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: | Length in cm | 5-9 | 10-14 | 15-19 | 20-24 | 25-29 | 30-34 | 35-39 | 40-44 | 45-49 | |---|---|---|---|---|---|---|---|---|---| | Number of leaves | 11 | 13 | 17 | 20 | 25 | 16 | 12 | 4 | 2 | 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 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 = f_i = 120. | Length (cm) | Frequency (f_i) | Midpoint (x_i) | f_i x_i | x_i^2 | f_i x_i^2 | |:-------------:|:-------------------:|:----------------:|:---------:|:-------:|:----------:| | 5-9 | 11 | 7 | 77 | 49 | 539 | | 10-14 | 13 | 12 | 156 | 144 | 1872 | | 15-19 | 17 | 17 | 289 | 289 | 4913 | | 20-24 | 20 | 22 | 440 | 484 | 9680 | | 25-29 | 25 | 27 | 675 | 729 | 18225 | | 30-34 | 16 | 32 | 512 | 1024 | 16384 | | 35-39 | 12 | 37 | 444 | 1369 | 16428 | | 40-44 | 4 | 42 | 168 | 1764 | 7056 | | 45-49 | 2 | 47 | 94 | 2209 | 4418 | | Total | 120 | | 2855 | | 79515 | Step 2: Calculate the mean (x). The formula for the mean of grouped data is x = ( f_i x_i)/( f_i). x = (2855)/(120) x ≈ 23.79166... x ≈ 23.79 cm (to 2 decimal places) Step 3: Calculate the variance (^2). The formula for the variance of grouped data is ^2 = ( f_i x_i^2)/( f_i) - (( f_i x_i)/( f_i))^2. ^2 = (79515)/(120) - ((2855)/(120))^2 ^2 = 662.625 - (23.79166...)^2 ^2 = 662.625 - 566.03576... ^2 = 96.58923... ^2 ≈ 96.59 cm^2 (to 2 decimal places) The mean is 23.79 cm and the variance is 96.59 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). | Length (cm) | Frequency (f) | Upper Class Boundary | Cumulative Frequency (CF) | |:-------------:|:-----------------:|:--------------------:|:-------------------------:| | 5-9 | 11 | 9.5 | 11 | | 10-14 | 13 | 14.5 | 24 | | 15-19 | 17 | 19.5 | 41 | | 20-24 | 20 | 24.5 | 61 | | 25-29 | 25 | 29.5 | 86 | | 30-34 | 16 | 34.5 | 102 | | 35-39 | 12 | 39.5 | 114 | | 40-44 | 4 | 44.5 | 118 | | 45-49 | 2 | 49.5 | 120 | 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 = (N)/(4) = (120)/(4) = 30^th value. Position of Q_3 = (3N)/(4) = (3 × 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 + ((N)/(4) - CF_beforef_Q1) × 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 + ((30 - 24)/(17)) × 5 Q_1 = 14.5 + ((6)/(17)) × 5 Q_1 = 14.5 + (30)/(17) ≈ 14.5 + 1.76 Q_1 ≈ 16.26 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 + ((3N)/(4) - CF_beforef_Q3) × 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 + ((90 - 86)/(16)) × 5 Q_3 = 29.5 + ((4)/(16)) × 5 Q_3 = 29.5 + (1)/(4) × 5 = 29.5 + 1.25 Q_3 = 30.75 cm Step 4: Calculate the Interquartile Range (IQR). IQR = Q_3 - Q_1 IQR = 30.75 - 16.26 IQR = 14.49 cm The estimated interquartile range is 14.49 cm. Drop the next question.