The table below gives the distribution of the number of letters per word in the first fifty words of an essay. i) If the average number of letter per word is 5, find the value of x and y. ii) Calculate, correct to two decimal places, the standard deviation of the distribution.

The table shows the number of letters ($x$) and the number of words ($f$).
The total number of words is 50, so $\sum f = 50$.
The average number of letters per word (mean) is 5, so $\bar{x} = 5$.

From the sum of frequencies:

$$1 + 3 + x + 10 + 12 + y + 6 + 4 + 1 = 50$$ $$37 + x + y = 50$$ $$x + y = 13 \quad (\text{Equation 1})$$

From the mean formula, $\bar{x} = \frac{\sum fx}{\sum f}$:

$$\sum fx = (1 \times 1) + (2 \times 3) + (3 \times x) + (4 \times 10) + (5 \times 12) + (6 \times y) + (7 \times 6) + (8 \times 4) + (9 \times 1)$$ $$\sum fx = 1 + 6 + 3x + 40 + 60 + 6y + 42 + 32 + 9$$ $$\sum fx = 190 + 3x + 6y$$

Since $\bar{x} = 5$ and $\sum f = 50$:

$$5 = \frac{190 + 3x + 6y}{50}$$ $$250 = 190 + 3x + 6y$$ $$60 = 3x + 6y$$

Divide by 3:

$$20 = x + 2y \quad (\text{Equation 2})$$

We have:

1. $x + y = 13$
2. $x + 2y = 20$

Subtract Equation 1 from Equation 2:

$$(x + 2y) - (x + y) = 20 - 13$$ $$y = 7$$

Substitute $y=7$ into Equation 1:

$$x + 7 = 13$$ $$x = 13 - 7$$ $$x = 6$$

The values are $\boxed{x=6, y=7}$.

Now that we know $x=6$ and $y=7$, the distribution is:

Number of letters (x)	Number of words (f)	fx	$x^2$	$fx^2$
1	1	1	1	1
2	3	6	4	12
3	6	18	9	54
4	10	40	16	160
5	12	60	25	300
6	7	42	36	252
7	6	42	49	294
8	4	32	64	256
9	1	9	81	81
Total	50	250		1410

From the table, $\sum f = 50$, $\sum fx = 250$, and $\sum fx^2 = 1410$.

The formula for variance is:

$$\sigma^2 = \frac{\sum fx^2}{\sum f} - \bar{x}^2$$

We know $\bar{x} = 5$.

$$\sigma^2 = \frac{1410}{50} - (5)^2$$ $$\sigma^2 = 28.2 - 25$$ $$\sigma^2 = 3.2$$

The standard deviation is the square root of the variance:

$$\sigma = \sqrt{3.2}$$ $$\sigma \approx 1.78885...$$

Rounding to two decimal places:

$$\sigma \approx 1.79$$

The standard deviation is $\boxed{\text{1.79}}$.