To check the quality of two brands of light bulbs, their life in burning hours was estimated for 100 bulbs of each brand: | Life (in hrs) | Brand A | Brand B | | --- | --- | --- | | 0–50 | 15 | 2 | | 50–100 | 20 | 8 | | 100–150 | 18 | 60 | | 150–200 | 25 | 25 | | 200–250 | 22 | 5 | (i) Which brand gives higher life? (ii) Which brand is more dependable?

To check the quality of two brands of light bulbs, their life in burning hours was estimated for 100 bulbs of

Exercises · Q8

To check the quality of two brands of light bulbs, their life in burning hours was estimated for 100 bulbs of each brand:

Life (in hrs)	Brand A	Brand B
0–50	15	2
50–100	20	8
100–150	18	60
150–200	25	25
200–250	22	5

(i) Which brand gives higher life? (ii) Which brand is more dependable?

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Mean life: A = 134.5, B = 136.5. C.V.: A ≈ 51.2%, B ≈ 27.3%. Brand B gives higher life and is more dependable.

Class mid-points are 25, 75, 125, 175, 225.

Brand A ( $f$ = 15, 20, 18, 25, 22; $\sum f = 100$ ). $\sum fm = 375 + 1500 + 2250 + 4375 + 4950 = 13450$ , so mean $= 134.5$ . Taking deviations from the mean, $\sum fd^2 = 473475$ , so $\sigma_A = \sqrt{\dfrac{473475}{100}} = \sqrt{4734.75} = 68.81$ and $C.V._A = \dfrac{68.81}{134.5}\times100 = 51.2\%$ .

Brand B ( $f$ = 2, 8, 60, 25, 5; $\sum f = 100$ ). $\sum fm = 50 + 600 + 7500 + 4375 + 1125 = 13650$ , so mean $= 136.5$ . Taking deviations from the mean, $\sum fd^2 = 139275$ , so $\sigma_B = \sqrt{\dfrac{139275}{100}} = \sqrt{1392.75} = 37.32$ and $C.V._B = \dfrac{37.32}{136.5}\times100 = 27.3\%$ . …

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