Examples of Arithmetic Mean
Example:
The following data shows distance covered by $$100$$ people to perform their routine jobs.
Distance (Km)

$$0 – 10$$

$$10 – 20$$

$$20 – 30$$

$$30 – 40$$

Number of People

$$10$$

$$20$$

$$40$$

$$30$$

Calculate the arithmetic mean by stepdeviation method; also explain why it is better than direct method in this particular case.
Solution:
The given distribution is grouped data and the variable involved is distance covered, while the number of people represents frequencies.
Distance Covered in (Km)

Number of People
$$f$$ 
Mid Points
$$x$$ 
$$u = \left( {\frac{{x – 5}}{{10}}} \right)$$

$$fu$$

$$0 – 10$$

$$10$$

$$5$$

$$ – 1$$

$$ – 10$$

$$10 – 20$$

$$20$$

$$15$$

$$0$$

$$0$$

$$20 – 30$$

$$40$$

$$25$$

$$ + 1$$

$$40$$

$$30 – 40$$

$$30$$

$$35$$

$$ + 2$$

$$60$$

Total

$$\sum f = 100$$

$$\sum fu = 90$$

Now we will find the arithmetic mean as $$\overline X = A + \frac{{\sum fu}}{{\sum f}} \times h$$
Here
$$A = 15$$, $$\sum fu = 90$$, $$\sum f = 100$$ and $$h = 10$$
$$\overline X = 15 + \frac{{90}}{{100}} \times 10 = 24$$ Km
Explanation:
Here from the mid points ($$x$$) it is very clear that each mid point is a multiple of $$5$$ and there is also a gap of $$10$$ from mid point to midpoint, i.e. class size or interval ($$h$$). Keeping this in mind, we should use the stepdeviation method instead of direct method.
Example:
The following frequency distribution showing the marks obtained by $$50$$ students in statistics at a certain college. Find the arithmetic mean using (1) direct method (2) shortcut method (3) stepdeviation.
Marks

$$20 – 29$$

$$30 – 39$$

$$40 – 49$$

$$50 – 59$$

$$60 – 69$$

$$70 – 79$$

$$80 – 89$$

Frequency

$$1$$

$$5$$

$$12$$

$$15$$

$$9$$

$$6$$

$$2$$

Solution:
Direct Method

ShortCut
Method 
StepDeviation
Method 

Marks

$$f$$

$$x$$

$$fx$$

$$D = x – A$$

$$fD$$

$$u = \frac{{x – A}}{h}$$

$$fu$$

$$20 – 29$$

$$1$$

$$24.5$$

$$24.5$$

$$ – 30$$

$$ – 30$$

$$ – 3$$

$$ – 3$$

$$30 – 39$$

$$5$$

$$34.5$$

$$172.5$$

$$ – 20$$

$$ – 100$$

$$ – 2$$

$$ – 10$$

$$40 – 49$$

$$12$$

$$44.5$$

$$534.5$$

$$ – 10$$

$$ – 120$$

$$ – 1$$

$$ – 12$$

$$50 – 59$$

$$15$$

$$54.5$$

$$817.5$$

$$0$$

$$0$$

$$0$$

$$0$$

$$60 – 69$$

$$9$$

$$64.5$$

$$580.5$$

$$10$$

$$90$$

$$1$$

$$9$$

$$70 – 79$$

$$6$$

$$74.5$$

$$447.5$$

$$20$$

$$120$$

$$2$$

$$12$$

$$80 – 89$$

$$2$$

$$84.5$$

$$169.5$$

$$30$$

$$60$$

$$3$$

$$6$$

Total

$$50$$

$$2745$$

$$20$$

$$2$$

(1) Direct Method:
$$\overline X = \frac{{\sum fx}}{{\sum f}} = \frac{{2745}}{{50}} = 54.9$$ or $$55$$ Marks
(2) ShortCut Method:
$$\overline X = A + \frac{{\sum fD}}{{\sum f}}$$ Where $$A = 54.5$$
$$ = 54.5 + \frac{{20}}{{50}} = 54.5 + 0.4 = 54.9$$Marks
(3) StepDeviation Method:
$$\overline X = A + \frac{{\sum fu}}{{\sum f}} \times h$$ Where $$A = 54.5$$ $$h = 10$$
$$ = 54.5 + \frac{2}{{50}} \times 10$$
$$ = 54.5 + 0.4 = 54.9$$ Marks
A.udayakumar
September 26 @ 5:29 pm
very nice explanation, and each and every example is very and clearly understood