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Use the display of data to find the standard deviation. A bar graph titled Score with a horizontal axis labeled from 5 to 11 in...Asked by VD
Use the display of data to find the standard deviation.
A bar graph titled Score with a horizontal axis labeled from 5 to 11 in increments of 1 and a vertical axis labeled Frequency from 0 to 12 in increments of 2 contains three vertical bars with labels and heights as follows, where the horizontal axis label is listed first and the height is listed second: 7, 7; 8, 5; 9, 7.
Score
5
6
7
8
9
10
11
0
2
4
6
8
10
12
Frequency
Question content area bottom
Part 1
The standard deviation is approximately
enter your response here.
(Do not round until the final answer. Then round to the nearest hundredth as needed.)
A bar graph titled Score with a horizontal axis labeled from 5 to 11 in increments of 1 and a vertical axis labeled Frequency from 0 to 12 in increments of 2 contains three vertical bars with labels and heights as follows, where the horizontal axis label is listed first and the height is listed second: 7, 7; 8, 5; 9, 7.
Score
5
6
7
8
9
10
11
0
2
4
6
8
10
12
Frequency
Question content area bottom
Part 1
The standard deviation is approximately
enter your response here.
(Do not round until the final answer. Then round to the nearest hundredth as needed.)
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To find the standard deviation based on the bar graph data provided, we first need to construct a frequency distribution and then calculate the standard deviation using the following steps:
1. **Data Collection:**
The data collected from the bar graph is:
- Score 7: Frequency 7
- Score 8: Frequency 5
- Score 9: Frequency 7
We also note that the scores 5, 6, 10, and 11 have a frequency of 0.
2. **Construct a Frequency Table:**
\[
\begin{array}{|c|c|}
\hline
\text{Score (x)} & \text{Frequency (f)} \\
\hline
5 & 0 \\
6 & 0 \\
7 & 7 \\
8 & 5 \\
9 & 7 \\
10 & 0 \\
11 & 0 \\
\hline
\end{array}
\]
3. **Calculate Mean (μ):**
The mean \( \mu \) is calculated as follows:
\[
\mu = \frac{\sum (x \cdot f)}{\sum f}
\]
where \( x \) is the score and \( f \) is the frequency.
- \(\sum (x \cdot f) = (5 \cdot 0) + (6 \cdot 0) + (7 \cdot 7) + (8 \cdot 5) + (9 \cdot 7) + (10 \cdot 0) + (11 \cdot 0)\)
- \(= 0 + 0 + 49 + 40 + 63 + 0 + 0 = 152\)
- \(\sum f = 0 + 0 + 7 + 5 + 7 + 0 + 0 = 19\)
So, the mean \( \mu \) is:
\[
\mu = \frac{152}{19} \approx 8
\]
4. **Calculate Variance (σ²):**
The variance is calculated as:
\[
\sigma^2 = \frac{\sum f(x - \mu)^2}{\sum f}
\]
We compute \( (x - \mu)^2 \) for each score:
- For \( x = 7 \):
\[
(7 - 8)^2 = (-1)^2 = 1
\]
- For \( x = 8 \):
\[
(8 - 8)^2 = (0)^2 = 0
\]
- For \( x = 9 \):
\[
(9 - 8)^2 = (1)^2 = 1
\]
Now we compute \( f(x - \mu)^2 \):
\[
\begin{align*}
f(x - \mu)^2 &= 0 \cdot 0 + 0 \cdot 0 + 7 \cdot 1 + 5 \cdot 0 + 7 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 \\
&= 0 + 0 + 7 + 0 + 7 + 0 + 0 = 14
\end{align*}
\]
Plugging it into the variance formula:
\[
\sigma^2 = \frac{14}{19} \approx 0.736842
\]
5. **Calculate Standard Deviation (σ):**
\[
\sigma = \sqrt{\sigma^2} = \sqrt{0.736842} \approx 0.859
\]
Finally, rounding to two decimal places, the standard deviation is approximately:
\[
\boxed{0.86}
\]
1. **Data Collection:**
The data collected from the bar graph is:
- Score 7: Frequency 7
- Score 8: Frequency 5
- Score 9: Frequency 7
We also note that the scores 5, 6, 10, and 11 have a frequency of 0.
2. **Construct a Frequency Table:**
\[
\begin{array}{|c|c|}
\hline
\text{Score (x)} & \text{Frequency (f)} \\
\hline
5 & 0 \\
6 & 0 \\
7 & 7 \\
8 & 5 \\
9 & 7 \\
10 & 0 \\
11 & 0 \\
\hline
\end{array}
\]
3. **Calculate Mean (μ):**
The mean \( \mu \) is calculated as follows:
\[
\mu = \frac{\sum (x \cdot f)}{\sum f}
\]
where \( x \) is the score and \( f \) is the frequency.
- \(\sum (x \cdot f) = (5 \cdot 0) + (6 \cdot 0) + (7 \cdot 7) + (8 \cdot 5) + (9 \cdot 7) + (10 \cdot 0) + (11 \cdot 0)\)
- \(= 0 + 0 + 49 + 40 + 63 + 0 + 0 = 152\)
- \(\sum f = 0 + 0 + 7 + 5 + 7 + 0 + 0 = 19\)
So, the mean \( \mu \) is:
\[
\mu = \frac{152}{19} \approx 8
\]
4. **Calculate Variance (σ²):**
The variance is calculated as:
\[
\sigma^2 = \frac{\sum f(x - \mu)^2}{\sum f}
\]
We compute \( (x - \mu)^2 \) for each score:
- For \( x = 7 \):
\[
(7 - 8)^2 = (-1)^2 = 1
\]
- For \( x = 8 \):
\[
(8 - 8)^2 = (0)^2 = 0
\]
- For \( x = 9 \):
\[
(9 - 8)^2 = (1)^2 = 1
\]
Now we compute \( f(x - \mu)^2 \):
\[
\begin{align*}
f(x - \mu)^2 &= 0 \cdot 0 + 0 \cdot 0 + 7 \cdot 1 + 5 \cdot 0 + 7 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 \\
&= 0 + 0 + 7 + 0 + 7 + 0 + 0 = 14
\end{align*}
\]
Plugging it into the variance formula:
\[
\sigma^2 = \frac{14}{19} \approx 0.736842
\]
5. **Calculate Standard Deviation (σ):**
\[
\sigma = \sqrt{\sigma^2} = \sqrt{0.736842} \approx 0.859
\]
Finally, rounding to two decimal places, the standard deviation is approximately:
\[
\boxed{0.86}
\]
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