Given assumed mean A = 59.5.

You're on a roll — 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 =