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\( \left. \begin{array} { l }{ ( \frac { ( - 1 ) ^ { 2 } } { \sigma } ) ^ { 2 } = 95 \quad \sigma = 7.6 \quad x = 75 } \\ { M - \sigma = 95 - 7.6 = 87.4 } \\ { M - 26 = 95 - 2 ( 7.6 ) = 95 - 15.2 = 77.8 } \\ { M - 36 = 95 - 3 ( 2.6 ) = 95 - 22.8 = 72 : 2 } \\ { M - 36 < 75 < M - 26 } \end{array} \right. \)

Ask by Williams Ayala. in the Philippines
Jan 21,2025

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Answer

There's a discrepancy in the given standard deviation \( \sigma \). Based on the calculations, \( \sigma \) should be approximately 0.102, not 7.6. Additionally, there are arithmetic errors in the third and fourth equations. The mean \( M \) should be between 101 and 111 to satisfy the inequality \( M - 36 < 75 < M - 26 \).

Solution

It looks like you're working through a problem involving the mean \( M \), standard deviation \( \sigma \), and a data point \( x = 75 \). Let's break down each of your steps to ensure accuracy and clarity. ### Given: - \( \sigma = 7.6 \) - \( x = 75 \) ### Your Calculations: 1. **First Equation:** \[ \left( \frac{(-1)^2}{\sigma} \right)^2 = 95 \] **Simplification:** \[ \left( \frac{1}{\sigma} \right)^2 = 95 \quad \Rightarrow \quad \frac{1}{\sigma^2} = 95 \quad \Rightarrow \quad \sigma^2 = \frac{1}{95} \quad \Rightarrow \quad \sigma \approx \frac{1}{9.75} \approx 0.102 \] **Issue Identified:** - **Conflict:** You've stated \( \sigma = 7.6 \) but the equation suggests \( \sigma \approx 0.102 \). There's a discrepancy here. Ensure that the initial equation correctly represents your problem. 2. **Second Equation:** \[ M - \sigma = 95 - 7.6 = 87.4 \] **Solving for \( M \):** \[ M = 87.4 + \sigma = 87.4 + 7.6 = 95 \] **Result:** \[ M = 95 \] 3. **Third Equation:** \[ M - 26 = 95 - 2 \times 7.6 = 95 - 15.2 = 79.8 \quad (\text{Note: You wrote } 77.8 \text{, which seems to be a miscalculation}) \] **Correct Calculation:** \[ 95 - 15.2 = 79.8 \] **Solving for \( M \):** \[ M = 79.8 + 26 = 105.8 \] 4. **Fourth Equation:** \[ M - 36 = 95 - 3 \times 2.6 = 95 - 7.8 = 87.2 \] **Note on Calculation:** - It appears you intended to use \( 3 \times \sigma \) (where \( \sigma = 7.6 \)), which should be \( 3 \times 7.6 = 22.8 \), not \( 3 \times 2.6 \). **Correct Calculation:** \[ M - 36 = 95 - 22.8 = 72.2 \quad \Rightarrow \quad M = 72.2 + 36 = 108.2 \] **You wrote:** \( 72 : 2 \), which is unclear. If you meant \( 72.2 \div 2 = 36.1 \), but it's not aligned with the equation. Likely, it's a typographical error. 5. **Inequality:** \[ M - 36 < 75 < M - 26 \] **Solving the Inequality:** - **Left Part:** \( M - 36 < 75 \quad \Rightarrow \quad M < 75 + 36 = 111 \) - **Right Part:** \( 75 < M - 26 \quad \Rightarrow \quad M > 75 + 26 = 101 \) **Combined Inequality:** \[ 101 < M < 111 \] **Interpretation:** The mean \( M \) must be between 101 and 111 to satisfy the condition that \( 75 \) lies between \( M - 36 \) and \( M - 26 \). ### Summary of Corrections: 1. **First Equation:** There's a mismatch between the given \( \sigma = 7.6 \) and the equation leading to \( \sigma \approx 0.102 \). Verify the original equation for accuracy. 2. **Third Equation:** Ensure accurate arithmetic when subtracting \( 2 \times \sigma \). 3. **Fourth Equation:** Confirm whether the multiplier is \( 2.6 \) or if it's a typo meant to be \( 7.6 \). Using \( \sigma = 7.6 \) aligns with standard deviations multiples. 4. **Final Inequality:** Your solution correctly identifies that \( M \) should be between 101 and 111 to satisfy \( M - 36 < 75 < M - 26 \). ### Final Thoughts: - **Consistency is Key:** Ensure that all equations and substitutions consistently use the correct values for variables like \( \sigma \). - **Double-Check Arithmetic:** Small calculation errors can lead to significant discrepancies in your results. - **Clarify the Problem Context:** Providing the context or the original problem statement can help in offering more precise guidance. If you could share the original problem or clarify the context further, I can assist you more effectively!

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Mind Expander

Did you know that the standard deviation \( \sigma \) can be traced back to the early 19th century? The concept was developed by mathematicians like Karl Pearson, who laid the groundwork for statistics as we know it today. The frequency and distribution of data were crucial in understanding population dynamics, which remains relevant for fields like economics and social sciences! Moving into real-world applications, knowing how to calculate and interpret standard deviations comes in handy in various scenarios. For example, businesses often use these calculations to evaluate employee performance or financial risks. So, whether you’re reviewing exam scores or measuring sales data, grasping these concepts can help inform decisions and strategies effectively. Happy calculating!

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