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Sagot :
Let's analyze the given data set and identify any outliers using the steps below:
1. List the data:
[tex]\(0, 0, 0, 417, 171, 355, 343, 212, 588, 474, 3393, 418, 375, 559, 0, 0, 113, 546, 394, 105\)[/tex]
2. Order the data:
[tex]\(0, 0, 0, 0, 0, 105, 113, 171, 212, 343, 355, 375, 394, 417, 418, 474, 546, 559, 588, 3393\)[/tex]
3. Determine the first quartile (Q1) and third quartile (Q3):
Q1 is the median of the first half of the data:
- First half: [tex]\(0, 0, 0, 0, 0, 105, 113, 171, 212, 343\)[/tex]
- Q1 position: [tex]\( (10 + 1) / 2 \)[/tex] = 5.5 ⇒ [tex]\(Q1\)[/tex] = [tex]\(\text{average of 5th and 6th values}\)[/tex] = (0 + 105) / 2 = 52.5
Q3 is the median of the second half of the data:
- Second half: [tex]\(343, 355, 375, 394, 417, 418, 474, 546, 559, 588, 3393\)[/tex]
- Q3 position: [tex]\((10 + 1) / 2\)[/tex] = 5.5 ⇒ [tex]\(Q3\)[/tex] = [tex]\(\text{average of 5th and 6th values}\)[/tex] = (417 + 418) / 2 = 417.5
4. Calculate the interquartile range (IQR):
[tex]\(IQR = Q3 - Q1 = 417.5 - 52.5 = 365\)[/tex]
5. Determine the lower and upper bounds for outliers:
- Lower bound: [tex]\(Q1 - 1.5 \times IQR = 52.5 - 1.5 \times 365\)[/tex] = 52.5 - 547.5 = -495
- Upper bound: [tex]\(Q3 + 1.5 \times IQR = 417.5 + 1.5 \times 365\)[/tex] = 417.5 + 547.5 = 965
Any data point less than -495 or greater than 965 is considered an outlier.
6. Identify the outliers:
In the ordered list of data:
- [tex]\(3393\)[/tex] is greater than 965.
Thus, the outlier is [tex]\(3393\)[/tex].
Summary:
(a) The outlier(s) is/are [tex]\(3393\)[/tex]
---
(b) Drawing a histogram:
To draw a histogram, divide the range of possible data values into intervals (bins) and count how many data points fall into each interval.
Binning could be as follows: [tex]\(0-199, 200-399, 400-599, 600-799, 800-\dots\)[/tex],
| Interval | Count |
|----------|-------|
| 0-199 | 8 |
| 200-399 | 5 |
| 400-599 | 6 |
| 600-799 | 0 |
| 800 and | 1 |
The histogram would have counts on the y-axis and intervals on the x-axis.
---
(c) Explanation:
The outliers likely represent unusual data points. In this case, [tex]\(3393\)[/tex] may indicate an unusually high income for a student, possibly due to a unique situation such as a high-paying job or an error in reporting data.
1. List the data:
[tex]\(0, 0, 0, 417, 171, 355, 343, 212, 588, 474, 3393, 418, 375, 559, 0, 0, 113, 546, 394, 105\)[/tex]
2. Order the data:
[tex]\(0, 0, 0, 0, 0, 105, 113, 171, 212, 343, 355, 375, 394, 417, 418, 474, 546, 559, 588, 3393\)[/tex]
3. Determine the first quartile (Q1) and third quartile (Q3):
Q1 is the median of the first half of the data:
- First half: [tex]\(0, 0, 0, 0, 0, 105, 113, 171, 212, 343\)[/tex]
- Q1 position: [tex]\( (10 + 1) / 2 \)[/tex] = 5.5 ⇒ [tex]\(Q1\)[/tex] = [tex]\(\text{average of 5th and 6th values}\)[/tex] = (0 + 105) / 2 = 52.5
Q3 is the median of the second half of the data:
- Second half: [tex]\(343, 355, 375, 394, 417, 418, 474, 546, 559, 588, 3393\)[/tex]
- Q3 position: [tex]\((10 + 1) / 2\)[/tex] = 5.5 ⇒ [tex]\(Q3\)[/tex] = [tex]\(\text{average of 5th and 6th values}\)[/tex] = (417 + 418) / 2 = 417.5
4. Calculate the interquartile range (IQR):
[tex]\(IQR = Q3 - Q1 = 417.5 - 52.5 = 365\)[/tex]
5. Determine the lower and upper bounds for outliers:
- Lower bound: [tex]\(Q1 - 1.5 \times IQR = 52.5 - 1.5 \times 365\)[/tex] = 52.5 - 547.5 = -495
- Upper bound: [tex]\(Q3 + 1.5 \times IQR = 417.5 + 1.5 \times 365\)[/tex] = 417.5 + 547.5 = 965
Any data point less than -495 or greater than 965 is considered an outlier.
6. Identify the outliers:
In the ordered list of data:
- [tex]\(3393\)[/tex] is greater than 965.
Thus, the outlier is [tex]\(3393\)[/tex].
Summary:
(a) The outlier(s) is/are [tex]\(3393\)[/tex]
---
(b) Drawing a histogram:
To draw a histogram, divide the range of possible data values into intervals (bins) and count how many data points fall into each interval.
Binning could be as follows: [tex]\(0-199, 200-399, 400-599, 600-799, 800-\dots\)[/tex],
| Interval | Count |
|----------|-------|
| 0-199 | 8 |
| 200-399 | 5 |
| 400-599 | 6 |
| 600-799 | 0 |
| 800 and | 1 |
The histogram would have counts on the y-axis and intervals on the x-axis.
---
(c) Explanation:
The outliers likely represent unusual data points. In this case, [tex]\(3393\)[/tex] may indicate an unusually high income for a student, possibly due to a unique situation such as a high-paying job or an error in reporting data.
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