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The following data were available for the past 10 years on demand and lead time.
Lead Time 6.5 4.3 6.9 6.0 6.9 6.9 5.8 7.3 4.5 6.3
Demand 103 83 116 97 112 104 106 109 92 96

Estimate correlation and co variance.

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Solutions:

$\overline{X_1} = 6.14$

$\overline{X_2} = 101.8$

$\overline{X_1 X_2} = 625.052$

$\sum^2_1 x_1 x_2 = 6328.5$

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$\sigma^2_2 = \frac{\sum^n_1 x^2_1 – n x^2_1}{n-1} = \frac{386.44 – 10 \times (6.14)^2}{9}$

$\sigma^2_2 = \frac{\sum^n_1 x^2_1 – n x^2_1}{n-1} = \frac{104520 – 10 \times (101.8)^2}{9}$

$= 98.62$

$\therefore$ $\sigma_1 = 1.02$ and $\sigma_2 = 9.93$

$co^v (x_1 x_2) = \frac{1}{1-n} /sum^n_1 x_1 x_2 – \bar{X_1 X_2}$

$= \frac{6328.5 – 6250.52}{10-1} = 8.66$

Corelation $\rho = \frac{ cov (x_1 x_2)}{ \sigma_1 \sigma_2}$

$= \frac{8.66}{(1.02) (9.93)} = 0.86$

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