QRS ₹ 15.

एक परीक्षामा विद्यार्थीहरूले प्राप्त गरेका अंकहरू तालिकामा दिइएका छन् (The marks obtained by students in an examination are as follows):

\begin{tabular}{|l|c|c|c|c|c|}
\hline
Marks obtained & [tex]$15-25$[/tex] & [tex]$25-35$[/tex] & [tex]$35-45$[/tex] & [tex]$45-55$[/tex] & [tex]$55-65$[/tex] \\
\hline
Frequency & 4 & 6 & 12 & m & 1 \\
\hline
\end{tabular}

a. यदि यो तथ्यांकको माध्य 39 भएमा m को मान निकाल्नुहोस् (If the mean of this data is 39, calculate the value of m.)

b. यस तथ्यांकको मध्यिका श्रेणी पत्ता लगाउनुहोस् (Find the median class of this data.)

c. रीत श्रेणी पत्ता लगाउनुहोस् (Find the modal class of this data.)

d. देखाउनुहोस् कि मध्यक अंकहरू माध्य अंकहरू भन्दा धेरै छन् (Show that the median marks are more than the mean marks.)



Answer :

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.