Answer :
To solve for the missing frequency corresponding to the age group [tex]\(40-45\)[/tex] given that the median age is 39, we follow a systematic approach using the cumulative frequency and the concept of median for grouped data.
### Step-by-Step Solution:
1. Determine the cumulative frequencies:
We'll start by calculating the cumulative frequencies up to each class interval. The given intervals and their respective frequencies are:
[tex]\[ \begin{aligned} & \text{Age Group} & \; & \text{Frequency} & \; & \text{Cumulative Frequency} \\ & 20 - 25 & \; & 50 & \; & 50 \\ & 25 - 30 & \; & 70 & \; & 120 \\ & 30 - 35 & \; & 100 & \; & 220 \\ & 35 - 40 & \; & 300 & \; & 520 \\ & 40 - 45 & \; & x & \; & 520 + x \\ & 45 - 50 & \; & 220 & \; & 740 + x \\ & 50 - 55 & \; & 70 & \; & 810 + x \\ & 55 - 60 & \; & 60 & \; & 870 + x \\ \end{aligned} \][/tex]
2. Find the median class:
The median of a dataset divides it into two equal halves. Therefore, we first calculate the median position. The total frequency [tex]\(N\)[/tex] is [tex]\(870 + x\)[/tex]. The median position is:
[tex]\[ \frac{N}{2} = \frac{870 + x}{2} \][/tex]
We know that the median age is 39, which falls into the age group [tex]\(35 - 40\)[/tex]. Hence, the median class is [tex]\(35 - 40\)[/tex].
3. Apply the formula for the median:
The formula for the median in grouped data is given by:
[tex]\[ \text{Median} = L + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h \][/tex]
Where:
- [tex]\(L\)[/tex] is the lower boundary of the median class.
- [tex]\(N\)[/tex] is the total frequency.
- [tex]\(CF\)[/tex] is the cumulative frequency before the median class.
- [tex]\(f\)[/tex] is the frequency of the median class.
- [tex]\(h\)[/tex] is the class width.
For the median class [tex]\(35 - 40\)[/tex]:
- [tex]\(L = 35\)[/tex]
- [tex]\(CF = 220\)[/tex]
- [tex]\(f = 300\)[/tex]
- [tex]\(h = 5\)[/tex]
Since the median is given as 39, substitute these values into the equation:
[tex]\[ 39 = 35 + \left(\frac{\frac{870 + x}{2} - 220}{300}\right) \times 5 \][/tex]
4. Solve for [tex]\(x\)[/tex]:
Simplify the equation step-by-step:
[tex]\[ 39 = 35 + \left(\frac{\frac{870 + x}{2} - 220}{300}\right) \times 5 \][/tex]
[tex]\[ 39 - 35 = \left(\frac{\frac{870 + x}{2} - 220}{300}\right) \times 5 \][/tex]
[tex]\[ 4 = \left(\frac{\frac{870 + x}{2} - 220}{300}\right) \times 5 \][/tex]
[tex]\[ \frac{4}{5} = \frac{\frac{870 + x}{2} - 220}{300} \][/tex]
[tex]\[ 0.8 = \frac{\frac{870 + x}{2} - 220}{300} \][/tex]
[tex]\[ 0.8 \times 300 = \frac{870 + x}{2} - 220 \][/tex]
[tex]\[ 240 = \frac{870 + x}{2} - 220 \][/tex]
[tex]\[ 240 + 220 = \frac{870 + x}{2} \][/tex]
[tex]\[ 460 = \frac{870 + x}{2} \][/tex]
[tex]\[ 460 \times 2 = 870 + x \][/tex]
[tex]\[ 920 = 870 + x \][/tex]
[tex]\[ x = 920 - 870 \][/tex]
[tex]\[ x = 50 \][/tex]
Therefore, the missing frequency for the age group [tex]\(40 - 45\)[/tex] is [tex]\(50\)[/tex].
### Step-by-Step Solution:
1. Determine the cumulative frequencies:
We'll start by calculating the cumulative frequencies up to each class interval. The given intervals and their respective frequencies are:
[tex]\[ \begin{aligned} & \text{Age Group} & \; & \text{Frequency} & \; & \text{Cumulative Frequency} \\ & 20 - 25 & \; & 50 & \; & 50 \\ & 25 - 30 & \; & 70 & \; & 120 \\ & 30 - 35 & \; & 100 & \; & 220 \\ & 35 - 40 & \; & 300 & \; & 520 \\ & 40 - 45 & \; & x & \; & 520 + x \\ & 45 - 50 & \; & 220 & \; & 740 + x \\ & 50 - 55 & \; & 70 & \; & 810 + x \\ & 55 - 60 & \; & 60 & \; & 870 + x \\ \end{aligned} \][/tex]
2. Find the median class:
The median of a dataset divides it into two equal halves. Therefore, we first calculate the median position. The total frequency [tex]\(N\)[/tex] is [tex]\(870 + x\)[/tex]. The median position is:
[tex]\[ \frac{N}{2} = \frac{870 + x}{2} \][/tex]
We know that the median age is 39, which falls into the age group [tex]\(35 - 40\)[/tex]. Hence, the median class is [tex]\(35 - 40\)[/tex].
3. Apply the formula for the median:
The formula for the median in grouped data is given by:
[tex]\[ \text{Median} = L + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h \][/tex]
Where:
- [tex]\(L\)[/tex] is the lower boundary of the median class.
- [tex]\(N\)[/tex] is the total frequency.
- [tex]\(CF\)[/tex] is the cumulative frequency before the median class.
- [tex]\(f\)[/tex] is the frequency of the median class.
- [tex]\(h\)[/tex] is the class width.
For the median class [tex]\(35 - 40\)[/tex]:
- [tex]\(L = 35\)[/tex]
- [tex]\(CF = 220\)[/tex]
- [tex]\(f = 300\)[/tex]
- [tex]\(h = 5\)[/tex]
Since the median is given as 39, substitute these values into the equation:
[tex]\[ 39 = 35 + \left(\frac{\frac{870 + x}{2} - 220}{300}\right) \times 5 \][/tex]
4. Solve for [tex]\(x\)[/tex]:
Simplify the equation step-by-step:
[tex]\[ 39 = 35 + \left(\frac{\frac{870 + x}{2} - 220}{300}\right) \times 5 \][/tex]
[tex]\[ 39 - 35 = \left(\frac{\frac{870 + x}{2} - 220}{300}\right) \times 5 \][/tex]
[tex]\[ 4 = \left(\frac{\frac{870 + x}{2} - 220}{300}\right) \times 5 \][/tex]
[tex]\[ \frac{4}{5} = \frac{\frac{870 + x}{2} - 220}{300} \][/tex]
[tex]\[ 0.8 = \frac{\frac{870 + x}{2} - 220}{300} \][/tex]
[tex]\[ 0.8 \times 300 = \frac{870 + x}{2} - 220 \][/tex]
[tex]\[ 240 = \frac{870 + x}{2} - 220 \][/tex]
[tex]\[ 240 + 220 = \frac{870 + x}{2} \][/tex]
[tex]\[ 460 = \frac{870 + x}{2} \][/tex]
[tex]\[ 460 \times 2 = 870 + x \][/tex]
[tex]\[ 920 = 870 + x \][/tex]
[tex]\[ x = 920 - 870 \][/tex]
[tex]\[ x = 50 \][/tex]
Therefore, the missing frequency for the age group [tex]\(40 - 45\)[/tex] is [tex]\(50\)[/tex].
