Using the data above, kindly help verify Chebyshev’s theorem
MATLAB: An Introduction with Applications
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Author:Amos Gilat
Publisher:Amos Gilat
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Hi, I’m confused on testing/verifying Chebyshev’s theorem, thought I did it in first pic but comment from my professor stated otherwise. Using the data above, kindly help verify Chebyshev’s theorem
![### Sample Data Analysis
#### Graph Description
The graph titled "Sample Data" is a vertical bar chart displaying 30 data points. The x-axis represents samples numbered from 1 to 30, while the y-axis represents their corresponding values, ranging from 0 to 120. The bars vary in height, showing the distribution of data values across these samples.
#### Statistical Analysis
From the graph given above, the mean is calculated as follows:
\[ \text{Mean} = \frac{1908.42}{30} = 63.6 \]
The standard deviation (Sd) is:
\[ \text{Sd} = 18.8 \]
#### Probability Calculations
Using Chebyshev's inequality, the probability \( P(|x - \mu| \geq k\sigma) \) satisfies:
\[ P(|x - \mu| \geq k\sigma) \leq \frac{1}{k^2} \]
Let’s assume \( k \) to be 1.5, 2, and 3:
1. **For \( k = 1.5 \):**
\[
\begin{align*}
63.6 - (1.5)(18.8) &= 63.6 - 28.2 = 35.4 \\
63.6 + (1.5)(18.8) &= 63.6 + 28.2 = 91.8 \\
\end{align*}
\]
2. **For \( k = 2 \):**
\[
\begin{align*}
63.6 - (2)(18.8) &= 63.6 - 37.6 = 26 \\
63.6 + (2)(18.8) &= 63.6 + 37.6 = 101.2 \\
\end{align*}
\]
3. **For \( k = 3 \):**
\[
\begin{align*}
63.6 - (3)(18.8) &= 63.6 - 56.4 = 7.2 \\
63.6 + (3)(18.8) &= 63.6 + 56.4 = 120 \\
\end{align*}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F79599c56-a340-49a0-b0ff-829b3947a798%2F0563503f-e93b-4445-ac53-7359981c15dd%2Fvhwc1ml_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Sample Data Analysis
#### Graph Description
The graph titled "Sample Data" is a vertical bar chart displaying 30 data points. The x-axis represents samples numbered from 1 to 30, while the y-axis represents their corresponding values, ranging from 0 to 120. The bars vary in height, showing the distribution of data values across these samples.
#### Statistical Analysis
From the graph given above, the mean is calculated as follows:
\[ \text{Mean} = \frac{1908.42}{30} = 63.6 \]
The standard deviation (Sd) is:
\[ \text{Sd} = 18.8 \]
#### Probability Calculations
Using Chebyshev's inequality, the probability \( P(|x - \mu| \geq k\sigma) \) satisfies:
\[ P(|x - \mu| \geq k\sigma) \leq \frac{1}{k^2} \]
Let’s assume \( k \) to be 1.5, 2, and 3:
1. **For \( k = 1.5 \):**
\[
\begin{align*}
63.6 - (1.5)(18.8) &= 63.6 - 28.2 = 35.4 \\
63.6 + (1.5)(18.8) &= 63.6 + 28.2 = 91.8 \\
\end{align*}
\]
2. **For \( k = 2 \):**
\[
\begin{align*}
63.6 - (2)(18.8) &= 63.6 - 37.6 = 26 \\
63.6 + (2)(18.8) &= 63.6 + 37.6 = 101.2 \\
\end{align*}
\]
3. **For \( k = 3 \):**
\[
\begin{align*}
63.6 - (3)(18.8) &= 63.6 - 56.4 = 7.2 \\
63.6 + (3)(18.8) &= 63.6 + 56.4 = 120 \\
\end{align*}
\]
Transcribed Image Text:You have not done the two examples I asked you to do, (k=2 and 3), to verify Chebyshev's theorem.
The example you did was a process to find the value of k when the probability of the inequality is 100% which is a good practice as long as you state the purpose of the example clearly.
Once we overcome the mathematical hurdle, we all move to data analysis.
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