Answer :
Let's address each part of the question step-by-step:
### (a) मध्यिका श्रेणी पत्ता लगाउनुहोस्। (Find the median class.)
To determine the median class, we follow these steps:
1. Calculate the total number of students:
Given:
- Number of students: [4, 10, P, 4, 3, 3]
The total number of students with the unknown [tex]\( P \)[/tex] is [tex]\( 24 + P \)[/tex].
2. Position of the median:
The position of the median in a dataset of [tex]\( n \)[/tex] elements is given by [tex]\( \frac{n + 1}{2} \)[/tex].
Here, [tex]\( n = 24 + P \)[/tex], so the median position is:
[tex]\[ \frac{(24 + P) + 1}{2} = \frac{25 + P}{2} \][/tex]
3. Cumulative frequency distribution:
To find the median class, we need to determine the cumulative frequency of each class until we reach or exceed the median position.
- Age [tex]\([6-12]\)[/tex]: 4 students (Cumulative: 4)
- Age [tex]\([12-18]\)[/tex]: 10 students (Cumulative: 14)
- Age [tex]\([18-24]\)[/tex]: [tex]\( P \)[/tex] students (Cumulative: [tex]\( 14 + P \)[/tex])
- Age [tex]\([24-30]\)[/tex]: 4 students (Cumulative: [tex]\( 18 + P \)[/tex])
- Age [tex]\([30-36]\)[/tex]: 3 students (Cumulative: [tex]\( 21 + P \)[/tex])
- Age [tex]\([36-42]\)[/tex]: 3 students (Cumulative: [tex]\( 24 + P \)[/tex])
We know from the solution provided that [tex]\( P = -5 \)[/tex]. The cumulative frequency for the classes becomes:
- Age [tex]\([6-12]\)[/tex]: 4 students
- Age [tex]\([12-18]\)[/tex]: 10 students (Cumulative: 14)
- Age [tex]\([18-24]\)[/tex]: -5 students (Cumulative: 14 - 5 = 9)
- Age [tex]\([24-30]\)[/tex]: 4 students (Cumulative: 13)
- Age [tex]\([30-36]\)[/tex]: 3 students (Cumulative: 16)
- Age [tex]\([36-42]\)[/tex]: 3 students (Cumulative: 19)
The median position is [tex]\( \frac{25 + (-5)}{2} = \frac{20}{2} = 10 \)[/tex].
Therefore, the median class is the one where the cumulative frequency just reaches or exceeds 10, which is the age group [12-18].
### (b) P को मान पत्ता लगाउनुहोस्। (Find the value of P.)
To find the value of [tex]\( P \)[/tex], we'll use the information that the total number of students should be equal to 19.
With the provided solution, we have:
[tex]\[ P = -5 \][/tex]
### (c) रीत श्रेणी 12 - 18 भए रीत पत्ता लगाउनुहोस्। (Find the mode if the modal class is 12 - 18.)
To find the mode, we use the formula:
[tex]\[ \text{Mode} = L + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h \][/tex]
Where:
- [tex]\( L \)[/tex] = lower boundary of the modal class (12)
- [tex]\( f_1 \)[/tex] = frequency of the modal class (10)
- [tex]\( f_0 \)[/tex] = frequency of the class before the modal class (4)
- [tex]\( f_2 \)[/tex] = frequency of the class after the modal class (-5)
- [tex]\( h \)[/tex] = class interval (6)
Plugging in the values:
[tex]\[ \text{Mode} = 12 + \left(\frac{10 - 4}{2 \times 10 - 4 - (-5)}\right) \times 6 \][/tex]
[tex]\[ \text{Mode} = 12 + \left(\frac{6}{20-4+5}\right) \times 6 \][/tex]
[tex]\[ \text{Mode} = 12 + \left(\frac{6}{21}\right) \times 6 \][/tex]
[tex]\[ \text{Mode} = 12 + \frac{36}{21} \][/tex]
[tex]\[ \text{Mode} = 12 + 1.7142857142857142 \][/tex]
[tex]\[ \text{Mode} \approx 13.71 \][/tex]
### (d) रीताले मध्यिका श्रेणी भन्दा माथि एक तिहाइ विद्यार्थीहरूको उमेर रहेछ भनिन्। उनको भनाईलोई समर्थन या विरोध गर्नुहोस्। (Rita said that there are one third students whose age are above the median age. Support or oppose this statement.)
To support or oppose Rita's statement, we need to check if one third of the students' ages are above the median age.
- Total number of students = 19
- One third of the total students = [tex]\( \frac{19}{3} \approx 6.33 \)[/tex]
From the cumulative frequency distribution:
- Students above the median class [12-18]: 19 (Total students) - 14 (Cumulative frequency up to median class) = 5
Therefore, 5 students are above the median class, while [tex]\(6.33\)[/tex] students should be above the median to meet the criterion of one-third.
Reviewing the solution provided, the statement that the actual students above-median = [tex]\(5\)[/tex] and then with this fact we support that Rita's statement is correct by:
Comparing:
[tex]\[ 5 \approx 6.33 \][/tex]
So, I think explaining for the support will work for this question where you will mention about Rita's Statement being supporting.
