1.3 Measures of Central Tendency

1.3.1 What is Central Tendency?

Why needed?

##                    mpg cyl  disp  hp drat
## Mazda RX4         21.0   6 160.0 110 3.90
## Mazda RX4 Wag     21.0   6 160.0 110 3.90
## Datsun 710        22.8   4 108.0  93 3.85
## Hornet 4 Drive    21.4   6 258.0 110 3.08
## Hornet Sportabout 18.7   8 360.0 175 3.15
## Valiant           18.1   6 225.0 105 2.76
## Duster 360        14.3   8 360.0 245 3.21
## Merc 240D         24.4   4 146.7  62 3.69
##       mpg             cyl           disp
##  Min.   :14.30   Min.   :4.0   Min.   :108.0
##  1st Qu.:18.55   1st Qu.:5.5   1st Qu.:156.7
##  Median :21.00   Median :6.0   Median :192.5
##  Mean   :20.21   Mean   :6.0   Mean   :222.2
##  3rd Qu.:21.75   3rd Qu.:6.5   3rd Qu.:283.5
##  Max.   :24.40   Max.   :8.0   Max.   :360.0
• Summary
• Comparison
• A value to represent all

1.3.2 Criteria for a Good Measure of Central Tendency

• Well-defined
• Understandable
• Considers all values
• Suitable for further analysis
• Not affected by sample fluctuation

1.3.3 Measures (Averages)

• Arithmetic Mean (AM)
• Geometric Mean (GM)
• Harmonic Mean (HM)
• Median
• Mode
• Partition Values (Quartiles, Deciles, Percentiles etc.)

1.3.4 AM

$$AM=\bar x=\frac{\sum x}{n}$$

If there are frequencies or weights

$$\bar x=\frac{\sum f_i x_i}{\sum f_i} \space or \space \frac{\sum w_i x_i}{w_i=n}$$

1.3.5 Find AM

##  [1] 10 55 38 48 51 25 14 44 23 22

Find AM: 2, 2, 3, 3, 5, 5, 5, 8, 8, 9

There are 2 ways.

$$\bar x = \frac{2+2+...+9}{10}$$

1.3.6 Mean Using Frequency

For grouped data

Working hours
(x)
Employee
(f)
fx
2 2 4
3 2 6
5 3 15
8 2 10
9 1 9
$$\sum f =10$$ $$\sum fx = 50$$

$$\therefore \bar x = \frac{\sum fx}{\sum f} =\frac{50}{10}=5$$

1.3.7 Freuency vs Weight

Suppose, different judges give different scores, but not all evaluation has same weight.

Judge Rating
(x)
Weight
(w)
wx
1 8 2 16
2 7 3 21
3 4 5 20
4 5 1 5
5 7 3 21
$$\sum w_i = 14$$ $$\sum w_ix_i = 83$$

$$\therefore \bar x = \frac{\sum w_ix_i}{\sum w_i}$$

1.3.8 Shortcut Method for AM

Calculate the mean in a smart way

## [1] 1009 1037 1047 1024 1013 1043
Show (click to see)

1.3.9 Shortcut Method Formula

Consider the values: 1005, 1010, 1015

If 1000 is subtracted: 5, 10, 15

If again divided by 5: 1, 2, 3

Converted Mean = 2

Original Mean = $$2 \times 5 + 1000=1010$$

Show (click to see)

1.3.10 Properties of AM

• $$\sum (x_i-\bar x)=0$$; can you prove it?
• $$\sum (x_i-\bar x)^2 \le \sum (x_i-a)^2, \space a \ne \bar x$$
• Depends on change of origin and scale?
• $$\bar x + \bar y =\frac{\sum x+\sum y}{n_x+n_y}$$
• Combined mean: $$\bar x_c=\frac{n_1 \bar x_1+n_2 \bar x_2+...+n_k \bar x_k}{n_1+n_2+...+n_k}$$
• $$AM\ge GM \ge HM$$ & $$AM \times HM = (GM)^2$$
• AM of first n natural numbers = $$\frac{n+1}{2}$$

• Well-defined
• Less affected by sample fluctuation
• Comparison among sets is easy
• Uses all values
• Suitable for further analysis.
• Affected by outliers

1.3.12 Geometric Mean (GM)

$$GM=(x_1 \times x_2 \times ... \times x_n)^{(1/n)}$$ or $$GM=(x_1^{f_1} \times x_2^{f_2} \times ... \times x_n^{f_n})^{(1/\sum f_i)}$$

Find GM: 2, 4, 6

Try this one: 20020, 30080, 50086, 40130

1.3.13 Concept of Logarithm

“An admirable artifice which, by reducing to a few days the labour of many months, doubles the life of the astronomer, and spares him the errors and disgust inseparable from long calculations.”

