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Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution 50.x: 1030507090y: 17f132f219 |
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Answer»
Given, mean = 50 \(\frac{\sum yx}{N}=50\) \(\frac{30f1+70f2+3480}{120}=50\) 30f1 + 70f2 + 3480 = 50 x 120 30f1 + 70f2 + 3480 = 6000 (i) Also, ∑y = 120 17 + f1 + 32 + f2 + 19 = 120 f1 + f2 = 52 f1 = 52 – f2 Substituting value of f1 in (i), we get 30 (52 – f2) + 70f2 + 3480 = 6000 40f2 = 960 f2 = 24 Hence, f1 = 52 – 24 = 28 Therefore, f1 = 28 and f2 = 24 |
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