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Answer :
The standard deviation [tex](\( \sigma \))[/tex] of the given distribution is approximately 5.08.
Calculate the Mean [tex](\( \bar{x} \))[/tex]
Midpoints (X):
[tex]\[ X_1 = 22.5, \ X_2 = 27.5, \ X_3 = 32.5, \ X_4 = 37.5, \ X_5 = 42.5, \ X_6 = 47.5 \]Frequency (\( f \)):\[ f_1 = 170, \ f_2 = 110, \ f_3 = 80, \ f_4 = 45, \ f_5 = 40, \ f_6 = 35 \]\[ \sum{X \cdot f} = (22.5 \cdot 170) + (27.5 \cdot 110) + (32.5 \cdot 80) + (37.5 \cdot 45) + (42.5 \cdot 40) + (47.5 \cdot 35) \]\[ \sum{X \cdot f} = 3825 + 3025 + 2600 + 1687.5 + 1700 + 1662.5 \]\[ \sum{X \cdot f} = 14400 \][/tex]
Total number of observations (N):
[tex]\[ N = 170 + 110 + 80 + 45 + 40 + 35 \]\[ N = 480 \]\[ \bar{x} = \frac{\sum{X \cdot f}}{N} = \frac{14400}{480} = 30 \][/tex]
Calculate the Standard Deviation [tex](\( \sigma \))[/tex]
[tex]\[ (X - \bar{x})^2 \][/tex]
[tex]\[ (22.5 - 30)^2 = 56.25, \ (27.5 - 30)^2 = 6.25, \ (32.5 - 30)^2 = 6.25, \ (37.5 - 30)^2 = 56.25, \ (42.5 - 30)^2 = 150.25, \ (47.5 - 30)^2 = 292.25 \][/tex]
[tex]\[ \sum{(X - \bar{x})^2 \cdot f} = (56.25 \cdot 170) + (6.25 \cdot 110) + (6.25 \cdot 80) + (56.25 \cdot 45) + (150.25 \cdot 40) + (292.25 \cdot 35) \]\[ \sum{(X - \bar{x})^2 \cdot f} = 9562.5 + 687.5 + 500 + 2531.25 + 6010 + 10228.75 \][/tex]
[tex]\[ \sum{(X - \bar{x})^2 \cdot f} = 35519 \]\[ \frac{\sum{(X - \bar{x})^2 \cdot f}}{N} = \frac{35519}{480} \]\[ \sigma = \sqrt{\frac{35519}{480}} \]\[ \sigma \approx 5.08 \][/tex]
Final Answer: The standard deviation [tex](\( \sigma \))[/tex] of the given distribution is approximately 5.08.
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