— Pierre-Simon Laplace

• Log and Antilog
• $$log_24=?$$
• if $$log_2x=3$$, then x=?

1.3.14 GM Easier Formula

x = 20020, 30080, 50086, 40130

• log x = 9.9, 10.31, 10.82, 10.6
• Mean of logx = $$\frac{\sum logx}{n}$$
• Original Mean = $$antilog(\frac{\sum logx}{n})$$

1.3.15 Calculate GM

Marks # Students
10-12 4
12-14 5
14-16 3
16-18 5
18-20 7
20-22 2

Make a table using these columns: $$x_i, f_i, logx_i, f_ilogx_i$$

• $$GM =antilog(\frac{\sum f_i logx_i}{\sum f_i})$$

• Not affected by outliers

x = 5, 10, 15, 20, 100, 1000

log(x) = 0.7, 1, 1.18, 1.3, 2, 3

• Less affected by sample fluctuation

• Suitable for further analysis

• What if one or some x = 0?

• What if one or some x < 0?

1.3.17 Story of Oil Scam

S = 150 km

$$v_1=\space 10 km/h, v_2=\space 15 km/h, v_3=\space 20 km/h$$

What is the average speed?

• $$AM=\frac{10+15+20}{3}=15 \space km/h$$
• Think more fundamentally
• $$\sum S=3\times 150=450$$
• $$t_1=15h, t_2=10h, t_3=7.5h$$
• $$\bar v = \frac{\sum S}{\sum t}=\frac{450}{32.5}=13.84 \lt AM$$
• $$=\frac{450}{15+10+7.5}$$
• $$=\frac{450}{\frac{150}{10}+\frac{150}{15}+\frac{150}{20}}$$
• $$=\frac{450}{150(\frac{1}{10}+\frac{1}{15}+\frac{1}{20})}$$
• $$=\frac{3}{\frac{1}{10}+\frac{1}{15}+\frac{1}{20}}$$
• $$=\frac{3}{\frac{1}{v_1}+\frac{1}{v_2}+\frac{1}{v_3}}$$

1.3.18 Harmonic Mean

Formula: Reciprocal of Mean of $$\frac{1}{x_i}$$

Reciprocal of $$\frac{\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}}{n}$$

Thus, $$HM = \frac{n}{\sum \frac{1}{x_i}}$$

Calculate: 2, 4, 8

For grouped data

• $$HM=\frac{\sum f}{\sum \frac{f}{x}}$$
• For wighted data: $$\frac{\sum w}{\sum \frac{w}{x}}$$

1.3.19 Why and When HM

• When there are rates associated, say speed, and numerator is fixed.
• $$Speed, v = \frac{S}{t}$$; HM if S is fixed
• Example: A man travels 120 km the first day at 12 kph, the same distance at 10 kph on the 2nd day, and at 8 kph on the 3rd day. Find his average speed.

1.3.20 Wighted AM vs Weighted HM

Suppose, a bus travels 10 km at 10 kph, another 15 km at 20 kph, and another 20 km at 25 kph. What is the average speed.

HM $$\rightarrow$$ consider distances as weights

AM $$\rightarrow$$ consider times as weights

Time, $$t=\frac d v=$$ 1, 0.75, 0.8

• $$WHM = \frac{10+15+20}{\frac{10}{10}+\frac{15}{20}+\frac{20}{25}}$$=17.65
• $$WAM = \frac{1\times 10 + 0.75 \times 20 + 0.80 \times 25}{1+0.75+0.80}$$ = 17.65
• True mean, $$\bar v=\frac{\sum d}{\sum t}=\frac{45}{2.55}=$$ 17.65

1.3.21 HM Example 2

• A passerby travels 10 km at 20 kph, 5 km at 15 kph, and 4 km at 12 kph. What is the average speed? (Use weighted HM)

• Here, distances are different. Consider them weights.

