Using the data above, kindly help verify Chebyshev’s theorem

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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 \( k = 3 \):**
   \[
   \begin{align*}
   63.6 - (3)(18.8) &= 63.6 - 56.4 = 7.2 \\
   63.6 + (3)(18.8) &= 63.6 + 56.4 = 120 \\
   \end{align*}
   \]
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 \( k = 3 \):** \[ \begin{align*} 63.6 - (3)(18.8) &= 63.6 - 56.4 = 7.2 \\ 63.6 + (3)(18.8) &= 63.6 + 56.4 = 120 \\ \end{align*} \]
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.
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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