Question

Computing Grades Going into the final exam, which will count as two-thirds of the final grade, Mike has test scores of \(86,80,84,\) and \(90 .\) What minimum score does Mike need on the final in order to earn a B, which requires an average score of \(80 ?\) What does he need to earn an A, which requires an average of \(90 ?\)

Short answer

Mike needs a minimum score of 77.5 for a B and 92.5 for an A.

Step-by-step solution

- Understand the weight of the grades

The final exam counts as two-thirds of the final grade. This means that the remaining one-third of the grade comes from the four test scores.

- Compute the average of the current test scores

First, calculate the average of Mike's test scores: \[ \text{Average}_{\text{tests}} = \frac{86 + 80 + 84 + 90}{4} = \frac{340}{4} = 85 \]

- Set up the equation for the overall average

Let \( x \) be the score Mike needs on the final exam. The overall average is given by:\[ \text{Final Grade} = \left(\frac{1}{3} \times \text{Average}_{\text{tests}} \right) + \left(\frac{2}{3} \times x \right) \]

- Solve for the minimum score needed for a B

A B requires an average score of 80. Set up the equation:\[ 80 = \left(\frac{1}{3} \times 85 \right) + \left(\frac{2}{3} \times x \right) \]Solve for \( x \):\[ 80 = 28.33 + \left(\frac{2}{3} \times x \right) \]\[ 80 - 28.33 = \left(\frac{2}{3} \times x \right) \]\[ 51.67 = \left(\frac{2}{3} \times x \right) \]\[ x = \frac{51.67 \times 3}{2} \approx 77.5 \]

- Solve for the minimum score needed for an A

An A requires an average score of 90. Set up the equation:\[ 90 = \left(\frac{1}{3} \times 85 \right) + \left(\frac{2}{3} \times x \right) \]Solve for \( x \):\[ 90 = 28.33 + \left(\frac{2}{3} \times x \right) \]\[ 90 - 28.33 = \left(\frac{2}{3} \times x \right) \]\[ 61.67 = \left(\frac{2}{3} \times x \right) \]\[ x = \frac{61.67 \times 3}{2} \approx 92.5 \]

Key concepts

weighted average

Understanding a weighted average is crucial for calculating grades when different components contribute differently to the final score.
In Mike's case, his final exam counts for two-thirds of his overall grade, and his previous test scores count for one-third.
This means the final exam has more impact on his final grade than his previous exams do.

To compute a weighted average, you multiply each component by its weight and then sum them up.

• Mike's test scores contribute one-third of the final grade, so we calculate their average: \text{Average}_{\text{tests}} = \frac{86 + 80 + 84 + 90}{4} = 85 \
• The final exam, denoted as \(x\), contributes two-thirds: \ \text{Final Grade} = \frac{1}{3}\times 85 + \frac{2}{3}\times x \
  The weights (one-third for tests and two-thirds for the final exam) are necessary to determine the overall grade.

solving equations

Equations help us find unknown values given certain conditions.
In our problem, we need to find the minimum score Mike needs on his final exam to achieve specific grades.
Let's look into the equations for a B and an A.

To earn a B, Mike's average grade must be 80.
This means we set our equation to 80 and solve for \(x\) (final exam score): \ 80 = \frac{1}{3}\times 85 + \frac{2}{3}\times x \ When you solve this: \ 80 - 28.33 = \frac{2}{3}\times x \ 51.67 = \frac{2}{3}\times x \ x = \frac{51.67\times 3}{2} = 77.5 \ Thus, the minimum score Mike needs on his final to get a B is approximately 77.5.
For an A, the average grade required is 90. We follow a similar process:
\ 90 = \frac{1}{3}\times 85 + \frac{2}{3}\times x \ Solve for \(x\): \ 90 - 28.33 = \frac{2}{3}\times x \ 61.67 = \frac{2}{3}\times x \ x = \frac{61.67\times 3}{2} = 92.5 \ So, to get an A, Mike needs to score roughly 92.5.

test scores analysis

Analyzing test scores involves calculating averages and understanding grade requirements.
Mike's average on his existing test scores is 85, which gives us a solid starting point.
Knowing that the final exam carries more weight (two-thirds), it significantly impacts Mike's overall grade.

Let's break down what we analyzed:

• For a B: With an average of 80 needed and tests averaging 85, the analysis showed that Mike's final exam score needs to be around 77.5.
• For an A: With a higher target (90 average required), Mike must score about 92.5 on his final exam.

This sort of analysis helps students understand how each score impacts their grade and emphasizes the importance of weighted averages in academic performance.
For students like Mike, focusing on the final exam becomes crucial because of its substantial weight in determining the final grade.