Question

information: weighted averages examine the table of student test scores for five tests they have taken.
test student a student b
1 95 76
2 74 88
3 82 90
4 92 81
5 81 72
ave.
grade

critical thinking questions

1. calculate the average grade for students a and b and enter the average in the table above.
2. if you know a students average grade can you tell what the students individual test scores were? explain.
3. suppose student c had an average of 83%. on each of his five tests he scored either 65% or 95%. which score occurred more often? explain.
4. what if the teacher decided that test five would count for 40% of the final grade and test four would count for 30% of the final grade and each of the other tests would count for 10%. calculate the new average for each student. note: this is called the weighted average ☺

student a new average student b new average

information: average atomic mass
on the periodic table you can find the average atomic mass for an element. this average is a weighted average and it tells you the average mass of all the isotopes of an element. the periodic table does not contain mass numbers for individual atoms, instead you can find the average mass of atoms. the average atomic mass is calculated just how you calculated the weighted average in question 4 above.

Explanation:

Step1: Calculate average for Student A (un - weighted)

Sum of scores: $95 + 74+82 + 92+81=424$. Average = $\frac{424}{5}=84.8$.

Step2: Calculate average for Student B (un - weighted)

Sum of scores: $76 + 88+90 + 81+72=407$. Average = $\frac{407}{5}=81.4$.

Step3: Answer to question 2

No. The average is a single - value summary of the data. Multiple combinations of individual scores can result in the same average. For example, $(80,80)$ and $(70,90)$ have the same average of 80.

Step4: Solve for Student C's scores

Let the number of 65% scores be $x$ and the number of 95% scores be $y$. We know $x + y=5$ and $65x + 95y=83\times5 = 415$. Substitute $x = 5 - y$ into $65x+95y = 415$: $65(5 - y)+95y=415$. Expand: $325-65y + 95y=415$. Combine like terms: $30y=90$, so $y = 3$ and $x = 2$. The 95% score occurred more often.

Step5: Calculate weighted average for Student A

Test 4 score: 92, Test 5 score: 81. Other tests: 95, 74, 82.
Weighted average=$0.1\times95+0.1\times74 + 0.1\times82+0.3\times92+0.4\times81$
$=9.5 + 7.4+8.2 + 27.6+32.4$
$=85.1$.

Step6: Calculate weighted average for Student B

Test 4 score: 81, Test 5 score: 72. Other tests: 76, 88, 90.
Weighted average=$0.1\times76+0.1\times88 + 0.1\times90+0.3\times81+0.4\times72$
$=7.6+8.8 + 9+24.3+28.8$
$=78.5$.

Answer:

1. Student A average: 84.8, Student B average: 81.4
2. No. Multiple score - combinations can give the same average.
3. 95% score occurred more often.
4. Student A New Average: 85.1, Student B New Average: 78.5