# Given Information - The boxplot shows the average number of apple pies consumed in eating contests across a particular state. - There are two outliers: - A higher outlier (from a professional, heavyweight contest). - A lower outlier (from a children’s eating contest). # What Have to Find - The effect of **removing all outliers** on the **mean** and **median** of the data set. # Definition or Concept Used - **Mean**: The average of the data set, affected by extreme values (outliers). - **Median**: The middle value of an ordered data set, less affected by outliers. - **Outliers**: Extreme values that are much higher or lower than the rest of the data. # Step-by-Step Solution - Removing **outliers** will eliminate both the very high and very low values from the data set. - The **mean** is sensitive to outliers. The high outlier increases the mean, while the low outlier decreases it. Removing both will generally bring the mean closer to the central values of the data. - Since both a high and a low outlier are being removed, the mean will move toward the center, but with the removal of an extreme low value, the mean will **increase**. - The **median** is the middle value and is not affected by the removal of extreme values at the ends, as long as the number of data points is sufficient and the removed values are not the actual median. # Summary The median will remain unchanged and the mean will increase.

# Given Information - The boxplot displays the average number of apple pies consumed in eating contests across a particular state. - There are two outliers: - A higher outlier (from a professional, heavyweight contest). - A lower outlier (from a children’s eating contest). # What Have to Find - The effect of **removing all outliers** on the **mean** and **median** of the data set. # Definition or Concept Used - **Mean**: The average of the data set, influenced by extreme values (outliers). - **Median**: The middle value of an ordered data set, less influenced by outliers. - **Outliers**: Values that differ significantly from other observations. # Step-by-Step Solution - Removing **both outliers** will eliminate the extreme high and low values from the data set. - The **mean** is affected by outliers; removing the high outlier will lower the mean, while removing the low outlier will have less impact, generally resulting in an increase in the mean. - The **median**, being the central value, will not change unless the outliers are the only values in the dataset affecting the middle point. - In this case, since both high and low outliers are removed, the central tendency of the remaining values will stabilize the median. # Summary The median will remain unchanged and the mean will decrease.

# Given Information - The boxplot illustrates the average number of apple pies consumed in eating contests across a particular state. - There are two outliers: - A low outlier (around 10). - A high outlier (around 55-60). - Most of the data is centered between approximately 20 and 40. - The median is around the middle of the box (about 28-30). # What Have to Find - The effect of **removing both outliers** on the **mean** and **median** of the data set. # Definition or Concept Used - **Mean**: The average value, sensitive to extreme values (outliers). - **Median**: The middle value of a sorted data set, less affected by outliers. - **Outliers**: Values significantly higher or lower than the rest of the data. # Step-by-Step Solution - The **median** is based on the middle position of the data. Removing extreme values (one very low and one very high) does not significantly alter the median, as they are located on opposite ends. - Therefore, the median will remain unchanged. - The **mean**, on the other hand, is influenced by extreme values. The high outlier pulls the mean upwards, while the low outlier pulls it downwards. However, the high outlier's effect is stronger due to its greater distance from the center of the data. - By removing both outliers, we eliminate the strong upward pull from the high outlier, which will lead to a decrease in the mean. # Summary The median will remain unchanged and the mean will decrease.

# Given Information - Torque (\( T \)): \( 3 \times 10^6 \, \text{N m} \) - Mean diameter (\( D \)): \( 20 \, \text{mm} = 0.02 \, \text{m} \) - Semi cone angle (\( \alpha \)): \( 12^\circ \) - Coefficient of friction (\( \mu \)): \( 0.25 \) - Maximum normal pressure (\( P_n \)): \( 1.4 \times 10^6 \, \text{N/m}^2 \) # What Have to Find - Necessary width of the contact surface (\( b \)). - Axial force (\( F_a \)) needed to hold the clutch surface together. # Definition or Concept Used 1. **Torque Relation**: \[ T = \frac{P_n \cdot b \cdot r^2 \cdot \mu}{\tan(\alpha)} \] where \( r \) is the mean radius and \( b \) is the width of the contact surface. 2. **Mean Radius**: \[ r = \frac{D}{2} = \frac{0.02}{2} = 0.01 \, \text{m} \] 3. **Axial Force**: \[ F_a = P_n \cdot b \cdot 2 \cdot r \] # Step-by-Step Solution - Calculate the mean radius: \[ r = 0.01 \, \text{m} \] - Substitute \( r \) into the torque relation: \[ 3 \times 10^6 = \frac{1.4 \times 10^6 \cdot b \cdot (0.01)^2 \cdot 0.25}{\tan(12^\circ)} \] - Calculate \( \tan(12^\circ) \): \[ \tan(12^\circ) \approx 0.2126 \] - Rearranging the equation gives: \[ b = \frac{3 \times 10^6 \cdot \tan(12^\circ)}{1.4 \times 10^6 \cdot (0.01)^2 \cdot 0.25} \] - Substituting the values: \[ b = \frac{3 \times 10^6 \cdot 0.2126}{1.4 \times 10^6 \cdot 0.0001 \cdot 0.25} \] \[ b = \frac{637800}{35000} \approx 18.22 \, \text{m} \] - Next, calculate the axial force: \[ F_a = 1.4 \times 10^6 \cdot b \cdot 2 \cdot 0.01 \] \[ F_a = 1.4 \times 10^6 \cdot 18.22 \cdot 0.02 \] \[ F_a \approx 51000 \, \text{N} \] # Summary - Necessary width of the contact surface: \( 18.22 \, \text{m} \) - Axial force needed to hold the clutch surface together: \( 51000 \, \text{N} \)