The table shows the length of 50 rods. Find: The frequency of the modal class. The actual mean, variance and standard deviation of the lengths. Take assumed mean A = 18.

From the table:

Length (cm)	No. of rods (frequency, $f$)
8-10	4
11-13	7
14-16	11
17-19	15
20-22	8
23-25	5

The highest frequency is 15, which corresponds to the class 17-19.
The frequency of the modal class is 15.

The formula for the actual mean using an assumed mean is:

$$\bar{x} = A + \frac{\sum fd}{\sum f}$$

Substitute the values:

$$\bar{x} = 18 + \frac{-57}{50}$$ $$\bar{x} = 18 - 1.14$$ $$\bar{x} = 16.86$$

The actual mean is $\boxed{\text{16.86 cm}}$.

The formula for variance using an assumed mean is:

$$\sigma^2 = \frac{\sum fd^2}{\sum f} - \left(\frac{\sum fd}{\sum f}\right)^2$$

Substitute the values:

$$\sigma^2 = \frac{927}{50} - \left(\frac{-57}{50}\right)^2$$ $$\sigma^2 = 18.54 - (-1.14)^2$$ $$\sigma^2 = 18.54 - 1.2996$$ $$\sigma^2 = 17.2404$$

The variance is $\boxed{\text{17.2404 cm}^2}$.

The formula for standard deviation is the square root of the variance:

$$\sigma = \sqrt{\sigma^2}$$

Substitute the variance:

$$\sigma = \sqrt{17.2404}$$ $$\sigma \approx 4.152156$$

Rounding to three decimal places:

$$\sigma \approx 4.152$$

The standard deviation is $\boxed{\text{4.152 cm}}$.