The problem requires us to calculate the coefficient of variation and Pearson's measure of skewness for the given grouped data representing the diameter of 250 round tablets. The data is provided in a cumulative frequency table.

1. Problem Description

The problem requires us to calculate the coefficient of variation and Pearson's measure of skewness for the given grouped data representing the diameter of 250 round tablets. The data is provided in a cumulative frequency table.

2. Solution Steps

First, we need to convert the cumulative frequency table to a frequency table.

| Diameter (mm) | Cumulative Frequency | Frequency | Midpoint ($x_i$) | $f_i x_i$ | $f_i x_i^2$ |

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

| 7-11 | 10 | 10 | 9 | 90 | 810 |

| 12-16 | 30 | 20 | 14 | 280 | 3920 |

| 17-21 | 58 | 28 | 19 | 532 | 10108 |

| 22-26 | 94 | 36 | 24 | 864 | 20736 |

| 27-31 | 146 | 52 | 29 | 1508 | 43732 |

| 32-36 | 184 | 38 | 34 | 1292 | 43928 |

| 37-41 | 216 | 32 | 39 | 1248 | 48672 |

| 42-46 | 237 | 21 | 44 | 924 | 40656 |

| 47-51 | 250 | 13 | 49 | 637 | 31213 |

| | | $\sum f_i = 250$ | | $\sum f_i x_i = 7375$ | $\sum f_i x_i^2 = 243775$ |

The frequency for each class interval is calculated by subtracting the cumulative frequency of the previous class interval from the cumulative frequency of the current class interval. For example, the frequency for the interval 12-16 is 30 - 10 =

2

0. The midpoint ($x_i$) is calculated as the average of the lower and upper limits of the class interval. For example, the midpoint for the interval 7-11 is (7+11)/2 =

9.

Next, calculate the mean:

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

$\bar{x} = \frac{7375}{250} = 29.5$

Next, calculate the standard deviation:

$s = \sqrt{\frac{\sum f_i x_i^2 - \frac{(\sum f_i x_i)^2}{\sum f_i}}{\sum f_i - 1}}$

$s = \sqrt{\frac{243775 - \frac{(7375)^2}{250}}{250 - 1}}$

$s = \sqrt{\frac{243775 - \frac{54390625}{250}}{249}}$

$s = \sqrt{\frac{243775 - 217562.5}{249}}$

$s = \sqrt{\frac{26212.5}{249}}$

$s = \sqrt{105.271084337}$

$s \approx 10.26$

I. Coefficient of Variation (CV)

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

$CV = \frac{10.26}{29.5} \times 100$

$CV \approx 34.78 \%$

II. Pearson's Measure of Skewness

We need to find the mode. Since the class 27-31 has the highest frequency (52), it is the modal class. Assume the mode is the midpoint of this class, which is

2

9. Pearson's coefficient of skewness = $\frac{3(\text{Mean} - \text{Median})}{\text{Standard Deviation}}$ or $\frac{\text{Mean} - \text{Mode}}{\text{Standard Deviation}}$. Since median is not known we use the second formula. Also, approximation of mode to the midpoint value will affect the calculation.

Skewness = $\frac{\bar{x} - \text{Mode}}{s}$

Skewness $\approx \frac{29.5 - 29}{10.26}$

Skewness $\approx \frac{0.5}{10.26}$

Skewness $\approx 0.0487$

3. Final Answer

I. Coefficient of Variation: Approximately 34.78%

II. Pearson's Measure of Skewness: Approximately 0.0487