The problem provides a dataset of 40 student's final examination marks. We are asked to: a. Create a grouped frequency distribution table with class intervals of 20-29, 30-39, and so on. b. Calculate the coefficient of variation (CV) and Pearson's coefficient of skewness for the given data.

1. Problem Description

The problem provides a dataset of 40 student's final examination marks. We are asked to:

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

b. Calculate the coefficient of variation (CV) and Pearson's coefficient of skewness for the given data.

2. Solution Steps

a. Grouped Frequency Distribution Table:

First, we define the class intervals:

20-29, 30-39, 40-49, 50-59, 60-69, 70-79, 80-89, 90-99

Next, we count the number of data points falling into each class:

* 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, 76, 78, 79, 79 = 8

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

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

The grouped frequency distribution table is as follows:

| Class Interval | Frequency (f) |

|----------------|----------------|

| 20-29 | 2 |

| 30-39 | 3 |

| 40-49 | 4 |

| 50-59 | 6 |

| 60-69 | 7 |

| 70-79 | 8 |

| 80-89 | 6 |

| 90-99 | 4 |

| Total | 40 |

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

First, let's calculate the mean ($x̄$) using the grouped data:

Midpoint of each class ($x_i$): 24.5, 34.5, 44.5, 54.5, 64.5, 74.5, 84.5, 94.5

$x̄ = \frac{Σ(f_i * x_i)}{Σf_i}$

$x̄ = \frac{(2*24.5) + (3*34.5) + (4*44.5) + (6*54.5) + (7*64.5) + (8*74.5) + (6*84.5) + (4*94.5)}{40}$

$x̄ = \frac{49 + 103.5 + 178 + 327 + 451.5 + 596 + 507 + 378}{40}$

$x̄ = \frac{2590}{40} = 64.75$

Next, calculate the standard deviation ($s$):

$s = \sqrt{\frac{Σ[f_i * (x_i - x̄)^2]}{n-1}}$

$s^2 = \frac{Σ[f_i * (x_i - x̄)^2]}{n-1}$

$s^2 = \frac{2*(24.5-64.75)^2 + 3*(34.5-64.75)^2 + 4*(44.5-64.75)^2 + 6*(54.5-64.75)^2 + 7*(64.5-64.75)^2 + 8*(74.5-64.75)^2 + 6*(84.5-64.75)^2 + 4*(94.5-64.75)^2}{40-1}$

$s^2 = \frac{2*1620.0625 + 3*915.0625 + 4*410.0625 + 6*105.0625 + 7*0.0625 + 8*95.0625 + 6*390.0625 + 4*885.0625}{39}$

$s^2 = \frac{3240.125 + 2745.1875 + 1640.25 + 630.375 + 0.4375 + 760.5 + 2340.375 + 3540.25}{39}$

$s^2 = \frac{15900}{39} \approx 407.6923$

$s = \sqrt{407.6923} \approx 20.1914$

Coefficient of Variation (CV):

$CV = \frac{s}{x̄} * 100$

$CV = \frac{20.1914}{64.75} * 100 \approx 31.18\%$

Pearson's Coefficient of Skewness (using mode):

First, estimate the mode:

The class with the highest frequency is 70-79, so the mode is likely within this interval. We can approximate the mode using the formula:

$Mode = L + \frac{f_m - f_1}{2f_m - f_1 - f_2} * w$

Where:

$L$ = lower boundary of the modal class (70)

$f_m$ = frequency of the modal class (8)

$f_1$ = frequency of the class before the modal class (7)

$f_2$ = frequency of the class after the modal class (6)

$w$ = class width (10)

$Mode = 70 + \frac{8 - 7}{2*8 - 7 - 6} * 10$

$Mode = 70 + \frac{1}{16 - 13} * 10$

$Mode = 70 + \frac{1}{3} * 10 = 70 + 3.33 = 73.33$

Pearson's coefficient of skewness:

$Skewness = \frac{3 * (Mean - Mode)}{Standard Deviation}$

$Skewness = \frac{3 * (64.75 - 73.33)}{20.1914} = \frac{3 * (-8.58)}{20.1914} = \frac{-25.74}{20.1914} \approx -1.275$

3. Final Answer

a. Grouped Frequency Distribution Table:

| Class Interval | Frequency (f) |

|----------------|----------------|

| 20-29 | 2 |

| 30-39 | 3 |

| 40-49 | 4 |

| 50-59 | 6 |

| 60-69 | 7 |

| 70-79 | 8 |

| 80-89 | 6 |

| 90-99 | 4 |

b. Coefficient of Variation (CV): approximately 31.18%

Pearson's Coefficient of Skewness: approximately -1.275