An article in Knee Surgery, Sports Traumatology, Arthroscopy, "Arthroscopic meniscal repair with an absorbable screw: results and surgical technique," (2005, Vol. 13, pp. 273-279) cites a success rate more than 90% for meniscal tears with a rim width of less than 3 mm, but only a 67% success rate for tears of 3-6 mm. If you are unlucky enough to suffer a meniscal tear of less than 3 mm on your left knee, and one of width 3-6 mm on your right knee, what are the mean and variance of the number of successful surgeries? Assume the surgeries are independent. Round your answers to three decimal places (e.g. 98.765). Mean = i Variance = i

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### Example Problem: Calculating Mean and Variance of Successful Surgeries

An article in **Knee Surgery, Sports Traumatology, Arthroscopy** titled "Arthroscopic meniscal repair with an absorbable screw: results and surgical technique" (2005, Vol. 13, pp. 273-279) cites a success rate of more than 90% for meniscal tears with a rim width of less than 3 mm, but only a 67% success rate for tears of 3-6 mm. 

If you are unlucky enough to suffer a meniscal tear of less than 3 mm on your left knee, and one of width 3-6 mm on your right knee, what are the mean and variance of the number of successful surgeries? Assume the surgeries are independent.

**Round your answers to three decimal places (e.g., 98.765).**

#### Inputs:
- **Mean =** [Input Box]
- **Variance =** [Input Box]

#### Explanation:
To solve this problem, the following steps should be undertaken:

1. **Identify Success Rates:**
   - The success rate for a meniscal tear of less than 3 mm is **more than 90%**. For the purpose of this example, assume it to be **0.90**.
   - The success rate for a meniscal tear of 3-6 mm is **0.67**.

2. **Calculate Mean and Variance for Independent Events:**
   - For two surgeries, we calculate both mean and variance of the number of successful surgeries. 
   - Let \( X_1 \) be the success indicator for the left knee surgery and \( X_2 \) for the right knee surgery.
   - \( X_1 \sim Bernoulli(p_1) \), where \( p_1 = 0.90 \)
   - \( X_2 \sim Bernoulli(p_2) \), where \( p_2 = 0.67 \)

3. **Mean Calculation:**
   - \( E(X_1) = p_1 = 0.90 \)
   - \( E(X_2) = p_2 = 0.67 \)
   - The total mean, \( E(X_1 + X_2) = E(X_1) + E(X_2) \)
   - Thus, \( \text{Mean} =
Transcribed Image Text:### Example Problem: Calculating Mean and Variance of Successful Surgeries An article in **Knee Surgery, Sports Traumatology, Arthroscopy** titled "Arthroscopic meniscal repair with an absorbable screw: results and surgical technique" (2005, Vol. 13, pp. 273-279) cites a success rate of more than 90% for meniscal tears with a rim width of less than 3 mm, but only a 67% success rate for tears of 3-6 mm. If you are unlucky enough to suffer a meniscal tear of less than 3 mm on your left knee, and one of width 3-6 mm on your right knee, what are the mean and variance of the number of successful surgeries? Assume the surgeries are independent. **Round your answers to three decimal places (e.g., 98.765).** #### Inputs: - **Mean =** [Input Box] - **Variance =** [Input Box] #### Explanation: To solve this problem, the following steps should be undertaken: 1. **Identify Success Rates:** - The success rate for a meniscal tear of less than 3 mm is **more than 90%**. For the purpose of this example, assume it to be **0.90**. - The success rate for a meniscal tear of 3-6 mm is **0.67**. 2. **Calculate Mean and Variance for Independent Events:** - For two surgeries, we calculate both mean and variance of the number of successful surgeries. - Let \( X_1 \) be the success indicator for the left knee surgery and \( X_2 \) for the right knee surgery. - \( X_1 \sim Bernoulli(p_1) \), where \( p_1 = 0.90 \) - \( X_2 \sim Bernoulli(p_2) \), where \( p_2 = 0.67 \) 3. **Mean Calculation:** - \( E(X_1) = p_1 = 0.90 \) - \( E(X_2) = p_2 = 0.67 \) - The total mean, \( E(X_1 + X_2) = E(X_1) + E(X_2) \) - Thus, \( \text{Mean} =
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