# 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