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 step-deviation 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 step-deviation 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) short-cut method (3) step-deviation.
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
|
Short-Cut
Method |
Step-Deviation
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) Short-Cut 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) Step-Deviation 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