15. The mean of the given data is 41, where \(a\) is the frequency of the class interval \((40-50)\).
[tex]\[
\begin{tabular}{|l|c|c|c|c|c|}
\hline \text{Class Interval} & 20-30 & 30-40 & 40-50 & 50-60 & 60-70 \\
\hline \text{Frequency} & 3 & 2 & \(a\) & 2 & 1 \\
\hline
\end{tabular}
\][/tex]

a) Write the formula to calculate the mean of a continuous series.

b) Find the value of \(a\) from the given data.

c) What is the class interval of the first quartile? Find it.

d) Pasang said, "First quartile class and modal class are the same." Justify his statement.



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.