The problem presents a statistical dataset of 180 authors categorized by the number of manuals they have written. The table provides the number of authors $n_i$ who have written $x_i$ manuals. The problem asks to identify the population and statistical unit, the observed character and its nature, the mode, median, quartiles Q1 and Q3, and to calculate the arithmetic, geometric, harmonic, and quadratic means, as well as the variance, standard deviation, and coefficient of variation for this series.

1. Problem Description

The problem presents a statistical dataset of 180 authors categorized by the number of manuals they have written. The table provides the number of authors $n_i$ who have written $x_i$ manuals. The problem asks to identify the population and statistical unit, the observed character and its nature, the mode, median, quartiles Q1 and Q3, and to calculate the arithmetic, geometric, harmonic, and quadratic means, as well as the variance, standard deviation, and coefficient of variation for this series.

2. Solution Steps

1. Population and Statistical Unit:

* The population studied is the set of 180 authors in the library.

* The statistical unit is each individual author.

2. Observed Character and its Nature:

* The observed character is the number of manuals written by each author.

* The nature of the character is quantitative discrete.

3. Mode, Median, and Quartiles:

* Mode: The mode is the value of $x_i$ with the highest frequency $n_i$. Here, the highest frequency is 52, corresponding to $x_i = 1$. So, the mode is

1.

* Median: The median is the middle value of the dataset. Since there are 180 authors, the median corresponds to the average of the 90th and 91st authors when the data is arranged in ascending order.

We can calculate the cumulative frequencies:

* 1 manual: 52 authors

* 2 manuals: 52 + 36 = 88 authors

* 3 manuals: 88 + 27 = 115 authors

Since the 90th and 91st authors fall within the category of 3 manuals, the median is

3.

* Quartile Q1: Q1 represents the 25th percentile. It is the value below which 25% of the data falls. $0.25 \times 180 = 45$. We need to find the value corresponding to the 45th author. From the cumulative frequencies, the 45th author falls within the category of 1 manual. Thus, Q1 =

1.

* Quartile Q3: Q3 represents the 75th percentile. It is the value below which 75% of the data falls. $0.75 \times 180 = 135$. We need to find the value corresponding to the 135th author.

* 1 manual: 52 authors

* 2 manuals: 52 + 36 = 88 authors

* 3 manuals: 88 + 27 = 115 authors

* 4 manuals: 115 + 45 = 160 authors

The 135th author falls within the category of 4 manuals. Thus, Q3 =

4.

4. Means, Variance, Standard Deviation, and Coefficient of Variation:

First, calculate the following:

$\sum n_i = 52 + 36 + 27 + 45 + 9 + 2 + 9 = 180$

$\sum x_i n_i = (1 \times 52) + (2 \times 36) + (3 \times 27) + (4 \times 45) + (5 \times 9) + (6 \times 2) + (7 \times 9) = 52 + 72 + 81 + 180 + 45 + 12 + 63 = 505$

* Arithmetic Mean ($\bar{x}$):

$\bar{x} = \frac{\sum x_i n_i}{\sum n_i} = \frac{505}{180} \approx 2.8056$

* Geometric Mean ($G$):

$G = \sqrt[N]{\prod x_i^{n_i}}$, where $N = \sum n_i$

$G = \sqrt[180]{1^{52} \times 2^{36} \times 3^{27} \times 4^{45} \times 5^{9} \times 6^{2} \times 7^{9}}$

$\ln(G) = \frac{1}{180}(52\ln(1) + 36\ln(2) + 27\ln(3) + 45\ln(4) + 9\ln(5) + 2\ln(6) + 9\ln(7))$

$\ln(G) = \frac{1}{180}(0 + 36(0.6931) + 27(1.0986) + 45(1.3863) + 9(1.6094) + 2(1.7918) + 9(1.9459))$

$\ln(G) = \frac{1}{180}(24.9516 + 29.6622 + 62.3835 + 14.4846 + 3.5836 + 17.5131) = \frac{1}{180}(152.5786) \approx 0.84766$

$G = e^{0.84766} \approx 2.334$

* Harmonic Mean ($H$):

$H = \frac{\sum n_i}{\sum \frac{n_i}{x_i}} = \frac{180}{\frac{52}{1} + \frac{36}{2} + \frac{27}{3} + \frac{45}{4} + \frac{9}{5} + \frac{2}{6} + \frac{9}{7}} = \frac{180}{52 + 18 + 9 + 11.25 + 1.8 + 0.3333 + 1.2857} = \frac{180}{93.669} \approx 1.9216$

* Quadratic Mean ($Q$):

$Q = \sqrt{\frac{\sum x_i^2 n_i}{\sum n_i}}$

$\sum x_i^2 n_i = (1^2 \times 52) + (2^2 \times 36) + (3^2 \times 27) + (4^2 \times 45) + (5^2 \times 9) + (6^2 \times 2) + (7^2 \times 9) = 52 + 144 + 243 + 720 + 225 + 72 + 441 = 1897$

$Q = \sqrt{\frac{1897}{180}} = \sqrt{10.5389} \approx 3.246$

* Variance ($\sigma^2$):

$\sigma^2 = \frac{\sum (x_i - \bar{x})^2 n_i}{\sum n_i} = \frac{\sum x_i^2 n_i}{\sum n_i} - \bar{x}^2 = \frac{1897}{180} - (\frac{505}{180})^2 = 10.5389 - (2.8056)^2 = 10.5389 - 7.8714 \approx 2.6675$

* Standard Deviation ($\sigma$):

$\sigma = \sqrt{\sigma^2} = \sqrt{2.6675} \approx 1.633$

* Coefficient of Variation ($CV$):

$CV = \frac{\sigma}{\bar{x}} = \frac{1.633}{2.8056} \approx 0.582$

3. Final Answer

* Population: 180 authors

* Statistical Unit: Each author

* Observed Character: Number of manuals written by an author

* Nature of the Character: Quantitative discrete

* Mode: 1

* Median: 3

* Q1: 1

* Q3: 4

* Arithmetic Mean: 2.8056

* Geometric Mean: 2.334

* Harmonic Mean: 1.9216

* Quadratic Mean: 3.246

* Variance: 2.6675

* Standard Deviation: 1.633

* Coefficient of Variation: 0.582