Distinguish between absolute and relative measure of dispersion. Two computer manufacturers A and B compete for profitable and prestigious contract. In their rivalry, each claim that their computer a consistent. For this it was decided to start execution of the same program simultaneously on 50 computers of each company and recorded the time as given below.

Time (in seconds) 0-2 2-4 4-6 6-8 8-10 10-12

Nature of computers manufactured by A 5 16 13 7 5 4

B 2 7 12 19 9 1

Which company’s computer is more consistent?

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Distinguish between absolute and relative measure of dispersion. Two computer manufacturers A and B compete for profitable and prestigious contract. In their rivalry, each claim that their computer a consistent. For this it was decided to start execution of the same program simultaneously on 50 computers of each company and recorded the time as given below.

Time (in seconds) 0-2 2-4 4-6 6-8 8-10 10-12

Nature of computers manufactured by A 5 16 13 7 5 4

B 2 7 12 19 9 1

Which company’s computer is more consistent?

Suresh Chand · Accepted Answer

Measure of dispersion is descriptive statistical measure used to measure the variation or spread or scatteriess in the data set. There are two types of measures of dispersion i.e absolute and relative measure of dispersion. The difference between them are: Absolute Measure Relative Measure A measure of dispersion is saied to be an absolute if it is expressed interms of original units of data. A measure of dispersion is saied to be relative if it is independent of original units of data and it is simply a cofficient. Some absolute measures: Range Quartile deviation Mean deviation Standard deviation Some relative measures are: Cofficient of range Cofficient of Quartile deviation Cofficient of Mean deviation Cofficient of variation Absolute measures are in fixed value Relative measures may be in ratio or percentage Solution: For Company A Time(x) No of Comp.(f) m fm fm2 0-2 5 1 5 5 2-4 16 3 48 144 4-6 13 5 65 325 6-8 7 7 49 343 8-10 5 9 45 405 10-12 4 11 44 484 N = 50 Σfm = 256 Σfm2 = 1706 Now, (bar{X} = frac{sum fm}{N} = frac{256}{50} = 5.12) Also, Standard deviation(σ) is (σ = sqrt{frac{sum fm^2}{N} - (frac{sum fm}{N})^2}) (= sqrt{frac{1706}{50} - (5.12)^2}) = 2.812 Also, Covariance (C.V) = (frac{σ}{bar{X}} times 100%) (= frac{2.812}{5.12} times 100%) = 54.92% For Company B Time(x) No of Comp.(f) m fm fm2 0-2 2 1 2 2 2-4 7 3 21 63 4-6 12 5 60 300 6-8 19 7 133 931 8-10 9 9 81 729 10-12 1 11 11 121 N = 50 Σfm = 308 Σfm2 = 2146 Now, (bar{X} = frac{sum fm}{N} = frac{308}{50} = 6.16) Also, Standard deviation(σ) is (σ = sqrt{frac{sum fm^2}{N} - (frac{sum fm}{N})^2}) (= sqrt{frac{2146}{50} - (6.16)^2}) = 1.725 Also, Covariance (C.V) = (frac{σ}{bar{X}} times 100%) (= frac{1.725}{6.16} times 100%) = 28% Since, the standard deviation and covarience of company B is less than that of A. So, Company B is considered more consistent.

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Time (in seconds)		0-2	2-4	4-6	6-8	8-10	10-12
Nature of computers manufactured by	A	5	16	13	7	5	4
Nature of computers manufactured by	B	2	7	12	19	9	1

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