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Sagot :
### Question 6
Let's determine which team has a higher Interquartile Range (IQR) and by how many home runs that team is higher.
#### Calculating IQR for Team 1:
1. Team 1 data: 18, 22, 21, 28, 30, 29, 32, 40, 33, 34, 28, 29, 22, 20
2. Q1 (First quartile): The value which separates the lowest 25% of the data. For Team 1, this is 22.
3. Q3 (Third quartile): The value which separates the highest 25% of the data. For Team 1, this is 32.
4. IQR (Interquartile Range): [tex]\( Q3 - Q1 \)[/tex]. For Team 1, this is [tex]\( 32 - 22 = 10 \)[/tex].
#### Calculating IQR for Team 2:
1. Team 2 data: 17, 24, 18, 35, 18, 25, 33, 38, 39, 25, 32, 30, 26, 25
2. Q1 (First quartile): For Team 2, this is 24.
3. Q3 (Third quartile): For Team 2, this is 33.
4. IQR (Interquartile Range): [tex]\( Q3 - Q1 \)[/tex]. For Team 2, this is [tex]\( 33 - 24 = 9 \)[/tex].
#### Higher IQR and Difference:
- Team 1 has an IQR of 10, and Team 2 has an IQR of 9.
- Therefore, Team 1 has a higher IQR by [tex]\( 10 - 9 = 1 \)[/tex] home run.
### Question 7
Now, let's find the mean, median, IQR, and standard deviation of the following dataset:
[tex]\[ \{169, 175, 170, 190, 175, 160, 165, 165, 155, 165, 185, 185\} \][/tex]
#### Mean:
The mean is calculated as the sum of all values divided by the number of values.
[tex]\[ \text{Mean} = \frac{169 + 175 + 170 + 190 + 175 + 160 + 165 + 165 + 155 + 165 + 185 + 185}{12} = 171.58 \][/tex]
#### Median:
The median is the middle value when the data set is ordered. For this dataset, the ordered values are:
[tex]\[ \{155, 160, 165, 165, 165, 169, 170, 175, 175, 185, 185, 190\} \][/tex]
With 12 values, the median will be the average of the 6th and 7th values:
[tex]\[ \text{Median} = \frac{169 + 170}{2} = 169.5 \][/tex]
#### IQR (Interquartile Range):
1. First Quartile (Q1): 25th percentile. For this data, [tex]\( Q1 = 163.75 \)[/tex].
2. Third Quartile (Q3): 75th percentile. For this data, [tex]\( Q3 = 176.25 \)[/tex].
3. IQR: [tex]\( Q3 - Q1 = 176.25 - 163.75 = 12.5 \)[/tex]
#### Standard Deviation:
The standard deviation is a measure of the amount of variation or dispersion in a set of values. For this dataset, the standard deviation is [tex]\( 10.29 \)[/tex].
### Summary of Question 7:
- Mean: 171.58
- Median: 169.5
- IQR: 12.5
- Standard Deviation: 10.29
Let's determine which team has a higher Interquartile Range (IQR) and by how many home runs that team is higher.
#### Calculating IQR for Team 1:
1. Team 1 data: 18, 22, 21, 28, 30, 29, 32, 40, 33, 34, 28, 29, 22, 20
2. Q1 (First quartile): The value which separates the lowest 25% of the data. For Team 1, this is 22.
3. Q3 (Third quartile): The value which separates the highest 25% of the data. For Team 1, this is 32.
4. IQR (Interquartile Range): [tex]\( Q3 - Q1 \)[/tex]. For Team 1, this is [tex]\( 32 - 22 = 10 \)[/tex].
#### Calculating IQR for Team 2:
1. Team 2 data: 17, 24, 18, 35, 18, 25, 33, 38, 39, 25, 32, 30, 26, 25
2. Q1 (First quartile): For Team 2, this is 24.
3. Q3 (Third quartile): For Team 2, this is 33.
4. IQR (Interquartile Range): [tex]\( Q3 - Q1 \)[/tex]. For Team 2, this is [tex]\( 33 - 24 = 9 \)[/tex].
#### Higher IQR and Difference:
- Team 1 has an IQR of 10, and Team 2 has an IQR of 9.
- Therefore, Team 1 has a higher IQR by [tex]\( 10 - 9 = 1 \)[/tex] home run.
### Question 7
Now, let's find the mean, median, IQR, and standard deviation of the following dataset:
[tex]\[ \{169, 175, 170, 190, 175, 160, 165, 165, 155, 165, 185, 185\} \][/tex]
#### Mean:
The mean is calculated as the sum of all values divided by the number of values.
[tex]\[ \text{Mean} = \frac{169 + 175 + 170 + 190 + 175 + 160 + 165 + 165 + 155 + 165 + 185 + 185}{12} = 171.58 \][/tex]
#### Median:
The median is the middle value when the data set is ordered. For this dataset, the ordered values are:
[tex]\[ \{155, 160, 165, 165, 165, 169, 170, 175, 175, 185, 185, 190\} \][/tex]
With 12 values, the median will be the average of the 6th and 7th values:
[tex]\[ \text{Median} = \frac{169 + 170}{2} = 169.5 \][/tex]
#### IQR (Interquartile Range):
1. First Quartile (Q1): 25th percentile. For this data, [tex]\( Q1 = 163.75 \)[/tex].
2. Third Quartile (Q3): 75th percentile. For this data, [tex]\( Q3 = 176.25 \)[/tex].
3. IQR: [tex]\( Q3 - Q1 = 176.25 - 163.75 = 12.5 \)[/tex]
#### Standard Deviation:
The standard deviation is a measure of the amount of variation or dispersion in a set of values. For this dataset, the standard deviation is [tex]\( 10.29 \)[/tex].
### Summary of Question 7:
- Mean: 171.58
- Median: 169.5
- IQR: 12.5
- Standard Deviation: 10.29
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