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Sagot :
To determine whether Madison's claim that two data sets with the same median will have the same variability is true or false, we need to examine both the median and the variability (specifically, the standard deviation) of the given data sets.
Let's analyze the three data sets provided:
1. Data Set 1: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
2. Data Set 2: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
3. Data Set 3: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
### Step-by-Step Solution:
1. Finding the Median:
- The median is the middle value of a data set when it is ordered in ascending order. If the data set has an even number of values, the median is the average of the two middle numbers.
- All three data sets are identical, so we only need to calculate the median once.
[tex]\[ \text{Data Set}: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] \][/tex]
- Since the data set has 12 values (an even number), the median is the average of the 6th and 7th values (which are 5 and 6).
[tex]\[ \text{Median} = \frac{5 + 6}{2} = 5.5 \][/tex]
Therefore, the median for all three data sets is 5.5.
2. Calculating the Standard Deviation:
- The standard deviation measures the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range.
- Since the data sets are identical, we can calculate the standard deviation just once.
[tex]\[ \text{Data Set}: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] \][/tex]
Using the standard deviation formula for a population:
[tex]\[ \text{Standard Deviation} = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i - \mu)^2} \][/tex]
Where:
- [tex]\( N \)[/tex] is the number of values in the data set (12 in this case).
- [tex]\( x_i \)[/tex] represents each value in the data set.
- [tex]\( \mu \)[/tex] is the mean of the data set.
The calculated standard deviation for all three data sets is approximately:
[tex]\[ 3.452 \][/tex]
### Conclusion:
- The medians for all three data sets are 5.5.
- The standard deviations for all three data sets are approximately 3.452.
Since all three data sets have the same median and the same variability (standard deviation), Madison's claim is supported by these data sets. These data sets provide a good example to support her claim.
Let's analyze the three data sets provided:
1. Data Set 1: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
2. Data Set 2: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
3. Data Set 3: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
### Step-by-Step Solution:
1. Finding the Median:
- The median is the middle value of a data set when it is ordered in ascending order. If the data set has an even number of values, the median is the average of the two middle numbers.
- All three data sets are identical, so we only need to calculate the median once.
[tex]\[ \text{Data Set}: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] \][/tex]
- Since the data set has 12 values (an even number), the median is the average of the 6th and 7th values (which are 5 and 6).
[tex]\[ \text{Median} = \frac{5 + 6}{2} = 5.5 \][/tex]
Therefore, the median for all three data sets is 5.5.
2. Calculating the Standard Deviation:
- The standard deviation measures the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range.
- Since the data sets are identical, we can calculate the standard deviation just once.
[tex]\[ \text{Data Set}: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] \][/tex]
Using the standard deviation formula for a population:
[tex]\[ \text{Standard Deviation} = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i - \mu)^2} \][/tex]
Where:
- [tex]\( N \)[/tex] is the number of values in the data set (12 in this case).
- [tex]\( x_i \)[/tex] represents each value in the data set.
- [tex]\( \mu \)[/tex] is the mean of the data set.
The calculated standard deviation for all three data sets is approximately:
[tex]\[ 3.452 \][/tex]
### Conclusion:
- The medians for all three data sets are 5.5.
- The standard deviations for all three data sets are approximately 3.452.
Since all three data sets have the same median and the same variability (standard deviation), Madison's claim is supported by these data sets. These data sets provide a good example to support her claim.
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