The problem presents a grouped frequency distribution table and asks us to compute several statistical measures. Specifically, we need to calculate the mean, variance, and standard deviation, and the coefficient of variation for the given data. Furthermore, we must determine the third quartile ($Q_3$), the seventh decile ($D_7$), and the seventy-fifth percentile ($P_{75}$). Finally, we need to assess the skewness of the data using Karl Pearson's coefficient of skewness formula.

1. Problem Description

The problem presents a grouped frequency distribution table and asks us to compute several statistical measures. Specifically, we need to calculate the mean, variance, and standard deviation, and the coefficient of variation for the given data. Furthermore, we must determine the third quartile ($Q_3$), the seventh decile ($D_7$), and the seventy-fifth percentile ($P_{75}$). Finally, we need to assess the skewness of the data using Karl Pearson's coefficient of skewness formula.

2. Solution Steps

a. Compute the mean, variance, and standard deviation of the data set.

First, we calculate the midpoint of each class interval.

The midpoints are: 15, 25, 35, 45, 55, 65, 75,

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Next, we multiply each midpoint by its corresponding frequency to get $f_ix_i$.

$15 \times 3 = 45$

$25 \times 7 = 175$

$35 \times 14 = 490$

$45 \times 20 = 900$

$55 \times 16 = 880$

$65 \times 10 = 650$

$75 \times 6 = 450$

$85 \times 4 = 340$

The sum of frequencies, $N = \sum f_i = 3 + 7 + 14 + 20 + 16 + 10 + 6 + 4 = 80$

The sum of $f_ix_i$, $\sum f_ix_i = 45 + 175 + 490 + 900 + 880 + 650 + 450 + 340 = 3930$

Mean, $\bar{x} = \frac{\sum f_ix_i}{N} = \frac{3930}{80} = 49.125$

Next, we calculate $(x_i - \bar{x})^2$.

$(15 - 49.125)^2 = 1164.265625$

$(25 - 49.125)^2 = 582.015625$

$(35 - 49.125)^2 = 200.640625$

$(45 - 49.125)^2 = 17.015625$

$(55 - 49.125)^2 = 34.515625$

$(65 - 49.125)^2 = 252.015625$

$(75 - 49.125)^2 = 669.265625$

$(85 - 49.125)^2 = 1287.015625$

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

$3 \times 1164.265625 = 3492.796875$

$7 \times 582.015625 = 4074.109375$

$14 \times 200.640625 = 2808.96875$

$20 \times 17.015625 = 340.3125$

$16 \times 34.515625 = 552.25$

$10 \times 252.015625 = 2520.15625$

$6 \times 669.265625 = 4015.59375$

$4 \times 1287.015625 = 5148.0625$

The sum of $f_i(x_i - \bar{x})^2$, $\sum f_i(x_i - \bar{x})^2 = 3492.796875 + 4074.109375 + 2808.96875 + 340.3125 + 552.25 + 2520.15625 + 4015.59375 + 5148.0625 = 22952.25$

Variance, $s^2 = \frac{\sum f_i(x_i - \bar{x})^2}{N - 1} = \frac{22952.25}{80 - 1} = \frac{22952.25}{79} = 290.5348101$

Standard deviation, $s = \sqrt{s^2} = \sqrt{290.5348101} = 17.04508174$

Coefficient of variation, $CV = \frac{s}{\bar{x}} \times 100 = \frac{17.04508174}{49.125} \times 100 = 34.70\%$

b. Determine the third quartile ($Q_3$), the seventh decile ($D_7$), and the seventy-fifth percentile ($P_{75}$) from the cumulative frequency distribution.

First, we construct the cumulative frequency distribution.

Class Interval | Frequency | Cumulative Frequency

10-20 | 3 | 3

20-30 | 7 | 10

30-40 | 14 | 24

40-50 | 20 | 44

50-60 | 16 | 60

60-70 | 10 | 70

70-80 | 6 | 76

80-90 | 4 | 80

To find the third quartile ($Q_3$), we need to find the value that corresponds to the 75th percentile.

$Q_3 = \frac{3N}{4} = \frac{3 \times 80}{4} = 60$

$Q_3$ lies in the class interval 50-

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0. $Q_3 = L + \frac{\frac{3N}{4} - cf}{f} \times h$

$Q_3 = 50 + \frac{60 - 44}{16} \times 10 = 50 + \frac{16}{16} \times 10 = 50 + 10 = 60$

To find the seventh decile ($D_7$), we need to find the value that corresponds to the 70th percentile.

$D_7 = \frac{7N}{10} = \frac{7 \times 80}{10} = 56$

$D_7$ lies in the class interval 50-

6

0. $D_7 = L + \frac{\frac{7N}{10} - cf}{f} \times h$

$D_7 = 50 + \frac{56 - 44}{16} \times 10 = 50 + \frac{12}{16} \times 10 = 50 + 7.5 = 57.5$

To find the seventy-fifth percentile ($P_{75}$), we need to find the value that corresponds to the 75th percentile.

$P_{75} = \frac{75N}{100} = \frac{75 \times 80}{100} = 60$

$P_{75}$ lies in the class interval 50-

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0. $P_{75} = L + \frac{\frac{75N}{100} - cf}{f} \times h$

$P_{75} = 50 + \frac{60 - 44}{16} \times 10 = 50 + \frac{16}{16} \times 10 = 50 + 10 = 60$

c. Assess the skewness of the data using Karl Pearson's coefficient of skewness formula.

Mode is the value with the highest frequency. From the original frequency distribution, we see that 40-50 has the highest frequency of

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0. Therefore, the mode class is 40-

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0. Mode $= L + \frac{f_m - f_1}{2f_m - f_1 - f_2} \times h = 40 + \frac{20 - 14}{2(20) - 14 - 16} \times 10 = 40 + \frac{6}{40 - 30} \times 10 = 40 + \frac{6}{10} \times 10 = 40 + 6 = 46$

Karl Pearson's coefficient of skewness $= \frac{\text{Mean} - \text{Mode}}{\text{Standard Deviation}} = \frac{49.125 - 46}{17.045} = \frac{3.125}{17.045} = 0.183$

3. Final Answer

a. Mean = 49.125, Variance = 290.535, Standard deviation = 17.045, Coefficient of variation = 34.70%

b. $Q_3$ = 60, $D_7$ = 57.5, $P_{75}$ = 60

c. Karl Pearson's coefficient of skewness = 0.183