MHB [ASK] Are the Frequency and the Score Switched?

  • Thread starter Thread starter Monoxdifly
  • Start date Start date
  • Tags Tags
    Frequency
Click For Summary
The discussion revolves around a table that presents frequency and score data, leading to confusion about whether these values are switched. The original calculations for the average score yielded results that did not match any provided answer options. Upon reassessing the data with the assumption that frequency and score might be reversed, a new mean score of approximately 7.17 was calculated. This new mean indicated that 20 students scored above average, yet the answer options still did not align with the findings. The thread highlights the challenge in interpreting the data correctly and the importance of verifying assumptions in statistical calculations.
Monoxdifly
MHB
Messages
288
Reaction score
0
Pay attention to the table below.
$$\begin{array}{|c|c|c|c|c|c|}\hline Frequency & 5 & 6 & 7 & 8 & 9 \\ \hline Score & 6 & 6 & 10 & 15 & 5\\ \hline \end{array}$$
The amount of students who get above average are...
A. 9 students
B. 17 students
C. 18 students
D. 26 students
I got the average as $$\frac{291}{35}$$, which is eight point something. So, I got 7 + 8 = 15 as the answer, but it was not in the option. I then assume that the frequency and score must be switched, but now I got $$\frac{291}{42}$$ which is six point something, so the answer should be 10 + 15 + 5 = 30 which was not in the options either. Where did I go wrong?
 
Mathematics news on Phys.org
Looking at the provided table, it appears the frequency and score are switched. Let's assume they are...

Using a weighted average, I get a mean score of:

$$\overline{x}=\frac{6\cdot5+6\cdot6+10\cdot7+15\cdot8+5\cdot9}{6+6+10+15+5}=\frac{301}{42}=\frac{43}{6}=7.1\overline{6}$$

It would appear 20 students scored higher than the mean. :D
 
That means, still no answer?
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
14
Views
2K
  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K