Given assumed mean A = 59.5.

To solve this problem, we first need to set up a frequency distribution table with additional columns for calculations using the assumed mean method.
Given assumed mean $A = 59.5$.
The class width $c = 10$ (e.g., $34.5 - 24.5 = 10$).

Step 1: Create a table for calculations.

Length (Class)	Frequency ($f$)	Midpoint ($x$)	$d = x - A$	$u = \frac{d}{c}$	$fu$	$fu^2$
25-34	9	29.5	-30	-3	-27	81
35-44	12	39.5	-20	-2	-24	48
45-54	55	49.5	-10	-1	-55	55
55-64	68	59.5	0	0	0	0
65-74	32	69.5	10	1	32	32
75-84	18	79.5	20	2	36	72
85-94	6	89.5	30	3	18	54
Total	$\sum f = 200$				$\sum fu = -20$	$\sum fu^2 = 342$

a) (i) The mean length
The formula for the mean using the assumed mean method with coding is:

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

Substitute the values from the table:

$$\bar{x} = 59.5 + 10 \times \frac{-20}{200}$$ $$\bar{x} = 59.5 + 10 \times (-0.1)$$ $$\bar{x} = 59.5 - 1$$ $$\bar{x} = 58.5$$

The mean length is $\boxed{\text{58.5 mm}}$.

a) (ii) The variance
The formula for the variance using the assumed mean method with coding is:

$$\sigma^2 = c^2 \left[ \frac{\sum fu^2}{\sum f} - \left(\frac{\sum fu}{\sum f}\right)^2 \right]$$

Substitute the values from the table:

$$\sigma^2 = 10^2 \left[ \frac{342}{200} - \left(\frac{-20}{200}\right)^2 \right]$$ $$\sigma^2 = 100 \left[ 1.71 - (-0.1)^2 \right]$$ $$\sigma^2 = 100 \left[ 1.71 - 0.01 \right]$$ $$\sigma^2 = 100 \left[ 1.70 \right]$$ $$\sigma^2 = 170$$

The variance is $\boxed{\text{170 mm}^2}$.

a) (iii) The standard deviation
The standard deviation is the square root of the variance:

$$\sigma = \sqrt{\sigma^2}$$ $$\sigma = \sqrt{170}$$ $$\sigma \approx 13.0384$$

Rounding to two decimal places:

$$\sigma \approx 13.04$$

The standard deviation is $\boxed{\text{13.04 mm}}$.

b) Find the percentage of leaves that has a length of 69.5 mm and above.
First, determine the class boundaries:
25-34 becomes 24.5 - 34.5
35-44 becomes 34.5 - 44.5
45-54 becomes 44.5 - 54.5
55-64 becomes 54.5 - 64.5
65-74 becomes 64.5 - 74.5
75-84 becomes 74.5 - 84.5
85-94 becomes 84.5 - 94.5

We need to find the number of leaves with a length of 69.5 mm and above.

• Leaves in the class 75-84 (74.5 - 84.5): 18 leaves
• Leaves in the class 85-94 (84.5 - 94.5): 6 leaves
• For the class 65-74 (64.5 - 74.5), which has 32 leaves, we need the portion from 69.5 mm to 74.5 mm. The class width is $74.5 - 64.5 = 10$ mm. The portion of the interval we are interested in is $74.5 - 69.5 =