Solution:
Given:
From the table of values derived above;
The mean for males is;
[tex]\begin{gathered} \bar{x}=\frac{\sum ^{}_{}fx}{n} \\ \bar{x}=\frac{5774.5}{141} \\ \bar{x}=40.95 \end{gathered}[/tex]
The standard deviation is;
Hence,
[tex]\begin{gathered} \sigma=\sqrt[]{\frac{300115.25-\frac{5774.5^2}{141}}{141}} \\ \sigma=\sqrt[]{\frac{300115.25-236488.2996}{141}} \\ \sigma=\sqrt[]{\frac{63626.95035}{141}} \\ \sigma=\sqrt[]{451.25497} \\ \sigma=21.243 \\ \\ To\text{ two decimal places,} \\ \sigma=21.24 \end{gathered}[/tex]
Therefore, the population standard deviation for males is 21.24
