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The number of values of \( a \in N \) such that the variance of \( 3,7,12, a, 43-a \) is a natural number is: |
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Answer» \(\bar x = \frac{\sum x_i}{5}=\frac{65}5=13\) \(\sum x_i^2\) = 9 + 49 + 144 + a2 + (43 - a)2 = 2a2 - 86a + 2051 \(\therefore\) variance = \(\frac1n\sum x_i^2-(\bar x)^2\) = \(\frac15(2a^2-86a+2051)-169\) = \(\frac15(2a^2-86a+2051-845)\) = \(\frac15(2a^2-86a+1206)\) = \(\frac25(a^2-43a+603)\) Variance is natural number if a2 - 43a + 603 is a multiple of 5. |
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