Answer :
Let's calculate the average of the given discrete series using the Short-cut Method with an assumed mean of 25.
### Step-by-Step Solution:
1. List the Sizes and their Frequencies:
[tex]\[ \begin{array}{|c|c|} \hline \text{Size} & \text{Frequency} (f) \\ \hline 30 & 2 \\ 29 & 4 \\ 28 & 5 \\ 27 & 3 \\ 26 & 2 \\ 25 & 7 \\ 24 & 1 \\ 23 & 4 \\ 22 & 5 \\ 21 & 7 \\ \hline \end{array} \][/tex]
2. Calculate the Deviations from the Assumed Mean (d = x - A):
Given the assumed mean [tex]\( A = 25 \)[/tex],
[tex]\[ \begin{array}{|c|c|} \hline \text{Size} & \text{Deviation} (d) \\ \hline 30 & 30 - 25 = 5 \\ 29 & 29 - 25 = 4 \\ 28 & 28 - 25 = 3 \\ 27 & 27 - 25 = 2 \\ 26 & 26 - 25 = 1 \\ 25 & 25 - 25 = 0 \\ 24 & 24 - 25 = -1 \\ 23 & 23 - 25 = -2 \\ 22 & 22 - 25 = -3 \\ 21 & 21 - 25 = -4 \\ \hline \end{array} \][/tex]
So the deviations [tex]\( d \)[/tex] are:
[tex]\[ [5, 4, 3, 2, 1, 0, -1, -2, -3, -4] \][/tex]
3. Compute the Products of Deviations and Frequencies (d \cdot f):
[tex]\[ \begin{array}{|c|c|c|} \hline \text{Size} & \text{Deviation} (d) & \text{Product} (d \cdot f) \\ \hline 30 & 5 & 5 \cdot 2 = 10 \\ 29 & 4 & 4 \cdot 4 = 16 \\ 28 & 3 & 3 \cdot 5 = 15 \\ 27 & 2 & 2 \cdot 3 = 6 \\ 26 & 1 & 1 \cdot 2 = 2 \\ 25 & 0 & 0 \cdot 7 = 0 \\ 24 & -1 & -1 \cdot 1 = -1 \\ 23 & -2 & -2 \cdot 4 = -8 \\ 22 & -3 & -3 \cdot 5 = -15 \\ 21 & -4 & -4 \cdot 7 = -28 \\ \hline \end{array} \][/tex]
Summing up the products:
[tex]\[ 10 + 16 + 15 + 6 + 2 + 0 - 1 - 8 - 15 - 28 = -3 \][/tex]
Sum of [tex]\( d \cdot f \)[/tex] values is [tex]\( -3 \)[/tex].
4. Sum of Frequencies:
[tex]\[ \sum f = 2 + 4 + 5 + 3 + 2 + 7 + 1 + 4 + 5 + 7 = 40 \][/tex]
Total frequency [tex]\( \sum f = 40 \)[/tex].
5. Calculate the Mean Using the Short-cut Method:
[tex]\[ \text{Mean} = A + \frac{\sum (d \cdot f)}{\sum f} \][/tex]
Plugging in the values:
[tex]\[ \text{Mean} = 25 + \frac{-3}{40} \][/tex]
[tex]\[ \text{Mean} = 25 - 0.075 \][/tex]
[tex]\[ \text{Mean} = 24.925 \][/tex]
### Final Answer:
The average of the given discrete series, using the Short-cut Method with an assumed mean of 25, is [tex]\( \boxed{24.925} \)[/tex].
### Step-by-Step Solution:
1. List the Sizes and their Frequencies:
[tex]\[ \begin{array}{|c|c|} \hline \text{Size} & \text{Frequency} (f) \\ \hline 30 & 2 \\ 29 & 4 \\ 28 & 5 \\ 27 & 3 \\ 26 & 2 \\ 25 & 7 \\ 24 & 1 \\ 23 & 4 \\ 22 & 5 \\ 21 & 7 \\ \hline \end{array} \][/tex]
2. Calculate the Deviations from the Assumed Mean (d = x - A):
Given the assumed mean [tex]\( A = 25 \)[/tex],
[tex]\[ \begin{array}{|c|c|} \hline \text{Size} & \text{Deviation} (d) \\ \hline 30 & 30 - 25 = 5 \\ 29 & 29 - 25 = 4 \\ 28 & 28 - 25 = 3 \\ 27 & 27 - 25 = 2 \\ 26 & 26 - 25 = 1 \\ 25 & 25 - 25 = 0 \\ 24 & 24 - 25 = -1 \\ 23 & 23 - 25 = -2 \\ 22 & 22 - 25 = -3 \\ 21 & 21 - 25 = -4 \\ \hline \end{array} \][/tex]
So the deviations [tex]\( d \)[/tex] are:
[tex]\[ [5, 4, 3, 2, 1, 0, -1, -2, -3, -4] \][/tex]
3. Compute the Products of Deviations and Frequencies (d \cdot f):
[tex]\[ \begin{array}{|c|c|c|} \hline \text{Size} & \text{Deviation} (d) & \text{Product} (d \cdot f) \\ \hline 30 & 5 & 5 \cdot 2 = 10 \\ 29 & 4 & 4 \cdot 4 = 16 \\ 28 & 3 & 3 \cdot 5 = 15 \\ 27 & 2 & 2 \cdot 3 = 6 \\ 26 & 1 & 1 \cdot 2 = 2 \\ 25 & 0 & 0 \cdot 7 = 0 \\ 24 & -1 & -1 \cdot 1 = -1 \\ 23 & -2 & -2 \cdot 4 = -8 \\ 22 & -3 & -3 \cdot 5 = -15 \\ 21 & -4 & -4 \cdot 7 = -28 \\ \hline \end{array} \][/tex]
Summing up the products:
[tex]\[ 10 + 16 + 15 + 6 + 2 + 0 - 1 - 8 - 15 - 28 = -3 \][/tex]
Sum of [tex]\( d \cdot f \)[/tex] values is [tex]\( -3 \)[/tex].
4. Sum of Frequencies:
[tex]\[ \sum f = 2 + 4 + 5 + 3 + 2 + 7 + 1 + 4 + 5 + 7 = 40 \][/tex]
Total frequency [tex]\( \sum f = 40 \)[/tex].
5. Calculate the Mean Using the Short-cut Method:
[tex]\[ \text{Mean} = A + \frac{\sum (d \cdot f)}{\sum f} \][/tex]
Plugging in the values:
[tex]\[ \text{Mean} = 25 + \frac{-3}{40} \][/tex]
[tex]\[ \text{Mean} = 25 - 0.075 \][/tex]
[tex]\[ \text{Mean} = 24.925 \][/tex]
### Final Answer:
The average of the given discrete series, using the Short-cut Method with an assumed mean of 25, is [tex]\( \boxed{24.925} \)[/tex].