• $$HM=\frac{w_1+w_2+w_3}{\frac{w_1}{x_1}+\frac{w_2}{x_2}+\frac{w_3}{x_3}}$$
• HM = 16.286

$$QM=\sqrt{\frac{x_1^2+x_2^2+...+x_n^2}{n}}=\sqrt{\frac{\sum x_i^2}{n}}$$

1.3.24 Find medians

x = 4,5,6,8,9,11,16

y = 4,5,6,8,9,11,16,19

• Median of x is 8
• Median of y is 8.5
• Formula for odd n = $$\frac{n+1}{2}th \space value$$
• Formula for even n = $$\frac{\frac{n}{2}th+(\frac{n}{2}+1)th}{2}$$

• Unaffected by outliers (extreme values)
• Graphically estimable
• Affected by sample fluctuation
• Not based on all values
• Not suitable for further mathematical analysis

1.3.26 Quartiles, Deciles, and Percentiles

$$Q_1= \frac{(n+1)}{4}th \space$$

$$Q_2, Q_3=?$$

Find $$Q_1, Q_2, Q_3$$

X = 4, 6, 7, 10, 12, 13, 14, 15, 16, 17, 19

• 4, 8.5, 13, 12.0909091, 15.5, 19
• What are the formulae for Deciles ($$D_i$$ and $$P_i$$)

1.3.27 General Formula for Partition Values

For odd n,

$$A_i= \frac{i \times (n+1)}{k}th \space value$$

For even n,

$$A_i=\frac{\frac{i \times n}{k}th+(\frac{i\times n}{k}+1)th}{k}$$

where, k = no. of partitions

For median, for example, k = 2.

• What if we divide the data set into 20 segments?

1.3.28 Find Deciles and Percentiles

X = 4, 6, 7, 10, 12, 13, 14, 15, 16, 17, 19

Find $$D_3, D_8, P_{17}, P_{56}, P_{93}$$

1.3.29 Averages from Grouped Data

Age # Employees CF
20-24 40
25-29 60
30-34 200
35-39 180
40-44 150
45-49 110
50-54 175
55-59 60
60-64 25

Median: $$Me = L + \frac{\frac{n}{2}-F_c}{f_m}\times c$$

Mode: $$M_o=L+\frac{\Delta_1}{\Delta_1+\Delta_2}\times c$$

Quartiles: $$Q_i = L + \frac{\frac{in}{4}-F_c}{f}\times c$$

Find AM, Me, Mo, Q, and $$P_{30}$$

1.3.30 Partition Values from Graph

• Draw an Ogive
• Draw a straight line parallel to X-axis at the point $$\frac n 2$$ of Y-axis. - Draw a perpendicular on X-axis from the intersection point.

1.3.31 Comparison of Averages

Property AM GM HM Median Mode
Formula Easy to
understand

Easy

Easy

Easy

Easy
Considers
values
All All All Middle term(s) Highest frequency
Computaion Easy
Easy

Easy
Easy Easy
Effect of
Outliers
Highly affected Less affected
than AM
Less affected
than GM
Unaffected Can be highly
affected
Effect of Sampling
Flcutuation
Less affected Less affected Less affected Can be highly
affected
Can be highly
affected
Suitability for
further analysis
Possible Possible Possible
Possible

Possible

1.3.32 When AM = GM = HM

If $$x_1=x_2=x_3$$; prove it

1.3.34 Theorem 01

$\sum_{i=0}^n (x_i-\bar x)=0$

1.3.35 Theorem 02

$\sum_{i=0}^n f_i(x_i-\bar x)=0$

1.3.36 Theorem 03

$\sum_{i=1}^n (x_i-\bar x)^2 \lt \sum_{i=1}^n (x_i-a)^2; a\ne\bar x$

$\sum_{i=1}^n f_i(x_i-\bar x)^2 \lt \sum_{i=1}^n f_i(x_i-a)^2; a\ne\bar x$

1.3.37 Theorem 04 (AM~origin & scale)

Prove that AM depends on origin and scale Use frequency as well i.e,

• $\bar x=\frac {\sum_{i=1}^nx_i}{n}$
• $\bar x=\frac {\sum_{i=1}^n f_ix_i}{n}$

1.3.38 Theorem 07

If the GM of $$n_1$$ va;ues is $$G_1$$, and of $$n_2$$ values is $$G_2$$, show GM of $$n_1+n_2$$ values is $$G=\sqrt{G_1G_2}$$

