Question 1193985: Test Obtained by Selected Grade 10 Students in Statistics
48 73 57 57 69 88 11 80 82 47
46 70 49 45 75 81 33 65 38 59
94 59 62 36 58 69 45 55 58 65
30 49 73 29 41 53 37 35 61 48
22 51 56 55 60 37 56 59 57 36
12 36 50 63 68 30 56 70 53 28
The Frequency Distribution of the Examination Results of Grade 10 Students
C.I Tally f
11-22
23-34
34-46
47-58
59-70
71-82
83-94
Using the frequency distribution above, determine the following
11. Size of the class interval
12. Median class
13. Modal class
14. Lower boundary of the median class
15. Upper boundary of the modal class
16. Upper limit of the modal class
17. Class midpoint of the modal class
18. The number of classes
19. Less than cumulative before the median class
20. The frequency of the modal class
21. fm1
22. fm2
23. the mean
24. the median
25. the mode
Answer by parmen(42)   (Show Source):
You can put this solution on YOUR website! **11. Size of the class interval:**
* Class Interval = Upper Limit - Lower Limit
* For the first class: 22 - 11 = 11
* **Size of the class interval: 11**
**12. Median class:**
* Find the total number of students (N): 60
* Find the median position: (N + 1) / 2 = (60 + 1) / 2 = 30.5
* The median class is the class containing the 30.5th observation.
* Looking at the cumulative frequencies, the median class is **47-58**.
**13. Modal class:**
* The modal class is the class with the highest frequency.
* **Modal class: 47-58**
**14. Lower boundary of the median class:**
* Lower boundary = Lower limit - 0.5
* Lower boundary of 47-58 = 47 - 0.5 = **46.5**
**15. Upper boundary of the modal class:**
* Upper boundary = Upper limit + 0.5
* Upper boundary of 47-58 = 58 + 0.5 = **58.5**
**16. Upper limit of the modal class:**
* **Upper limit of the modal class: 58**
**17. Class midpoint of the modal class:**
* Class midpoint = (Lower limit + Upper limit) / 2
* Class midpoint of 47-58 = (47 + 58) / 2 = 52.5
**18. The number of classes:**
* **Number of classes: 7**
**19. Less than cumulative before the median class:**
* Calculate the cumulative frequencies for each class.
* Less than cumulative before the median class (47-58) is the sum of frequencies of all classes before it.
* **Less than cumulative before the median class: 23**
**20. The frequency of the modal class:**
* **Frequency of the modal class (47-58): 17**
**21. fm1:**
* fm1 is the frequency of the class preceding the modal class.
* **fm1 = 12** (frequency of the 34-46 class)
**22. fm2:**
* fm2 is the frequency of the class succeeding the modal class.
* **fm2 = 10** (frequency of the 59-70 class)
**23. The mean:**
* To calculate the mean, you would need the midpoint of each class and their respective frequencies.
* Use the formula:
Mean = Σ(f * x) / Σf
where 'f' is the frequency of each class and 'x' is the midpoint of each class.
**24. The median:**
* Median = L + [(N/2 - cf) / fm] * c
where:
L = Lower boundary of the median class
N = Total number of observations
cf = Cumulative frequency before the median class
fm = Frequency of the median class
c = Class width
* Median = 46.5 + [(30.5 - 23) / 17] * 11
* Median = 46.5 + [7.5 / 17] * 11
* Median ≈ 46.5 + 4.91
* **Median ≈ 51.41**
**25. The mode:**
* The mode is the midpoint of the modal class.
* **Mode = 52.5**
**Note:**
* This analysis assumes that the data within each class is evenly distributed.
* For more precise calculations of the mean and other statistics, you would need the original raw data.
I hope this helps! Let me know if you have any other questions.
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