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Exercise: Variance and Standard Devia... · Q15

Q.Find the variance and standard deviation of the first 10 natural numbers: 1,2,3,4,5,6,7,8,9,101, 2, 3, 4, 5, 6, 7, 8, 9, 10.

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For 1,2,…,101,2,\ldots,10 (n=10n=10), the mean is xˉ=1+2+⋯+1010=n+12=5.5\bar x=\dfrac{1+2+\cdots+10}{10}=\dfrac{n+1}{2}=5.5. The deviations are −4.5,−3.5,−2.5,−1.5,−0.5,0.5,1.5,2.5,3.5,4.5-4.5,-3.5,-2.5,-1.5,-0.5,0.5,1.5,2.5,3.5,4.5, with squares 20.25,12.25,6.25,2.25,0.25,0.25,2.25,6.25,12.25,20.2520.25,12.25,6.25,2.25,0.25,0.25,2.25,6.25,12.25,20.25, summing to 82.582.5. So σ2=82.5/10=8.25\sigma^2=82.5/10=8.25. This matches the standard closed-form result for the first nn natural numbers, σ2=n2−112=100−112=9912=8.25\sigma^2=\dfrac{n^2-1}{12}=\dfrac{100-1}{12}=\dfrac{99}{12}=8.25, confirming the direct computation. Therefore σ=8.25≈2.87\sigma=\sqrt{8.25}\approx2.87. [!ANSWER] Mean =5.5=5.5; Variance =8.25=8.25; Standard Deviation ≈2.87\approx2.87.

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