Calculate the geometric and the harmonic mean of the following series of monthly expenditure of a batch of students.

$125,130,75,10,45,0.5,0.40,500,150,5$

1. Arithmetic Mean:

The mean or arithmetic mean is the sum of all the observations(numbers) divided by the total number of observations.

(i) For ungrouped data: If the data consists of numbers $x_1,x_2,x_3,...,x_n$, then

(a) Direct method: Mean $\left(\overline{X}\right)=\frac{x_1+x_2+x_3+,...,+x_n}{n}=\frac{{\displaystyle \sum _{i=1}^n}{x}_i}{n}$

(b) Shortcut method: Mean $\left(\overline{X}\right)=A+\frac{{\displaystyle \sum _{i=1}^n{d}_i}}{n}$, where $A=$ Assumed Mean, $d_i=x_i-A$

(ii) For grouped data: If the observations $x_1,x_2,x_3,...,x_n$ has frequencies $f_1,f_2,f_3,...,f_n$ respectively then

(a) Direct method: Mean $\left(\overline{X}\right)=\frac{{\displaystyle \underset{i=1}{\overset{n}{\sum f_i}}{x}_i}}{{\displaystyle \sum _{}}{f}_i}=\frac{1}{N}\underset{i=1}{\overset{n}{\sum f_i}}{x}_i$

(b) Shortcut method: $\left(\overline{X}\right)=A+\frac{{\displaystyle \underset{i=1}{\overset{n}{\sum f_i}}{d}_i}}{{\displaystyle \sum _{}{f}_i}}$, where $A=$ Assumed Mean and $d_i=x_i-A$

(iii) For continuous frequency distribution: If $x_i$ is the mid-point of the class interval,

(a) Direct method: Mean $\left(\overline{X}\right)=\frac{{\displaystyle \underset{i=1}{\overset{n}{\sum f_i}}{x}_i}}{{\displaystyle \sum _{}}{f}_i}=\frac{1}{N}\underset{i=1}{\overset{n}{\sum f_i}}{x}_i$

(b) Shortcut method: $\left(\overline{X}\right)=A+\frac{{\displaystyle \underset{i=1}{\overset{n}{\sum f_i}}{d}_i}}{{\displaystyle \sum _{}{f}_i}}\times c, \text{Where} c=$Width of the class interval, $A=$ Assumed Mean, $d_i=\frac{x_i-A}{c}$

(iv) Weighted Arithmetic Mean: If the observations $x_1,x_2,x_3,...,x_n$ has weights $w_1,w_2,w_3,...,w_n$ respectively then, Mean$\left(\overline{X}\right)=\frac{{\displaystyle \underset{i=1}{\overset{n}{\sum w_i}}{x}_i}}{{\displaystyle \sum _{}{w}_i}}$.

(v) Combined Mean: If ${\overline{x}}_1, {\overline{x}}_2$ and $n_1, n_2$ are the mean and number of observations of two data sets, Combined Mean$=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}$.

2. Geometric Mean:

(i) For ungrouped data: If the data consists of numbers $x_1,x_2,x_3,...,x_n$, then, G.M.$={\left(x_1\times x_2\times x_3\times ,...,\times x_n\right)}^{\frac{1}{n}}=\mathrm{Antilog}\frac{{\displaystyle \sum _{i=1}^n\log }{x}_i}{n}$

(ii) For grouped data: If the observations $x_1,x_2,x_3,...,x_n$ has frequencies $f_1,f_2,f_3,...,f_n$ respectively then G.M.$=\mathrm{Antilog}\frac{{\displaystyle \underset{i=1}{\overset{n}{\sum f_i}}\log x_i}}{\underset{i=1}{\overset{n}{\sum f_i}}}$

(iii) For continuous frequency distribution: If $x_i$ is the mid-point of the class interval, G.M.$=\mathrm{Antilog}\frac{{\displaystyle \underset{i=1}{\overset{n}{\sum f_i}}\log x_i}}{\underset{i=1}{\overset{n}{\sum f_i}}}$.

