Statistics - Apply weighted average in practical comparative scenarios

A student's final grade is calculated using a weighted average. The Internal Assessment (weight $30\%$) score is $80$, and the Final Examination (weight $70\%$) score is $90$. Calculate the final weighted average score.

Given: Weight for Internal ($w_1$) = $30$, Score ($x_1$) = $80$ Weight for Final ($w_2$) = $70$, Score ($x_2$) = $90$

Using the formula: $\bar{x}_w = \frac{w_1x_1 + w_2x_2}{w_1 + w_2}$

$\bar{x}_w = \frac{30 \times 80 + 70 \times 90}{30 + 70}$

Sum calculation: $\begin{array}{r} 2400 \\ + 6300 \\ \hline 8700 \end{array}$

$\bar{x}_w = \frac{8700}{100} = 87$

The final weighted average score is $87$.

We multiply each score by its corresponding percentage weight and divide by the total weight ($100\%$). This ensures the final exam has a greater impact on the result than the internal assessment.

Section A of Grade 9 has $20$ students with an average height of $150\text{ cm}$. Section B has $30$ students with an average height of $160\text{ cm}$. Find the combined average height of both sections.

Given: $n_1 = 20, \bar{x}_1 = 150$ $n_2 = 30, \bar{x}_2 = 160$

Combined Mean formula: $\bar{X}_{combined} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}$

$\bar{X}_{combined} = \frac{20 \times 150 + 30 \times 160}{20 + 30}$

Calculation of Numerator: $20 \times 150 = 3000$ $30 \times 160 = 4800$ $\begin{array}{r} 3000 \\ + 4800 \\ \hline 7800 \end{array}$

Calculation of Denominator: $20 + 30 = 50$

$\bar{X}_{combined} = \frac{7800}{50} = 156$

The combined average height is $156\text{ cm}$.

Since the number of students in Section B is higher, the combined average is pulled closer to Section B's average ($160$) than Section A's average ($150$).