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Sagot :
Let's solve each part step-by-step:
### Part (a): Calculate the value of [tex]\( m \)[/tex]
Given classes and frequencies are:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Marks obtained} & 15-25 & 25-35 & 35-45 & 45-55 & 55-65 \\ \hline \text{Frequency} & 4 & 6 & 12 & m & 1 \\ \hline \end{array} \][/tex]
The mean of the data is given as 39.
1. Calculate the midpoints of each class interval:
- For [tex]\(15-25\)[/tex]: [tex]\( \frac{15+25}{2} = 20 \)[/tex]
- For [tex]\(25-35\)[/tex]: [tex]\( \frac{25+35}{2} = 30 \)[/tex]
- For [tex]\(35-45\)[/tex]: [tex]\( \frac{35+45}{2} = 40 \)[/tex]
- For [tex]\(45-55\)[/tex]: [tex]\( \frac{45+55}{2} = 50 \)[/tex]
- For [tex]\(55-65\)[/tex]: [tex]\( \frac{55+65}{2} = 60 \)[/tex]
Therefore, the midpoints are [tex]\(20, 30, 40, 50, 60\)[/tex].
2. Calculate the total frequency [tex]\( N \)[/tex]:
[tex]\[ N = 4 + 6 + 12 + m + 1 = 23 + m \][/tex]
3. Using the formula for the mean:
[tex]\[ \text{Mean} = \frac{\sum f_i x_i}{N} \][/tex]
where [tex]\(f_i\)[/tex] is the frequency and [tex]\(x_i\)[/tex] is the midpoint. Given the mean is 39:
[tex]\[ 39 = \frac{4 \cdot 20 + 6 \cdot 30 + 12 \cdot 40 + m \cdot 50 + 1 \cdot 60}{23 + m} \][/tex]
4. Calculate the sum of the products of frequency and midpoints for the known frequencies:
[tex]\[ \sum f_i x_i (\text{known}) = 4 \cdot 20 + 6 \cdot 30 + 12 \cdot 40 + 1 \cdot 60 = 80 + 180 + 480 + 60 = 800 \][/tex]
5. Substitute in the equation and solve for [tex]\(m\)[/tex]:
[tex]\[ 39 = \frac{800 + 50m}{23 + m} \][/tex]
[tex]\[ 39(23 + m) = 800 + 50m \][/tex]
[tex]\[ 897 + 39m = 800 + 50m \][/tex]
[tex]\[ 897 - 800 = 50m - 39m \][/tex]
[tex]\[ 97 = 11m \][/tex]
[tex]\[ m = \frac{97}{11} \approx 8.82 \][/tex]
Since [tex]\(m\)[/tex] must be a whole number in the context of frequency, we approximate [tex]\(m\)[/tex] to the nearest integer, which is [tex]\( m = 9 \)[/tex].
### Part (b): Find the median class
1. Calculate the cumulative frequencies:
[tex]\[ \begin{array}{c|c} \text{Class Interval} & \text{Cumulative Frequency} \\ \hline 15-25 & 4 \\ 25-35 & 4 + 6 = 10 \\ 35-45 & 10 + 12 = 22 \\ 45-55 & 22 + 9 = 31 \\ 55-65 & 31 + 1 = 32 \\ \end{array} \][/tex]
2. Calculate [tex]\( N \)[/tex]:
[tex]\[ N = 23 + 9 = 32 \][/tex]
3. Find the median position [tex]\((\frac{N}{2})\)[/tex]:
[tex]\[ \frac{32}{2} = 16 \][/tex]
4. The cumulative frequency just greater than 16 is 22, which corresponds to the class interval [tex]\(35-45\)[/tex].
Therefore, the median class is [tex]\( 35-45 \)[/tex].
### Part (c): Find the modal class
The modal class is the class with the highest frequency. In our frequency distribution, the class interval [tex]\(35-45\)[/tex] has the highest frequency of 12.
Therefore, the modal class is [tex]\( 35-45 \)[/tex].
### Part (d): Show that median marks are more than the median of the marks distribution
1. The median class was found to be [tex]\(35-45\)[/tex].
2. For exact calculation of median marks:
[tex]\[ \text{Median} = L + \left(\frac{\frac{N}{2} - CF}{f}\right) \cdot h \][/tex]
where:
- [tex]\(L\)[/tex] is the lower boundary of the median class,
- [tex]\(CF\)[/tex] is the cumulative frequency of the class before the median class,
- [tex]\(f\)[/tex] is the frequency of the median class,
- [tex]\(h\)[/tex] is the class width.
Given [tex]\(L = 35\)[/tex], [tex]\(CF = 10\)[/tex], [tex]\(f = 12\)[/tex], and [tex]\(h = 10\)[/tex]:
[tex]\[ \text{Median} = 35 + \left(\frac{16 - 10}{12}\right) \cdot 10 \][/tex]
[tex]\[ \text{Median} = 35 + \left(\frac{6}{12}\right) \cdot 10 \][/tex]
[tex]\[ \text{Median} = 35 + 5 = 40 \][/tex]
The median marks (40) are indeed larger than the mean marks (39).
Thus, we have shown that median marks are more than the mean marks.
