The problem provides a dataset of final examination marks of 40 students. The task is to: a. Prepare a grouped frequency distribution table using class intervals of 20-29, 30-39, and so on. b. Compute the Coefficient of Variation (CV) and the skewness using Pearson's measure.

1. Problem Description

The problem provides a dataset of final examination marks of 40 students. The task is to:

a. Prepare a grouped frequency distribution table using class intervals of 20-29, 30-39, and so on.

b. Compute the Coefficient of Variation (CV) and the skewness using Pearson's measure.

2. Solution Steps

a. Grouped Frequency Distribution:

First, create the class intervals. The data ranges from 24 to

9

5. Thus the class intervals would be:

20-29

30-39

40-49

50-59

60-69

70-79

80-89

90-99

Next, count the number of data points falling within each class interval (frequency):

20-29: 24, 26 = 2

30-39: 31, 35, 35 = 3

40-49: 40, 42, 40, 43 = 4

50-59: 50, 53, 53, 56, 53, 58 = 6

60-69: 60, 61, 62, 64, 64, 67, 69 = 7

70-79: 72, 73, 75, 75, 72, 76, 78, 79, 79 = 9

80-89: 80, 84, 80, 85, 87, 87 = 6

90-99: 90, 90, 92, 95 = 4

b. Coefficient of Variation (CV) and Skewness using Pearson's measure:

(i) Calculate the mean ($\bar{x}$):

To find the mean of the grouped data, we need to find the midpoint of each class interval ($x_i$), multiply it by the frequency ($f_i$) for that interval, sum the results, and then divide by the total number of observations (n=40).

Midpoints ($x_i$): 24.5, 34.5, 44.5, 54.5, 64.5, 74.5, 84.5, 94.5

$\bar{x} = \frac{\sum{f_i x_i}}{n}$

$\sum{f_i x_i} = (2*24.5) + (3*34.5) + (4*44.5) + (6*54.5) + (7*64.5) + (9*74.5) + (6*84.5) + (4*94.5)$

$\sum{f_i x_i} = 49 + 103.5 + 178 + 327 + 451.5 + 670.5 + 507 + 378 = 2664.5$

$\bar{x} = \frac{2664.5}{40} = 66.6125$

(ii) Calculate the standard deviation (s):

$s = \sqrt{\frac{\sum{f_i (x_i - \bar{x})^2}}{n-1}}$ (Using n-1 since we are dealing with a sample)

First calculate $(x_i - \bar{x})^2$ for each interval:

$(24.5-66.6125)^2 = 1777.82$

$(34.5-66.6125)^2 = 1031.34$

$(44.5-66.6125)^2 = 488.91$

$(54.5-66.6125)^2 = 146.72$

$(64.5-66.6125)^2 = 4.46$

$(74.5-66.6125)^2 = 62.22$

$(84.5-66.6125)^2 = 320.05$

$(94.5-66.6125)^2 = 777.79$

Next, calculate $f_i*(x_i - \bar{x})^2$:

$2*1777.82 = 3555.64$

$3*1031.34 = 3094.02$

$4*488.91 = 1955.64$

$6*146.72 = 880.32$

$7*4.46 = 31.22$

$9*62.22 = 559.98$

$6*320.05 = 1920.3$

$4*777.79 = 3111.16$

$\sum{f_i (x_i - \bar{x})^2} = 3555.64 + 3094.02 + 1955.64 + 880.32 + 31.22 + 559.98 + 1920.3 + 3111.16 = 15108.28$

$s = \sqrt{\frac{15108.28}{40-1}} = \sqrt{\frac{15108.28}{39}} = \sqrt{387.39} = 19.68$

(iii) Calculate the Coefficient of Variation (CV):

$CV = \frac{s}{\bar{x}} * 100$

$CV = \frac{19.68}{66.6125} * 100 = 0.2954 * 100 = 29.54\%$

(iv) Estimate the Mode:

The mode lies within the class with the highest frequency, which is 70-79 (frequency = 9).

Since the mean is 66.6125, we know that the mode will likely be in the 70-79 group. We can approximate the mode using the formula:

$Mode = L + \frac{f_m - f_1}{2f_m - f_1 - f_2} * w$, where L is the lower limit of the modal class (70), $f_m$ is the frequency of the modal class (9), $f_1$ is the frequency of the class before the modal class (7), $f_2$ is the frequency of the class after the modal class (6), and w is the class width (10).

$Mode = 70 + \frac{9-7}{2*9 - 7 - 6} * 10 = 70 + \frac{2}{18-13} * 10 = 70 + \frac{2}{5} * 10 = 70 + 4 = 74$

(v) Calculate Pearson's coefficient of skewness:

Pearson's first coefficient of skewness:

$Skewness = \frac{Mean - Mode}{Standard Deviation}$

$Skewness = \frac{66.6125 - 74}{19.68} = \frac{-7.3875}{19.68} = -0.375$

Since the value is negative, the distribution is negatively skewed.

3. Final Answer

a. Grouped Frequency Distribution:

20-29: 2

30-39: 3

40-49: 4

50-59: 6

60-69: 7

70-79: 9

80-89: 6

90-99: 4

b. Coefficient of Variation (CV): 29.54%

Skewness (Pearson's Measure): -0.375

The distribution is negatively skewed.