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Sagot :
### Solution:
#### Part (a)
To determine the correction factor for the given data with class boundaries like [tex]$10-19, 20-29, 30-39, \ldots$[/tex], we must recognize that here, the difference (width) between class boundaries is consistent. The correction factor for such uniformly spaced classes is denoted by the multiplication factor 1. Thus, the correct answer is:
(ii) 1
#### Part (b)
The formula to find the mid-value (or class mark) of a class interval is given by:
[tex]\[ \text{Mid-value} = \frac{\text{Lower Bound} + \text{Upper Bound}}{2} \][/tex]
#### Part (c)
To construct a cumulative frequency table:
[tex]\[ \begin{array}{|c|c|c|} \hline \text{Marks} & \text{Frequency} & \text{Cumulative Frequency} \\ \hline 0-10 & 3 & 3 \\ 10-20 & 8 & 11 \\ 20-30 & 10 & 21 \\ 30-40 & 17 & 38 \\ 40-50 & 6 & 44 \\ 50-60 & 2 & 46 \\ 60-70 & 4 & 50 \\ \hline \end{array} \][/tex]
where the cumulative frequency is calculated by successively adding each frequency from the previous cumulative sum.
#### Part (d)
To draw ogives (cumulative frequency curves):
- Less than Ogive: Plot the cumulative frequency at the upper boundary of each class.
[tex]\[ \begin{array}{|c|c|} \hline \text{Upper Bound} & \text{Cumulative Frequency} \\ \hline 10 & 3 \\ 20 & 11 \\ 30 & 21 \\ 40 & 38 \\ 50 & 44 \\ 60 & 46 \\ 70 & 50 \\ \hline \end{array} \][/tex]
- More than Ogive: Plot the reverse cumulative frequency (starting from the highest class downwards).
[tex]\[ \begin{array}{|c|c|} \hline \text{Lower Bound} & \text{More than Cumulative Frequency} \\ \hline 0 & 50 \\ 10 & 47 \\ 20 & 39 \\ 30 & 29 \\ 40 & 12 \\ 50 & 6 \\ 60 & 4 \\ \hline \end{array} \][/tex]
Both ogives can be plotted on the same coordinate axes with marks on the x-axis and cumulative frequency on the y-axis.
#### Part (e)
To find the median of the distribution:
1. Calculate the total number of students, [tex]\( N \)[/tex]:
[tex]\[ N = 50 \][/tex]
2. Determine the median position:
[tex]\[ \text{Median Position} = \frac{N + 1}{2} = \frac{51}{2} = 25.5 \][/tex]
3. Locate the median class:
We find which cumulative frequency just meets or exceeds the median position of 25.5. Observing the cumulative frequency table, this occurs in the class 30-40, with a cumulative frequency of 38.
4. Class parameters for the median class:
- Lower bound ([tex]\( L \)[/tex]) of the median class: 30
- Cumulative frequency before the median class ([tex]\( CF \)[/tex]): 21
- Frequency of the median class ([tex]\( f \)[/tex]): 17
- Class width ([tex]\( h \)[/tex]): 10
5. Use the median formula:
[tex]\[ \text{Median} = L + \left( \frac{\frac{N}{2} - CF}{f} \right) \cdot h \][/tex]
Substituting the values:
[tex]\[ \text{Median} = 30 + \left( \frac{25 - 21}{17} \right) \cdot 10 \][/tex]
Simplifying further:
[tex]\[ \text{Median} = 30 + \left( \frac{4}{17} \right) \cdot 10 = 30 + 2.35294117647059 = 32.35294117647059 \][/tex]
So, the median mark is approximately 32.35.
#### Part (a)
To determine the correction factor for the given data with class boundaries like [tex]$10-19, 20-29, 30-39, \ldots$[/tex], we must recognize that here, the difference (width) between class boundaries is consistent. The correction factor for such uniformly spaced classes is denoted by the multiplication factor 1. Thus, the correct answer is:
(ii) 1
#### Part (b)
The formula to find the mid-value (or class mark) of a class interval is given by:
[tex]\[ \text{Mid-value} = \frac{\text{Lower Bound} + \text{Upper Bound}}{2} \][/tex]
#### Part (c)
To construct a cumulative frequency table:
[tex]\[ \begin{array}{|c|c|c|} \hline \text{Marks} & \text{Frequency} & \text{Cumulative Frequency} \\ \hline 0-10 & 3 & 3 \\ 10-20 & 8 & 11 \\ 20-30 & 10 & 21 \\ 30-40 & 17 & 38 \\ 40-50 & 6 & 44 \\ 50-60 & 2 & 46 \\ 60-70 & 4 & 50 \\ \hline \end{array} \][/tex]
where the cumulative frequency is calculated by successively adding each frequency from the previous cumulative sum.
#### Part (d)
To draw ogives (cumulative frequency curves):
- Less than Ogive: Plot the cumulative frequency at the upper boundary of each class.
[tex]\[ \begin{array}{|c|c|} \hline \text{Upper Bound} & \text{Cumulative Frequency} \\ \hline 10 & 3 \\ 20 & 11 \\ 30 & 21 \\ 40 & 38 \\ 50 & 44 \\ 60 & 46 \\ 70 & 50 \\ \hline \end{array} \][/tex]
- More than Ogive: Plot the reverse cumulative frequency (starting from the highest class downwards).
[tex]\[ \begin{array}{|c|c|} \hline \text{Lower Bound} & \text{More than Cumulative Frequency} \\ \hline 0 & 50 \\ 10 & 47 \\ 20 & 39 \\ 30 & 29 \\ 40 & 12 \\ 50 & 6 \\ 60 & 4 \\ \hline \end{array} \][/tex]
Both ogives can be plotted on the same coordinate axes with marks on the x-axis and cumulative frequency on the y-axis.
#### Part (e)
To find the median of the distribution:
1. Calculate the total number of students, [tex]\( N \)[/tex]:
[tex]\[ N = 50 \][/tex]
2. Determine the median position:
[tex]\[ \text{Median Position} = \frac{N + 1}{2} = \frac{51}{2} = 25.5 \][/tex]
3. Locate the median class:
We find which cumulative frequency just meets or exceeds the median position of 25.5. Observing the cumulative frequency table, this occurs in the class 30-40, with a cumulative frequency of 38.
4. Class parameters for the median class:
- Lower bound ([tex]\( L \)[/tex]) of the median class: 30
- Cumulative frequency before the median class ([tex]\( CF \)[/tex]): 21
- Frequency of the median class ([tex]\( f \)[/tex]): 17
- Class width ([tex]\( h \)[/tex]): 10
5. Use the median formula:
[tex]\[ \text{Median} = L + \left( \frac{\frac{N}{2} - CF}{f} \right) \cdot h \][/tex]
Substituting the values:
[tex]\[ \text{Median} = 30 + \left( \frac{25 - 21}{17} \right) \cdot 10 \][/tex]
Simplifying further:
[tex]\[ \text{Median} = 30 + \left( \frac{4}{17} \right) \cdot 10 = 30 + 2.35294117647059 = 32.35294117647059 \][/tex]
So, the median mark is approximately 32.35.
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