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Sagot :
Alright, let's systematically determine if the given data set contains any outliers. Here's a step-by-step explanation:
1. List the Available Data:
The table shows the lengths of fish in a pond:
[tex]\[ 15, 18, 8, 22, 7, 15, 10, 18 \][/tex]
2. Sort the Data:
First, we sort the data in ascending order:
[tex]\[ 7, 8, 10, 15, 15, 18, 18, 22 \][/tex]
3. Determine the Quartiles:
- First Quartile (Q1): This is the median of the first half of the data (not including the median if the number of data points is odd).
[tex]\[ \text{First half of data: } 7, 8, 10, 15 \][/tex]
The median or Q1 of this subset is halfway between 8 and 10:
[tex]\[ Q1 = 9.5 \][/tex]
- Third Quartile (Q3): This is the median of the second half of the data (not including the median if the number of data points is odd).
[tex]\[ \text{Second half of data: } 15, 18, 18, 22 \][/tex]
The median or Q3 of this subset is halfway between 18 and 18:
[tex]\[ Q3 = 18.0 \][/tex]
4. Calculate the Interquartile Range (IQR):
[tex]\[ IQR = Q3 - Q1 = 18.0 - 9.5 = 8.5 \][/tex]
5. Determine the Bounds for Outliers:
- Lower Bound: Any data point below the lower bound is considered an outlier.
[tex]\[ \text{Lower Bound} = Q1 - 1.5 \times IQR = 9.5 - 1.5 \times 8.5 = -3.25 \][/tex]
Since the length of fish cannot be negative, any reasonable positive fish length will not be considered a lower outlier in this context.
- Upper Bound: Any data point above the upper bound is considered an outlier.
[tex]\[ \text{Upper Bound} = Q3 + 1.5 \times IQR = 18.0 + 1.5 \times 8.5 = 30.75 \][/tex]
6. Identify the Outliers:
We now check the fish lengths against these bounds:
[tex]\[ \text{Fish lengths: } 7, 8, 10, 15, 15, 18, 18, 22 \][/tex]
None of these values lie below the lower bound of -3.25 or above the upper bound of 30.75.
Therefore, there are no outliers among the fish lengths in the given dataset.
The correct answer is:
```
There are no outliers.
```
1. List the Available Data:
The table shows the lengths of fish in a pond:
[tex]\[ 15, 18, 8, 22, 7, 15, 10, 18 \][/tex]
2. Sort the Data:
First, we sort the data in ascending order:
[tex]\[ 7, 8, 10, 15, 15, 18, 18, 22 \][/tex]
3. Determine the Quartiles:
- First Quartile (Q1): This is the median of the first half of the data (not including the median if the number of data points is odd).
[tex]\[ \text{First half of data: } 7, 8, 10, 15 \][/tex]
The median or Q1 of this subset is halfway between 8 and 10:
[tex]\[ Q1 = 9.5 \][/tex]
- Third Quartile (Q3): This is the median of the second half of the data (not including the median if the number of data points is odd).
[tex]\[ \text{Second half of data: } 15, 18, 18, 22 \][/tex]
The median or Q3 of this subset is halfway between 18 and 18:
[tex]\[ Q3 = 18.0 \][/tex]
4. Calculate the Interquartile Range (IQR):
[tex]\[ IQR = Q3 - Q1 = 18.0 - 9.5 = 8.5 \][/tex]
5. Determine the Bounds for Outliers:
- Lower Bound: Any data point below the lower bound is considered an outlier.
[tex]\[ \text{Lower Bound} = Q1 - 1.5 \times IQR = 9.5 - 1.5 \times 8.5 = -3.25 \][/tex]
Since the length of fish cannot be negative, any reasonable positive fish length will not be considered a lower outlier in this context.
- Upper Bound: Any data point above the upper bound is considered an outlier.
[tex]\[ \text{Upper Bound} = Q3 + 1.5 \times IQR = 18.0 + 1.5 \times 8.5 = 30.75 \][/tex]
6. Identify the Outliers:
We now check the fish lengths against these bounds:
[tex]\[ \text{Fish lengths: } 7, 8, 10, 15, 15, 18, 18, 22 \][/tex]
None of these values lie below the lower bound of -3.25 or above the upper bound of 30.75.
Therefore, there are no outliers among the fish lengths in the given dataset.
The correct answer is:
```
There are no outliers.
```
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