The number of complaint calls received per day by a customer-care centre, recorded over 50 days, is given belo

Worked Examples · Example 4

The number of complaint calls received per day by a customer-care centre, recorded over 50 days, is given below. Find the Mean Deviation about the Mean and its Coefficient.

Calls ( $x$ )	0	1	2	3	4
Number of days ( $f$ )	6	12	18	9	5

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Step 1 — Find the mean.

$x$	$f$	$fx$
0	6	0
1	12	12
2	18	36
3	9	27
4	5	20
Total	50	95

$\bar{x} = 95/50 = 1.9$ .

Step 2 — Find $|x-1.9|$ and $f|x-1.9|$ for each row.

| $x$ | $f$ | $|x-1.9|$ | $f|x-1.9|$ |

|---|---|---|---|

| 0 | 6 | 1.9 | 11.4 |

| 1 | 12 | 0.9 | 10.8 |

| 2 | 18 | 0.1 | 1.8 |

| 3 | 9 | 1.1 | 9.9 |

| 4 | 5 | 2.1 | 10.5 |

| Total | 50 | | 44.4 |

Step 3 — Divide, then find the coefficient.

$\text{MD}(\bar{x}) = \dfrac{44.4}{50} = 0.888$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2022Set ANNUAL1 markMCQ
$Q. From the following the one which is a relative measure of dispersion is :$
(a) Range
(b) Quartile deviation
(c) Coefficient of mean deviation
(d) Standard deviation
›Reveal solutionSolution
Relative measures are ratios/coefficients; here that is the coefficient of mean deviation.

Measures of dispersion are of two kinds:

Absolute — expressed in the original units (Range, Quartile deviation, Mean deviation, Standard deviation).

Relative — a unit-free ratio obtained by dividing an absolute measure by an average (Coefficient of range, Coefficient of quartile deviation, Coefficient of mean deviation, Coefficient of variation). …
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CBSE 2022Set ANNUAL1 markMCQ
$Q. The mean deviation of 10, 12, 20 and 30, is :$
(a) $7$
(b) $18$
(c) $14$
(d) $6$
›Reveal solutionSolution

$\bar{x}=18$ , $\text{M.D.}=\frac{8+6+2+12}{4}=7$ .

Mean:

$\bar{x}=\frac{10+12+20+30}{4}=\frac{72}{4}=18.$

Absolute deviations from the mean:

$|10-18|=8,\;|12-18|=6,\;|20-18|=2,\;|30-18|=12.$

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. The formula for calculating coefficient of mean deviation is :$
(a) $\frac{\text{Mean}}{\text{Mean deviation}}$
(b) $\frac{\text{Mean deviation}}{\text{Mean}}$
(c) $\text{Mean deviation} \times \text{Mean}$
(d) $\frac{\text{Median}}{\text{Mean deviation}}$
›Reveal solutionSolution

Coefficient of M.D. $=\dfrac{\text{Mean deviation}}{\text{average used}}$ .

A relative measure of dispersion is formed by dividing the absolute measure by the corresponding average. For mean deviation taken about the mean,

$\text{Coefficient of M.D.}=\frac{\text{Mean deviation}}{\text{Mean}}.$

…

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