Question

Bio Chemical Oxygen Demand(BOD) concentration depends upon
velocity of flow. Work out the appropriate mean value from the data of
velocity of flow ( m/s) and BOD values given below.
5.0, 9.0, 6.0, 40.0, 2.0, 6.0, 3.0, 10.0, 5.0, 3.0, 7.0 BOD ( mg/l;)
0.8 1.0 1.2 1.4 1.0 1.3 1.0 1.4 1.2 1.0 0.9 Vel( m/S)
What statistic will you compute?

Answer 1

Answer on Question #63775 – Math – Statistics and Probability

Question

Bio Chemical Oxygen Demand(BOD) concentration depends upon velocity of flow. Work out the appropriate mean value from the data of velocity of flow ( m/s) and BOD values given below.

5.0. 9.0. 6.0. 40.0. 2.0. 6.0. 3.0. 10.0. 5.0. 3.0. 7.0 BOD ( mg/l;)

0.8 1.0 1.2 1.4 1.0 1.3 1.0 1.4 1.2 1.0 0.9 Vel( m/S)

What statistic will you compute?

Solution

Mean velocity of flow is

$$\bar{x} = \frac{\sum x}{n} = \frac{0.8 + 0.9 + 1 + 1 + 1 + 1 + 1.2 + 1.2 + 1.3 + 1.4 + 1.4}{11} = \frac{12.2}{11} = 1.11 \frac{m}{s}$$

Mean BOD is

$$\bar{y} = \frac{\sum y}{n} = \frac{5 + 7 + 9 + 2 + 3 + 3 + 6 + 5 + 6 + 40 + 10}{11} = \frac{96}{11} = 8.73 \frac{mg}{l}$$

We need to compute the correlation coefficient between velocity of flow and BOD:

$$r = \frac{n \sum xy - \sum x \sum y}{\sqrt{(n \sum x^2 - (\sum x)^2)((n \sum y^2 - (\sum y)^2)}} = \frac{11 \cdot 118.3 - 12.2 \cdot 96}{\sqrt{(11 \cdot 13.94 - (12.2)^2)((11 \cdot 1974 - (96)^2)}} = 0.60$$

$r = 0.60$ means that there is the positive strong linear relationship between velocity of flow and BOD.

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