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Question 1164691: The data recorded in following Table 1 shows the length of medical checkup conducted by a health & safety team, in meters.
20 25 33 38 39 41 21
28 33 32 38 41 47 49
59 60 31 38 48 36 60
Required:
a) Construct frequency distribution with number of classes = 5.
i. Class intervals ii. Mid points
iii. Class boundaries iv. Frequency
v. Cumulative frequency
b) Find the mean and median of the minutes.
c) Find standard deviation also comment on the result.
d) Sketch histogram and frequency polygon
Found 2 solutions by CPhill, greenestamps:
Answer by CPhill(2189)   (Show Source):
You can put this solution on YOUR website! To organize and analyze the medical checkup data, we first need to identify the basic statistics of the raw data:
* **Sample Size ():** 18
* **Minimum value:** 20
* **Maximum value:** 60
* **Range:**
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### a) Frequency Distribution Table
With 5 classes, the class width is . To ensure all data fits comfortably, we will use a width of 9 (e.g., 20 - 28).
| Class Intervals | Class Boundaries | Midpoint () | Frequency () | Cumulative Frequency () |
| --- | --- | --- | --- | --- |
| 20 - 28 | 19.5 - 28.5 | 24 | 4 | 4 |
| 29 - 37 | 28.5 - 37.5 | 33 | 4 | 8 |
| 38 - 46 | 37.5 - 46.5 | 42 | 6 | 14 |
| 47 - 55 | 46.5 - 55.5 | 51 | 2 | 16 |
| 56 - 64 | 55.5 - 64.5 | 60 | 2 | 18 |
| **Total** | | | **18** | |
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### b) Mean and Median
Using the raw data for precision:
**Mean ():**
Sum of all values = 692.
**Median:**
Arranging the 18 values in ascending order:
20, 21, 25, 28, 31, 32, 33, 33, 38, 38, 38, 39, 41, 41, 47, 49, 59, 60.
The median is the average of the 9th and 10th values:
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### c) Standard Deviation
First, we find the variance () of the raw data:
1. Subtract the mean from each value and square the result.
2. Sum those squares: .
3.
**Standard Deviation ():**
**Comment:** The standard deviation of 11.09 suggests that most checkup lengths fall within approximately 11 units of the average. Since the mean (38.44) and median (38) are very close, the distribution is roughly symmetrical, though slightly pulled toward the higher end by the values in the 50s and 60s.
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### d) Histogram and Frequency Polygon
The histogram uses the **Class Boundaries** on the x-axis and **Frequency** on the y-axis. The frequency polygon is created by connecting the midpoints of the tops of the histogram bars.
* **Histogram:** Bars will be highest in the 37.5 - 46.5 range (frequency of 6).
* **Frequency Polygon:** Start the line at the midpoint of an imaginary class before the first (15) and end it at the midpoint of an imaginary class after the last (69).
Would you like me to calculate the **Coefficient of Variation** to further analyze the consistency of these checkup times?
Answer by greenestamps(13295)  (Show Source):
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