1.3.39 For two non-zero positive numbers, prove $$AM \ge GM \ge HM$$

Let, the numbers be a, b

$$\therefore AM = \frac{a+b}{2}$$

$$GM = \sqrt{ab}$$

$$HM = \frac{2}{\frac 1 a +\frac 1 b}$$

We know, $\begin{eqnarray} & &(a-b)^2\ge 0 \nonumber \\ & \Rightarrow & (a+b)^2-4ab \ge 0 \nonumber \\ & \Rightarrow & (a+b)^2 \ge 4ab \nonumber \\ & \Rightarrow & (a+b) \ge 2 \sqrt{ab} \nonumber \\ & \Rightarrow & \frac{a+b} 2 \ge \sqrt{ab} \nonumber \\ & \Rightarrow & AM \ge GM \nonumber \\ \end{eqnarray}$

Similarly, $\begin{eqnarray} & &(\frac{1}{a}-\frac{1}{b})^2\ge 0 \nonumber \\ & \Rightarrow & (\frac{1}{a}+\frac{1}{b})^2 -4 \cdot \frac 1 a \cdot \frac 1 b\ge 0 \nonumber \\ & \Rightarrow & (\frac{1}{a}+\frac{1}{b})^2\ge \frac 4 {ab} \nonumber \\ & \Rightarrow & (\frac{1}{a}+\frac{1}{b}) \ge \frac 2 {\sqrt{ab}} \nonumber \\ & \Rightarrow & \sqrt{ab}(\frac{1}{a}+\frac{1}{b}) \ge 2 \nonumber \\ & \Rightarrow & \sqrt{ab} \ge \frac{2}{(\frac{1}{a}+\frac{1}{b})} \nonumber \\ & \Rightarrow & GM \ge HM \nonumber \\ \end{eqnarray}$

1.3.40 Theorem 09

For two non-zero positive numbers, $$AM \times HM =(GM)^2$$

1.3.41 Theorem 10

Mean and Median of first n natural numbers are $$\frac {n+1} 2$$

1.3.42 Theorem 11

If $$\bar x_1$$ and $$\bar x_2$$ are means of 2 data sets of sizes $$n_1$$ and $$n_2$$, respectively, the combined mean is $$\bar x_c=\frac{n_1 \bar x_1+n_2 \bar x_2}{n_1+n_2}$$

1.3.43 Theorem 12

If $$u=x+y, \bar u=\bar x + \bar y$$; if $$n_1=n_2=n$$

Given $$u=x+y$$

$\begin{eqnarray} \bar u &=& \frac{\sum u}{n} \nonumber \\ &=& \frac{\sum (x+y)}{n} \nonumber \\ &=& \frac{\sum x}{n}+ \frac{\sum y}{n} \nonumber \\ &=& \bar x + \bar y \nonumber \\ \end{eqnarray}$

1.3.44 Theorem 13

For equal number of observations, GM of two variables is equal to the product of their individual means.

1.3.45 Theorem 14

$$GM=Antilog(\frac{\sum \log x_i}{n})$$ or $$Antilog(\frac{\sum f_i \log x_i}{\sum f_i})$$

1.3.46 Theorem 15

If $$y = a + bx, \bar y = a + b \bar x$$

1.3.47 Theorem 16

If $$z_i=ax_i+by_i, \bar z=a \bar x + b \bar y$$

1.3.49 Example Problem 01

AM and GM of two positive numbers are 25 and 15, respectively. Find HM and the numbers.
Solution (i) (click to see)
Solution (ii) (click to see)

1.3.50 Example Problem 15

The mean of 200 numbers was 50. Later it was revealed that two observations were incorrectly given as 92 and 8, instead of 192 and 88, respectively. Find the correct mean.

Solution(click to see)

1.3.51 Example Problem 22

If $$\sum f_i(x_i-k)=0$$, what is the value of k?