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The variance of five positive observations is 3.6. If four of the observations are 2, 2, 4, 5 then what is the remaining observation ?
Concept:\(\displaystyle Mean=? x=\frac{Sum\ of \ observation\ }{Number\ of \ observation}\)?2 ?= \(\displaystyle \frac{?_{i=1}^5(x_i?? x)^2 }{N}\) (a + b)2 = a2 + 2ab + b2Calculation:Given: N = 5 and ?2 ?= 3.6Let 'a' be the fifth observation.We know that \(\displaystyle Mean=? x=\frac{Sum\ of \ observation\ }{Number\ of \ observation}\)? \(\displaystyle \bar x=\frac{2\ +\ 2\ +\ 4\ +\ 5\ +\ a }{5}\)? 5x? = 13 + a? a = 5x? - 13 ------(i)Also, we know that, ?2 ?= \(\displaystyle \frac{?_{i=1}^5(x_i?? x)^2 }{N}\) = 3.6? \(\displaystyle \frac{?_{i=1}^5(x_i?? x)^2 }{N}\) = (3.6)? (2 ? x?)2 + (2 ? x?)2 + (4 ? x?)2 + (5 ? x?)2 + (? ? x?)2 = 18? 4 + x?2 - 4x? + 4 + x?2 - 4x? + 16 + x?2 - 8x? + 25 + x?2 - 10x? + a2 + x?2 - 2ax? = 18 ? 49 + 5x?2 - 26x? + a2 - 2ax? = 18Putting the value of equation (i),? 49 + 5x?2 - 26x? + (5x? - 13)2 - 2(5x? - 13)x? = 18? 49 + 5x?2 - 26x? + 25x?2 - 130x? + 169 - 10x?2 + 26x? =18? 218 + 20x?2 - 130x? = 18? 20x?2 - 130x? + 200 = 0? 2x?2 - 13x? + 20 = 0? 2x?2 - 8x? - 5x? + 20 = 0? 2x?(x? - 4) - 5(x? - 4) = 0? (x? - 4) (2x? - 5) = 0? x? = 4, x? = 5/2Putting the value of x? in equation (i),we get,a = 5 × 4 - 13 = 7 and a = 5 × 2.5 - 13 = - 0.5As the observations are positive, a = - 0.5 is not possible.? The required number is 7.
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