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\title{Question 1: Statistical Analysis of a Sample Population}
\author{}
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\section*{Problem Statement}
A sample consists of 120 males and 80 females. The table below shows the distribution of individuals by gender and age.
\bigskip
\section*{Part 1: Total Number of Individuals Younger Than 20}
Let the number of males ($n_{\mathrm{male}}$) and females ($n_{\mathrm{female}}$) in the sample be:
\[
\begin{aligned}
n_{\mathrm{male}} &= 120, \\
n_{\mathrm{female}} &= 80.
\end{aligned}
\]
Thus, the total sample size is:
\[
n = n_{\mathrm{male}} + n_{\mathrm{female}} = 200.
\]
Let the number of age groups be:
\[
K = 4.
\]
Define the relative frequencies of males ($p_{\mathrm{male}_k}$) and females ($p_{\mathrm{female}_k}$) in each group:
\[
\begin{alignedat}{4}
p_{\mathrm{male}_1} &= 10\%,\quad p_{\mathrm{male}_2} &= 10\%,\quad \dots,\quad p_{\mathrm{male}_K} &= 50\%, \\
p_{\mathrm{female}_1} &= 20\%,\quad p_{\mathrm{female}_2} &= 20\%,\quad \dots,\quad p_{\mathrm{female}_K} &= 30\%.
\end{alignedat}
\]
The absolute frequency of males and females in each group is given by:
\[
\begin{alignedat}{2}
N_{\mathrm{male}_k} &= n_{\mathrm{male}} p_{\mathrm{male}_k},\quad &k \in \{1,2,\dots,K\}, \\
N_{\mathrm{female}_k} &= n_{\mathrm{female}} p_{\mathrm{female}_k},\quad &k \in \{1,2,\dots,K\}.
\end{alignedat}
\]
The total absolute frequency is:
\[
N_k = N_{\mathrm{male}_k} + N_{\mathrm{female}_k},\quad k \in \{1,2,\dots,K\}.
\]
The number of people younger than 20 is:
\[
N_{\#\{y/o \in \{0,1,\dots,19\}\}} = N_1 = 28.
\]
\textbf{Answer for Part 1:}
\[
\boxed{N_{\#\{y/o \in \{0,1,\dots,19\}\}} = 28}.
\]
\bigskip
\section*{Part 2: Percentage of Individuals Aged 50 or Older}
The relative frequency for the total population is:
\[
p_k = \frac{N_k}{n},\quad k \in \{1,2,\dots,K\}.
\]
The percentage of individuals aged 50 or older is:
\[
p_{\frac{\#\{y/o \in \{50,51,\dots,89\}\}}{\#\{y/o \in \{0,1,\dots,89\}\}}} = p_4 = 42\%.
\]
\textbf{Answer for Part 2:}
\[
\boxed{p_{\frac{\#\{y/o \in \{50,51,\dots,89\}\}}{\#\{y/o \in \{0,1,\dots,89\}\}}} = 42\%}.
\]
\bigskip
\section*{Part 3: Number of Males Aged 30 or Older}
The number of males aged 30 years or older is:
\[
N_{\mathrm{male}_{\#\{y/o \in \{30,31,\dots,89\}\}}} = \sum_{k=3}^{K} N_{\mathrm{male}_k} = 96.
\]
\textbf{Answer for Part 3:}
\[
\boxed{N_{\mathrm{male}_{\#\{y/o \in \{30,31,\dots,89\}\}}} = 96}.
\]
\bigskip
\section*{Part 4: Median Age Calculation}
Define the age class intervals:
\[
\begin{aligned}
z_1 &= y/o \in \{0,1,\dots,19\}, \\
z_2 &= y/o \in \{20,21,\dots,29\}, \\
&\dots \\
z_K &= y/o \in \{50,51,\dots,89\},\quad k \in \{1,2,\dots,K\}.
\end{aligned}
\]
The median position is at:
\[
p_m = 50\%.
\]
Since:
\[
\sum_{k=1}^{2} p_k = 28\% < p_m < 58\% = \sum_{k=1}^{3} p_k,
\]
the median falls within $z_3$. The lower boundary of $z_3$ is:
\[
L_{z_3} = \frac{29+30}{2} = 29.5.
\]
The class width is:
\[
C_{z_3} = \frac{(49+50)-(29+30)}{2} = 20.
\]
Assuming a uniform distribution within $z_3$, the median is calculated as:
\[
m = L_{z_3} + \left(\frac{p_m - \sum_{k=1}^{2} p_k}{p_3}\right) C_{z_3}.
\]
Substituting values:
\[
m = 29.5 + \left(\frac{50\% - 28\%}{30\%}\right) \times 20 = \frac{265}{6} \approx 44.167.
\]
\textbf{Answer for Part 4:}
\[
\boxed{m = \frac{265}{6} \approx 44.167}.
\]
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