Weighted Measurement – Problem 4

1st: Compute for the summation of W

\[ \sum W = W_1 + W_2 + W_3 + W_4 \] \[ \sum W = 2 + 5 + 1 + 8 \] \[ \sum W = 16 \] 2nd: Multiply W with the distance

\[ X = W \times DE \] \[ X_1 = 2 \times 521.62 \] \[ X_1 = \frac{26081}{25} \] \[ X_2 = 5 \times 521.51 \] \[ X_2 = \frac{52151}{20} \] \[ X_3 = 1 \times 521.13 \] \[ X_3 = 521.13 \] \[ X_4 = 8 \times 521.93 \] \[ X_4 = \frac{104386}{25} \] 3rd: Compute for the summation of X

\[ \sum X = X_1 + X_2 + X_3 + X_4 \] \[ \sum X = \frac{26081}{25} + \frac{52151}{20} + 521.13 + \frac{104386}{25} \] \[ \sum X = \frac{208684}{25} \] 4th:
\[ MPV = \frac{\sum X}{\sum W} \] \[ MPV = \frac{\frac{208684}{25}}{16} \] \[ MPV = \frac{52171}{100} \] \[ \therefore MPV = 521.71 m. \leftarrow answer \]