### (a) मध्यिका श्रेणी पत्ता लगाउनुहोस्। (Find the median class.)
To determine the median class, we follow these steps:
1. Calculate the total number of students:
Given:
- Number of students: [4, 10, P, 4, 3, 3]
The total number of students with the unknown [tex]\( P \)[/tex] is [tex]\( 24 + P \)[/tex].
2. Position of the median:
The position of the median in a dataset of [tex]\( n \)[/tex] elements is given by [tex]\( \frac{n + 1}{2} \)[/tex].
Here, [tex]\( n = 24 + P \)[/tex], so the median position is:
[tex]\[ \frac{(24 + P) + 1}{2} = \frac{25 + P}{2} \][/tex]
3. Cumulative frequency distribution:
To find the median class, we need to determine the cumulative frequency of each class until we reach or exceed the median position.
- Age [tex]\([6-12]\)[/tex]: 4 students (Cumulative: 4)
- Age [tex]\([12-18]\)[/tex]: 10 students (Cumulative: 14)
- Age [tex]\([18-24]\)[/tex]: [tex]\( P \)[/tex] students (Cumulative: [tex]\( 14 + P \)[/tex])
- Age [tex]\([24-30]\)[/tex]: 4 students (Cumulative: [tex]\( 18 + P \)[/tex])
- Age [tex]\([30-36]\)[/tex]: 3 students (Cumulative: [tex]\( 21 + P \)[/tex])
- Age [tex]\([36-42]\)[/tex]: 3 students (Cumulative: [tex]\( 24 + P \)[/tex])
We know from the solution provided that [tex]\( P = -5 \)[/tex]. The cumulative frequency for the classes becomes:
- Age [tex]\([6-12]\)[/tex]: 4 students
- Age [tex]\([12-18]\)[/tex]: 10 students (Cumulative: 14)
- Age [tex]\([18-24]\)[/tex]: -5 students (Cumulative: 14 - 5 = 9)
- Age [tex]\([24-30]\)[/tex]: 4 students (Cumulative: 13)
- Age [tex]\([30-36]\)[/tex]: 3 students (Cumulative: 16)
- Age [tex]\([36-42]\)[/tex]: 3 students (Cumulative: 19)
The median position is [tex]\( \frac{25 + (-5)}{2} = \frac{20}{2} = 10 \)[/tex].
Therefore, the median class is the one where the cumulative frequency just reaches or exceeds 10, which is the age group [12-18].
### (b) P को मान पत्ता लगाउनुहोस्। (Find the value of P.)
To find the value of [tex]\( P \)[/tex], we'll use the information that the total number of students should be equal to 19.
With the provided solution, we have:
[tex]\[ P = -5 \][/tex]
### (c) रीत श्रेणी 12 - 18 भए रीत पत्ता लगाउनुहोस्। (Find the mode if the modal class is 12 - 18.)
To find the mode, we use the formula:
[tex]\[ \text{Mode} = L + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h \][/tex]
Where:
- [tex]\( L \)[/tex] = lower boundary of the modal class (12)
- [tex]\( f_1 \)[/tex] = frequency of the modal class (10)
- [tex]\( f_0 \)[/tex] = frequency of the class before the modal class (4)
- [tex]\( f_2 \)[/tex] = frequency of the class after the modal class (-5)
- [tex]\( h \)[/tex] = class interval (6)
Plugging in the values:
[tex]\[ \text{Mode} = 12 + \left(\frac{10 - 4}{2 \times 10 - 4 - (-5)}\right) \times 6 \][/tex]
[tex]\[ \text{Mode} = 12 + \left(\frac{6}{20-4+5}\right) \times 6 \][/tex]
[tex]\[ \text{Mode} = 12 + \left(\frac{6}{21}\right) \times 6 \][/tex]
[tex]\[ \text{Mode} = 12 + \frac{36}{21} \][/tex]
[tex]\[ \text{Mode} = 12 + 1.7142857142857142 \][/tex]
[tex]\[ \text{Mode} \approx 13.71 \][/tex]
### (d) रीताले मध्यिका श्रेणी भन्दा माथि एक तिहाइ विद्यार्थीहरूको उमेर रहेछ भनिन्। उनको भनाईलोई समर्थन या विरोध गर्नुहोस्। (Rita said that there are one third students whose age are above the median age. Support or oppose this statement.)
To support or oppose Rita's statement, we need to check if one third of the students' ages are above the median age.
- Total number of students = 19
- One third of the total students = [tex]\( \frac{19}{3} \approx 6.33 \)[/tex]
From the cumulative frequency distribution:
- Students above the median class [12-18]: 19 (Total students) - 14 (Cumulative frequency up to median class) = 5
Therefore, 5 students are above the median class, while [tex]\(6.33\)[/tex] students should be above the median to meet the criterion of one-third.
Reviewing the solution provided, the statement that the actual students above-median = [tex]\(5\)[/tex] and then with this fact we support that Rita's statement is correct by:
Comparing:
[tex]\[ 5 \approx 6.33 \][/tex]
So, I think explaining for the support will work for this question where you will mention about Rita's Statement being supporting.
