Answer :
Let's solve the problem step by step:
### Part (a): Formula to Calculate Mean of a Continuous Series
To calculate the mean of a continuous series, you can use the formula:
[tex]\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \][/tex]
where \( f_i \) are the frequencies of the class intervals and \( x_i \) are the midpoints of the class intervals.
### Part (b): Find the value of \( a \) from the given data
Given data:
- Mean (\(\bar{x}\)) = 41
- Frequencies: \( f_i = [3, 2, a, 2, 1] \)
- Class intervals: \( [20-30], [30-40], [40-50], [50-60], [60-70] \)
- Midpoints (\( x_i \)) of the class intervals: \( [25, 35, 45, 55, 65] \)
The formula for the mean is:
[tex]\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \][/tex]
Using the given mean value:
[tex]\[ 41 = \frac{3 \cdot 25 + 2 \cdot 35 + a \cdot 45 + 2 \cdot 55 + 1 \cdot 65}{3 + 2 + a + 2 + 1} \][/tex]
Calculate the sum of \( f_i x_i \) (numerator) with known frequencies:
[tex]\[ 3 \cdot 25 + 2 \cdot 35 + 2 \cdot 55 + 1 \cdot 65 = 75 + 70 + 110 + 65 = 320 \][/tex]
Thus, the equation becomes:
[tex]\[ 41 = \frac{320 + 45a}{8 + a} \][/tex]
Multiplying both sides by the denominator:
[tex]\[ 41(8 + a) = 320 + 45a \][/tex]
[tex]\[ 328 + 41a = 320 + 45a \][/tex]
[tex]\[ 328 - 320 = 45a - 41a \][/tex]
[tex]\[ 8 = 4a \][/tex]
[tex]\[ a = 2 \][/tex]
So, the value of \( a \) is 2.
### Part (c): Find the Class Interval of the First Quartile (Q1)
To find the first quartile, we need to calculate \( N \) (total frequency):
[tex]\[ N = 3 + 2 + 2 + 2 + 1 = 10 \][/tex]
The position of the first quartile (\( Q1 \)) is given by:
[tex]\[ Q1 = \frac{N+1}{4} = \frac{10+1}{4} = \frac{11}{4} = 2.75 \][/tex]
Now, identify the class interval (group) where the cumulative frequency includes 2.75.
Cumulative frequencies:
- \( 20-30 \): \( 3 \)
- \( 30-40 \): \( 3 + 2 = 5 \)
- \( 40-50 \): \( 5 + 2 = 7 \) (we stop here because \( 2.75 \) lies within this group)
Thus, the class interval for the first quartile (\( Q1 \)) is \( [30-40] \).
### Part (d): Justify Pasang's Statement
First, identify the modal class interval. The modal class is the class with the highest frequency.
Given frequencies: \( 3, 2, 2, 2, 1 \)
The highest frequency is \( 3 \), which corresponds to the class interval \( [20-30] \).
Pasang said, "First quartile class and modal class are the same."
However, from our calculations:
- The first quartile falls in the class interval \( [30-40] \).
- The modal class (class with the highest frequency) is \( [20-30] \).
Therefore, Pasang's statement is incorrect. The first quartile class [tex]\( [30-40] \)[/tex] and the modal class [tex]\( [20-30] \)[/tex] are not the same.
### Part (a): Formula to Calculate Mean of a Continuous Series
To calculate the mean of a continuous series, you can use the formula:
[tex]\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \][/tex]
where \( f_i \) are the frequencies of the class intervals and \( x_i \) are the midpoints of the class intervals.
### Part (b): Find the value of \( a \) from the given data
Given data:
- Mean (\(\bar{x}\)) = 41
- Frequencies: \( f_i = [3, 2, a, 2, 1] \)
- Class intervals: \( [20-30], [30-40], [40-50], [50-60], [60-70] \)
- Midpoints (\( x_i \)) of the class intervals: \( [25, 35, 45, 55, 65] \)
The formula for the mean is:
[tex]\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \][/tex]
Using the given mean value:
[tex]\[ 41 = \frac{3 \cdot 25 + 2 \cdot 35 + a \cdot 45 + 2 \cdot 55 + 1 \cdot 65}{3 + 2 + a + 2 + 1} \][/tex]
Calculate the sum of \( f_i x_i \) (numerator) with known frequencies:
[tex]\[ 3 \cdot 25 + 2 \cdot 35 + 2 \cdot 55 + 1 \cdot 65 = 75 + 70 + 110 + 65 = 320 \][/tex]
Thus, the equation becomes:
[tex]\[ 41 = \frac{320 + 45a}{8 + a} \][/tex]
Multiplying both sides by the denominator:
[tex]\[ 41(8 + a) = 320 + 45a \][/tex]
[tex]\[ 328 + 41a = 320 + 45a \][/tex]
[tex]\[ 328 - 320 = 45a - 41a \][/tex]
[tex]\[ 8 = 4a \][/tex]
[tex]\[ a = 2 \][/tex]
So, the value of \( a \) is 2.
### Part (c): Find the Class Interval of the First Quartile (Q1)
To find the first quartile, we need to calculate \( N \) (total frequency):
[tex]\[ N = 3 + 2 + 2 + 2 + 1 = 10 \][/tex]
The position of the first quartile (\( Q1 \)) is given by:
[tex]\[ Q1 = \frac{N+1}{4} = \frac{10+1}{4} = \frac{11}{4} = 2.75 \][/tex]
Now, identify the class interval (group) where the cumulative frequency includes 2.75.
Cumulative frequencies:
- \( 20-30 \): \( 3 \)
- \( 30-40 \): \( 3 + 2 = 5 \)
- \( 40-50 \): \( 5 + 2 = 7 \) (we stop here because \( 2.75 \) lies within this group)
Thus, the class interval for the first quartile (\( Q1 \)) is \( [30-40] \).
### Part (d): Justify Pasang's Statement
First, identify the modal class interval. The modal class is the class with the highest frequency.
Given frequencies: \( 3, 2, 2, 2, 1 \)
The highest frequency is \( 3 \), which corresponds to the class interval \( [20-30] \).
Pasang said, "First quartile class and modal class are the same."
However, from our calculations:
- The first quartile falls in the class interval \( [30-40] \).
- The modal class (class with the highest frequency) is \( [20-30] \).
Therefore, Pasang's statement is incorrect. The first quartile class [tex]\( [30-40] \)[/tex] and the modal class [tex]\( [20-30] \)[/tex] are not the same.
