Answer question 2 in English

here's the solution for Question 2.

Question 2: The lengths of some randomly selected leaves from a certain fruit tree were measured and recorded as follows:

Note: The sum of the frequencies in the table is $11+13+17+20+25+16+12+4+2 = 120$. This contradicts the problem statement which mentions "110 leaves". We will proceed with the sum of frequencies from the table, $N=120$, for all calculations.

a) State the modal class of this distribution of lengths.

Step 1: Identify the class with the highest frequency. The highest frequency in the table is 25. Step 2: Determine the class interval corresponding to this frequency. The class interval corresponding to a frequency of 25 is 25-29 cm.

The modal class is $\boxed{\text{25-29 cm}}$.

b) Calculate the mean and variance of the lengths of the leaves, to 2 decimal places.

Step 1: Create a table to calculate midpoints ($x_i$), $f_i x_i$, and $f_i x_i^2$. The total number of leaves is $N = \sum f_i = 120$.

Step 2: Calculate the mean ($\bar{x}$). The formula for the mean of grouped data is $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$. $\bar{x} = \frac{2855}{120}$ $\bar{x} \approx 23.79166...$ $\bar{x} \approx 23.79 \text{ cm (to 2 decimal places)}$

Step 3: Calculate the variance ($\sigma^2$). The formula for the variance of grouped data is $\sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - \left(\frac{\sum f_i x_i}{\sum f_i}\right)^2$. $\sigma^2 = \frac{79515}{120} - \left(\frac{2855}{120}\right)^2$ $\sigma^2 = 662.625 - (23.79166...)^2$ $\sigma^2 = 662.625 - 566.03576...$ $\sigma^2 = 96.58923...$ $\sigma^2 \approx 96.59 \text{ cm}^2 \text{ (to 2 decimal places)}$

The mean is $\boxed{23.79 \text{ cm}}$ and the variance is $\boxed{96.59 \text{ cm}^2}$.

c) Construct the cumulative frequency table and draw the cumulative frequency curve for these data.

Step 1: Construct the cumulative frequency table. We use the upper class boundaries for plotting the cumulative frequency curve. The class boundaries are adjusted to be continuous (e.g., 5-9 becomes 4.5-9.5).

Step 2: Describe how to draw the cumulative frequency curve (ogive).

• Plot the points (Upper Class Boundary, Cumulative Frequency) from the table.
• Start the curve at (4.5, 0) on the x-axis (lower boundary of the first class with 0 cumulative frequency).
• Plot the points: (9.5, 11), (14.5, 24), (19.5, 41), (24.5, 61), (29.5, 86), (34.5, 102), (39.5, 114), (44.5, 118), (49.5, 120).
• Join these points with a smooth curve. The x-axis represents the length in cm, and the y-axis represents the cumulative frequency.

d) Use your curve to estimate the interquartile range.

Step 1: Determine the positions of the first quartile ($Q_1$) and third quartile ($Q_3$). Total number of leaves $N = 120$. Position of $Q_1 = \frac{N}{4} = \frac{120}{4} = 30^{th}$ value. Position of $Q_3 = \frac{3N}{4} = \frac{3 \times 120}{4} = 90^{th}$ value.

Step 2: Estimate $Q_1$ using linear interpolation (as a curve cannot be drawn here). $Q_1$ is the $30^{th}$ value. From the cumulative frequency table, the $30^{th}$ value falls in the 15-19 cm class (CF 24 at 14.5 cm, CF 41 at 19.5 cm). Using the formula $Q_1 = L + \left(\frac{\frac{N}{4} - CF_{before}}{f_{Q1}}\right) \times w$: $L = 14.5$ (lower class boundary of the 15-19 class) $CF_{before} = 24$ (cumulative frequency before the 15-19 class) $f_{Q1} = 17$ (frequency of the 15-19 class) $w = 5$ (class width) $Q_1 = 14.5 + \left(\frac{30 - 24}{17}\right) \times 5$ $Q_1 = 14.5 + \left(\frac{6}{17}\right) \times 5$ $Q_1 = 14.5 + \frac{30}{17} \approx 14.5 + 1.76$ $Q_1 \approx 16.26 \text{ cm}$

Step 3: Estimate $Q_3$ using linear interpolation. $Q_3$ is the $90^{th}$ value. From the cumulative frequency table, the $90^{th}$ value falls in the 30-34 cm class (CF 86 at 29.5 cm, CF 102 at 34.5 cm). Using the formula $Q_3 = L + \left(\frac{\frac{3N}{4} - CF_{before}}{f_{Q3}}\right) \times w$: $L = 29.5$ (lower class boundary of the 30-34 class) $CF_{before} = 86$ (cumulative frequency before the 30-34 class) $f_{Q3} = 16$ (frequency of the 30-34 class) $w = 5$ (class width) $Q_3 = 29.5 + \left(\frac{90 - 86}{16}\right) \times 5$ $Q_3 = 29.5 + \left(\frac{4}{16}\right) \times 5$ $Q_3 = 29.5 + \frac{1}{4} \times 5 = 29.5 + 1.25$ $Q_3 = 30.75 \text{ cm}$

Step 4: Calculate the Interquartile Range (IQR). $IQR = Q_3 - Q_1$ $IQR = 30.75 - 16.26$ $IQR = 14.49 \text{ cm}$

The estimated interquartile range is $\boxed{14.49 \text{ cm}}$.