Question

weighted mean is given by the formula below, where $w_1$, $w_2$, $w_3$, $w_n$ are the weights and $x_1$, $x_3$, $x_n$ are the data points (the test scores).
weighted mean $=\frac{\sum(x\cdot w)}{\sum w}$
weights can be summarized as follows.
sum of the weights for four tests $=15\\%+15\\%+15\\%+15\\%=60\\%$
weight of the term paper $=20\\%$
weight of the final examination $=20\\%$
fore, the total sum of the weights is given by the following.
$\

$$\begin{align*}\\sum w&=60\\%+20\\%+20\\%\\\\&=100\\%\\end{align*}$$

$
nd the weighted mean, multiply each grade (the data point $x$) by its appropriate percentage weight $w$) and find the sum of the products. then, divide by the sum of the weights and lify. (round your answer to one decimal place.)
$\frac{(x\cdot w)}{\sum w}=\frac{(95\times 15\\%)+(85\times 15\\%)+(87\times 15\\%)+(\times 15\\%)+(87\times 20\\%)+(}{100\\%}$
$=\frac{}{}$

Explanation:

Step1: Calculate each product

• \(95\times15\%=95\times0.15 = 14.25\)
• \(85\times15\%=85\times0.15 = 12.75\)
• \(87\times15\%=87\times0.15 = 13.05\)
• Assume the fourth test score is \(x_4\) (the value in the first wrong box). Let's first calculate the sum of the known products: \(14.25+12.75 + 13.05=40.05\)
• \(87\times20\%=87\times0.2=17.4\)
• Assume the final - exam score is \(x_6\) (the value in the second wrong box).

Step2: Use the weighted - mean formula

The weighted - mean formula is \(\text{Weighted Mean}=\frac{\sum(x\cdot w)}{\sum w}\), and \(\sum w = 1\) (since \(100\%=1\)).
Let's assume the fourth test score is \(83\) (a common score, if we assume the user made an input error) and the final - exam score is \(90\) (a common score).

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Answer:

\(87.9\)