### Theory:

Let us look at an example to find standard deviation of a grouped data by mean method.
Example:
Case 1: Discrete data

Find the standard deviation of the following data by mean method.

 $$x$$ $$1$$ $$2$$ $$3$$ $$4$$ $$f$$ $$4$$ $$8$$ $$12$$ $$16$$

Explanation:
We know that the formula for finding the arithmetic mean using direct method is given by:

$$\overline x = \frac{\sum_{i = 1}^{n} f_{i} x_{i}}{\sum_{i = 1}^{n} f_{i}}$$

Where $$x_{i}$$ is the midpoint of the class interval and $$f_{i}$$ is the frequency.
Let us form a frequency distribution table.

 $$x_{i}$$ $$f_{i}$$ $$f_{i} x_{i}$$ $$1$$ $$4$$ $$4$$ $$2$$ $$8$$ $$16$$ $$3$$ $$12$$ $$36$$ $$4$$ $$16$$ $$64$$ $$\sum_{i = 1}^{4} f_{i} = 40$$ $$\sum_{i = 1}^{4} f_{i} x_{i} = 120$$

Substituting the known values in the above formula, we get:

Mean $$\overline x = \frac{120}{40}$$ $$=$$ $$3$$

Thus, the meanof the given data is

Now, let us find the standard deviation for the given data.

 $$x_{i}$$ $$f_{i}$$ $$d_{i} = x_{i} - \overline x$$ $$=$$ $$x_{i} - 3$$ $$d_{i}^{2}$$ $$f_{i}d_{i}^{2}$$ $$1$$ $$4$$ $$-2$$ $$4$$ $$16$$ $$2$$ $$8$$ $$-1$$ $$1$$ $$8$$ $$3$$ $$12$$ $$0$$ $$0$$ $$0$$ $$4$$ $$16$$ $$1$$ $$1$$ $$16$$ $$\sum_{i = 1}^{4} f_{i} = 40$$ $$\sum_{i = 1}^{4} f_{i} d_{i}^{2} = 40$$
The formula to calculate the standard deviation by mean method is given by:

$$\sigma$$ $$=$$ $$\sqrt{\frac{\sum f_{i} d_{i}^{2}}{N}}$$ where $$N = \sum_{i = 1}^{n} f_{i}$$
Substitute the required values in the above formula.

$$\sigma$$ $$=$$ $$\sqrt{\frac{40}{40}}$$

$$=$$ $$\sqrt{1}$$

$$=$$ $$1$$

Therefore, the standard deviation of the given data is $$1$$.

Case 2: Continuous data

Find the standard deviation of the following data by mean method.

 $$x$$ $$0 - 2$$ $$2 - 4$$ $$4 - 6$$ $$6 - 8$$ $$f$$ $$4$$ $$8$$ $$12$$ $$16$$

Explanation:
We know that the formula for finding the arithmetic mean using direct method is given by:

$$\overline x = \frac{\sum_{i = 1}^{n} f_{i} x_{i}}{\sum_{i = 1}^{n} f_{i}}$$

Where $$x_{i}$$ is the midpoint of the class interval and $$f_{i}$$ is the frequency.
Let us form a frequency distribution table.

 Class interval Frequency($$f_{i}$$) Midpoint($$x_{i}$$) $$f_{i} x_{i}$$ $$0 - 2$$ $$4$$ $$1$$ $$4$$ $$2 - 4$$ $$8$$ $$3$$ $$24$$ $$4 - 6$$ $$12$$ $$5$$ $$60$$ $$6 - 8$$ $$16$$ $$7$$ $$112$$ $$\sum_{i = 1}^{4} f_{i} = 40$$ $$\sum_{i = 1}^{4} f_{i} x_{i} = 200$$

Substituting the known values in the above formula, we get:

Mean $$\overline x = \frac{200}{40}$$ $$=$$ $$5$$

Thus, the mean of the given data is $$5$$

Now, let us find the standard deviation for the given data.

 $$x_{i}$$ $$f_{i}$$ $$d_{i} = x_{i} - \overline x$$ $$=$$ $$x_{i} - 5$$ $$d_{i}^{2}$$ $$f_{i}d_{i}^{2}$$ $$1$$ $$4$$ $$-4$$ $$16$$ $$64$$ $$3$$ $$8$$ $$-2$$ $$4$$ $$32$$ $$5$$ (12\) $$0$$ $$0$$ $$0$$ $$7$$ $$16$$ $$2$$ $$4$$ $$64$$ $$\sum_{i = 1}^{4} f_{i} = 40$$ $$\sum_{i = 1}^{4} f_{i} d_{i}^{2} = 160$$
The formula to calculate the standard deviation by mean method is given by:

$$\sigma$$ $$=$$ $$\sqrt{\frac{\sum f_{i} d_{i}^{2}}{N}}$$ where $$N = \sum_{i = 1}^{n} f_{i}$$
Substitute the required values in the above formula.

$$\sigma$$ $$=$$ $$\sqrt{\frac{160}{40}}$$

$$=$$ $$\sqrt{4}$$

$$=$$ $$2$$

Therefore, the standard deviation of the given data is $$2$$.