Questions: Find the standard deviation for the grouped data. College Units Frequency 0-19 4 20-39 3 40-59 6 60-79 3 80-99 5 100-119 9 s= (Do not round until the final answer. Then round to the nearest tenth as needed.)

Solution Steps

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:

$$s = \sqrt{1128.8778} \approx 33.6$$

Final Answer