### Part (a): Calculate the value of [tex]\( m \)[/tex]
Given classes and frequencies are:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Marks obtained} & 15-25 & 25-35 & 35-45 & 45-55 & 55-65 \\ \hline \text{Frequency} & 4 & 6 & 12 & m & 1 \\ \hline \end{array} \][/tex]
The mean of the data is given as 39.
1. Calculate the midpoints of each class interval:
- For [tex]\(15-25\)[/tex]: [tex]\( \frac{15+25}{2} = 20 \)[/tex]
- For [tex]\(25-35\)[/tex]: [tex]\( \frac{25+35}{2} = 30 \)[/tex]
- For [tex]\(35-45\)[/tex]: [tex]\( \frac{35+45}{2} = 40 \)[/tex]
- For [tex]\(45-55\)[/tex]: [tex]\( \frac{45+55}{2} = 50 \)[/tex]
- For [tex]\(55-65\)[/tex]: [tex]\( \frac{55+65}{2} = 60 \)[/tex]
Therefore, the midpoints are [tex]\(20, 30, 40, 50, 60\)[/tex].
2. Calculate the total frequency [tex]\( N \)[/tex]:
[tex]\[ N = 4 + 6 + 12 + m + 1 = 23 + m \][/tex]
3. Using the formula for the mean:
[tex]\[ \text{Mean} = \frac{\sum f_i x_i}{N} \][/tex]
where [tex]\(f_i\)[/tex] is the frequency and [tex]\(x_i\)[/tex] is the midpoint. Given the mean is 39:
[tex]\[ 39 = \frac{4 \cdot 20 + 6 \cdot 30 + 12 \cdot 40 + m \cdot 50 + 1 \cdot 60}{23 + m} \][/tex]
4. Calculate the sum of the products of frequency and midpoints for the known frequencies:
[tex]\[ \sum f_i x_i (\text{known}) = 4 \cdot 20 + 6 \cdot 30 + 12 \cdot 40 + 1 \cdot 60 = 80 + 180 + 480 + 60 = 800 \][/tex]
5. Substitute in the equation and solve for [tex]\(m\)[/tex]:
[tex]\[ 39 = \frac{800 + 50m}{23 + m} \][/tex]
[tex]\[ 39(23 + m) = 800 + 50m \][/tex]
[tex]\[ 897 + 39m = 800 + 50m \][/tex]
[tex]\[ 897 - 800 = 50m - 39m \][/tex]
[tex]\[ 97 = 11m \][/tex]
[tex]\[ m = \frac{97}{11} \approx 8.82 \][/tex]
Since [tex]\(m\)[/tex] must be a whole number in the context of frequency, we approximate [tex]\(m\)[/tex] to the nearest integer, which is [tex]\( m = 9 \)[/tex].
### Part (b): Find the median class
1. Calculate the cumulative frequencies:
[tex]\[ \begin{array}{c|c} \text{Class Interval} & \text{Cumulative Frequency} \\ \hline 15-25 & 4 \\ 25-35 & 4 + 6 = 10 \\ 35-45 & 10 + 12 = 22 \\ 45-55 & 22 + 9 = 31 \\ 55-65 & 31 + 1 = 32 \\ \end{array} \][/tex]
2. Calculate [tex]\( N \)[/tex]:
[tex]\[ N = 23 + 9 = 32 \][/tex]
3. Find the median position [tex]\((\frac{N}{2})\)[/tex]:
[tex]\[ \frac{32}{2} = 16 \][/tex]
4. The cumulative frequency just greater than 16 is 22, which corresponds to the class interval [tex]\(35-45\)[/tex].
Therefore, the median class is [tex]\( 35-45 \)[/tex].
### Part (c): Find the modal class
The modal class is the class with the highest frequency. In our frequency distribution, the class interval [tex]\(35-45\)[/tex] has the highest frequency of 12.
Therefore, the modal class is [tex]\( 35-45 \)[/tex].
### Part (d): Show that median marks are more than the median of the marks distribution
1. The median class was found to be [tex]\(35-45\)[/tex].
2. For exact calculation of median marks:
[tex]\[ \text{Median} = L + \left(\frac{\frac{N}{2} - CF}{f}\right) \cdot h \][/tex]
where:
- [tex]\(L\)[/tex] is the lower boundary of the median class,
- [tex]\(CF\)[/tex] is the cumulative frequency of the class before the median class,
- [tex]\(f\)[/tex] is the frequency of the median class,
- [tex]\(h\)[/tex] is the class width.
Given [tex]\(L = 35\)[/tex], [tex]\(CF = 10\)[/tex], [tex]\(f = 12\)[/tex], and [tex]\(h = 10\)[/tex]:
[tex]\[ \text{Median} = 35 + \left(\frac{16 - 10}{12}\right) \cdot 10 \][/tex]
[tex]\[ \text{Median} = 35 + \left(\frac{6}{12}\right) \cdot 10 \][/tex]
[tex]\[ \text{Median} = 35 + 5 = 40 \][/tex]
The median marks (40) are indeed larger than the mean marks (39).
Thus, we have shown that median marks are more than the mean marks.
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