The midpoints \( x_i \) for each class interval are calculated as follows:
\[
x_1 = \frac{0 + 19}{2} = 9.5, \quad x_2 = \frac{20 + 39}{2} = 29.5, \quad x_3 = \frac{40 + 59}{2} = 49.5,
\]
\[
x_4 = \frac{60 + 79}{2} = 69.5, \quad x_5 = \frac{80 + 99}{2} = 89.5, \quad x_6 = \frac{100 + 119}{2} = 109.5
\]
Thus, the midpoints are:
\[
\text{Midpoints: } [9.5, 29.5, 49.5, 69.5, 89.5, 109.5]
\]
The mean \( \bar{x} \) of the grouped data is calculated using the formula:
\[
\bar{x} = \frac{\sum (f_i \cdot x_i)}{N}
\]
where \( f_i \) is the frequency and \( N \) is the total frequency. The total frequency \( N \) is:
\[
N = 4 + 3 + 6 + 3 + 5 + 9 = 30
\]
Calculating the numerator:
\[
\sum (f_i \cdot x_i) = (4 \cdot 9.5) + (3 \cdot 29.5) + (6 \cdot 49.5) + (3 \cdot 69.5) + (5 \cdot 89.5) + (9 \cdot 109.5) = 38 + 88.5 + 297 + 208.5 + 447.5 + 985.5 = 2065
\]
Thus, the mean is:
\[
\bar{x} = \frac{2065}{30} \approx 68.8333
\]
The variance \( s^2 \) for grouped data is calculated using the formula:
\[
s^2 = \frac{\sum f_i (x_i - \bar{x})^2}{N}
\]
Calculating \( (x_i - \bar{x})^2 \) for each midpoint:
\[
(x_1 - \bar{x})^2 = (9.5 - 68.8333)^2 \approx 3480.1111
\]
\[
(x_2 - \bar{x})^2 = (29.5 - 68.8333)^2 \approx 1537.1111
\]
\[
(x_3 - \bar{x})^2 = (49.5 - 68.8333)^2 \approx 372.1111
\]
\[
(x_4 - \bar{x})^2 = (69.5 - 68.8333)^2 \approx 0.4444
\]
\[
(x_5 - \bar{x})^2 = (89.5 - 68.8333)^2 \approx 426.1111
\]
\[
(x_6 - \bar{x})^2 = (109.5 - 68.8333)^2 \approx 1640.1111
\]
Now, calculating the variance:
\[
\sum f_i (x_i - \bar{x})^2 = (4 \cdot 3480.1111) + (3 \cdot 1537.1111) + (6 \cdot 372.1111) + (3 \cdot 0.4444) + (5 \cdot 426.1111) + (9 \cdot 1640.1111)
\]
\[
= 13920.4444 + 4611.3333 + 2232.6667 + 1.3332 + 2130.5555 + 14760.9999 \approx 33966.3333
\]
Thus, the variance is:
\[
s^2 = \frac{33966.3333}{30} \approx 1128.8778
\]
The standard deviation \( s \) is the square root of the variance:
\[
s = \sqrt{1128.8778} \approx 33.6
\]
The standard deviation for the grouped data is:
\[
\boxed{s = 35.6}
\]
Answered
