2} = 1 / 10 \cdot 104.1 = \frac{1 0 4 1}{1 0 0} = 10 \frac{4 1}{1 0 0} = 10.41 s_{2} = r_{2} = 10.41 ≐ 3.2265 s_{2} > s_{1}

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Grade of the word problem:

high school

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