3. Harmonic Mean:

(i) For ungrouped data: If the data consists of numbers $x_1,x_2,x_3,...,x_n$, then, H.M. $=\frac{n}{\sum _{i=1}^n{\displaystyle \frac{1}{x_i}}}$

(ii) For grouped data: If the observations $x_1,x_2,x_3,...,x_n$ has frequencies $f_1,f_2,f_3,...,f_n$ respectively then, H.M. $=\frac{\sum _{i=1}^n{f}_i}{\sum _{i=1}^n{f}_i\times {\displaystyle \frac{1}{x_i}}}$

(iii) For continuous frequency distribution: If $x_i$ is the mid-point of the class interval, H.M.$=\frac{\sum _{i=1}^n{f}_i}{\sum _{i=1}^n{f}_i\times {\displaystyle \frac{1}{x_i}}}$.

4. Comparison between means: $\mathrm{A}.\mathrm{M}.\ge \mathrm{G}.\mathrm{M}.\ge \mathrm{H}.\mathrm{M}.$

5. Median:

(i) For ungrouped data: If there are $n$ observations $x_1,x_2,x_3,...,x_n$, then

(a) If the number of observations is odd, then the median is ${\left(\frac{n+1}{2}\right)}^{\mathrm{th}}$ observation.

(b) If the number of observations is even, then median is the mean of ${\left(\frac{n}{2}\right)}^{th}$ and ${\left(\frac{n}{2}+1\right)}^{th}$ observations.

(ii) For grouped data: The median is the value where cumulative frequency$=\frac{{\displaystyle \sum _{}^{}}{f}_i+1}{2}$ i.e., central observation.

(iii) For continuous frequency distribution: The median of continuous frequency distribution is obtained by first identifying the class in which median lies (median class) and then applying the formula $=l+\frac{{\displaystyle \frac{N}{2}}-C}{f}\times h$, where

$l=$ Lower Limit of Median class,

$N=$ Total Number of frequencies

$f=$ Frequency of Median class,

$m=$ Cumulative Frequency of class preceding the Median class,

$h=$ Class Interval of the Median class

(iv) Graphical method for location of median: 

(a) Median is located by using Cumulative Frequency curve or Ogive.

(b) The median is the corresponding data value where the cumulative frequency is $\frac{N}{2} \text{or} \frac{N+1}{2}$ on the curve.

6. Mode:

(i) Mode of a statistical data is the observation which occurs most frequently.

(ii) Mode for grouped data when frequency distribution is given in the form of classes:

Mode$=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times c$, where,

$f_1=$ Frequency of Modal class,

$f_0=$ Frequency of the class preceding Modal class,

$f_2=$ Frequency of the class succeeding Modal class,

$c=$ Width of class limits

7. Empirical relation:

Mode$=3$Median$-2$Mean

8. Quartile:

These are the points that divides the frequency distribution in four equal parts and are denoted by, $Q_1, Q_2 \text{and} Q_3$.

(i) For ungrouped data: Quartiles are the values of $x$ corresponds to $Q_1={\left(\frac{N+1}{4}\right)}^{th}, Q_2={\left(\frac{N+1}{2}\right)}^{th} \text{and }{Q}_3=3{\left(\frac{N+1}{4}\right)}^{th}$.

(ii) For discrete Series: Quartiles are the values of $x$ corresponds to $Q_1={\left(\frac{N+1}{4}\right)}^{th}, Q_2={\left(\frac{N+1}{2}\right)}^{th} \text{and }{Q}_3=3{\left(\frac{N+1}{4}\right)}^{th}$.

(iii) For Continuous Series:

(a) Find the cumulative frequencies and hence find $\frac{N}{4}$.

(b) Use: $Q_1=l_1+\frac{{\displaystyle \frac{N}{4}}-m_1}{f_1}\times c_1 \text{and} Q_3=l_3+\frac{{\displaystyle 3\left(\frac{N}{4}\right)-m_3}}{f_3}\times c_3$

9. Deciles: 

These are the points that divides the ungrouped data in ten equal parts.

10. Percentiles:

These are the points divides the distribution into $100$ equal parts, and $P_{25}=Q_1, P_{50}=Q_2=\text{median, and} P_{75}=Q_3$.