The given data points are:

$$\bar{x} = \frac{2755}{50} = 55.1$$

The mean is $\boxed{\text{55.1}}$.

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The median position is $\frac{N}{2} = \frac{50}{2} = 25$.
Looking at the cumulative frequency column, the 25th value falls within the class where the cumulative frequency first exceeds 25. This is the 50-59 class (cf = 29).
The median class is $\boxed{\text{50-59}}$.

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The modal class is the class with the highest frequency. From the table, the highest frequency is 9, which corresponds to the 50-59 class.
The modal class is $\boxed{\text{50-59}}$.

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First, we calculate $|x - \bar{x}|$ and $f|x - \bar{x}|$ for each class. $\bar{x} = 55.1$.

| Class Mark (x) | $|x - \bar{x}|$ | Frequency (f) | $f|x - \bar{x}|$ | | :------------- | :-------------- | :------------ | :--------------- | | 14.5 | 40.6 | 1 | 40.6 | | 24.5 | 30.6 | 5 | 153.0 | | 34.5 | 20.6 | 7 | 144.2 | | 44.5 | 10.6 | 7 | 74.2 | | 54.5 | 0.6 | 9 | 5.4 | | 64.5 | 9.4 | 8 | 75.2 | | 74.5 | 19.4 | 6 | 116.4 | | 84.5 | 29.4 | 5 | 147.0 | | 94.5 | 39.4 | 2 | 78.8 | | Total | | $\sum f = 50$ | $\sum f|x - \bar{x}| = 834.8$ |

$$MD = \frac{834.8}{50} = 16.696 \approx 16.70$$

The mean deviation is $\boxed{\text{16.70}}$.

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First, find the first quartile ($Q_1$) and third quartile ($Q_3$).
$Q_1$ position: $\frac{N}{4} = \frac{50}{4} = 12.5^{th}$ value. This falls in the 30-39 class.
$Q_3$ position: $\frac{3N}{4} = \frac{3 \times 50}{4} = 37.5^{th}$ value. This falls in the 70-79 class.

The formula for a quartile is $Q_k = L + \left(\frac{\frac{kN}{4} - cf_b}{f_m}\right)c$, where L is the lower class boundary, $cf_b$ is the cumulative frequency before the quartile class, $f_m$ is the frequency of the quartile class, and c is the class width.

For $Q_1$:
$L = 29.5$, $cf_b = 6$, $f_m = 7$, $c = 10$.

$$Q_1 = 29.5 + \left(\frac{12.5 - 6}{7}\right) \times 10 = 29.5 + \left(\frac{6.5}{7}\right) \times 10 \approx 29.5 + 9.2857 = 38.7857$$

For $Q_3$:
$L = 69.5$, $cf_b = 37$, $f_m = 6$, $c = 10$.

$$Q_3 = 69.5 + \left(\frac{37.5 - 37}{6}\right) \times 10 = 69.5 + \left(\frac{0.5}{6}\right) \times 10 \approx 69.5 + 0.8333 = 70.3333$$

Now, calculate SIQR:

$$SIQR = \frac{Q_3 - Q_1}{2} = \frac{70.3333 - 38.7857}{2} = \frac{31.5476}{2} \approx 15.7738$$

The semi-interquartile range is $\boxed{\text{15.77}}$.

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First, we calculate $(x - \bar{x})^2$ and $f(x - \bar{x})^2$ for each class. $\bar{x} = 55.1$.

Class Mark (x)	$(x - \bar{x})$	$(x - \bar{x})^2$	Frequency (f)	$f(x - \bar{x})^2$
14.5	-40.6	1648.36	1	1648.36
24.5	-30.6	936.36	5	4681.80
34.5	-20.6	424.36	7	2970.52
44.5	-10.6	112.36	7	786.52
54.5	-0.6	0.36	9	3.24
64.5	9.4	88.36	8	706.88
74.5	19.4	376.36	6	2258.16
84.5	29.4	864.36	5	4321.80
94.5	39.4	1552.36	2	3104.72
Total			$\sum f = 50$	$\sum f(x - \bar{x})^2 = 20481.00$

$$s = \sqrt{\frac{20481.00}{50}} = \sqrt{409.62} \approx 20.2391$$

The standard deviation is $\boxed{\text{20.24}}$.