Math  /  Data & Statistics

QuestionThis question: 1 point(s) possible Subm
Find the mean for the data items in the given frequency distribution. \begin{tabular}{|l|l|l|l|l|l|l|l|l|} \hline Score, x\mathbf{x} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline Frequency, f\mathbf{f} & 3 & 6 & 2 & 5 & 3 & 1 & 3 & 2 \\ \hline \end{tabular}
The mean is \square (Round to 3 decimal places as needed.)

Studdy Solution

STEP 1

What is this asking? We need to find the average score, given how many times each score appears. Watch out! Don't just add the scores and divide by the number of scores; we need to account for how often each score occurs!

STEP 2

1. Calculate the total occurrences of all scores.
2. Calculate the weighted sum of the scores.
3. Calculate the mean.

STEP 3

Let's **add up** all the frequencies to find the **total number** of scores.
This tells us how many scores we have in total!
We're doing this because the mean is calculated by dividing the sum of all scores by the total number of scores.

STEP 4

3+6+2+5+3+1+3+2=253 + 6 + 2 + 5 + 3 + 1 + 3 + 2 = \mathbf{25} So, we have a total of **25** scores.
Woohoo!

STEP 5

Now, we need to find the sum of all the scores.
Since some scores occur multiple times, we **multiply** each score by its frequency.
This is called a *weighted sum* because each score is weighted by how many times it appears.
It's like giving each score the importance it deserves!

STEP 6

(13)+(26)+(32)+(45)+(53)+(61)+(73)+(82)(1 \cdot 3) + (2 \cdot 6) + (3 \cdot 2) + (4 \cdot 5) + (5 \cdot 3) + (6 \cdot 1) + (7 \cdot 3) + (8 \cdot 2) =3+12+6+20+15+6+21+16= 3 + 12 + 6 + 20 + 15 + 6 + 21 + 16 =99= \mathbf{99}The **weighted sum** of the scores is **99**.
Awesome!

STEP 7

Finally, to find the mean, we **divide** the weighted sum of the scores by the total number of scores.
This gives us the average score, taking into account how often each score appears.

STEP 8

9925=3.96\frac{\mathbf{99}}{\mathbf{25}} = 3.96 Rounding to three decimal places, we get 3.960\mathbf{3.960}.

STEP 9

The mean of the data is 3.960\mathbf{3.960}.

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