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The mean and Standard deviation of a sample were found to be 9.5 and 2.5, respectively. Later, an additional observation 15 was added to the original data. Find the S.D. of the 11 observation.(a) 2.6(b) 2.8(c) 2.86(d) 3.24This question was posed to me in exam.My question is from Statistics topic in division Statistics of Mathematics – Class 11 |
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Answer» RIGHT answer is (c) 2.86 The EXPLANATION: \(\frac{1}{n}\) (Σ xi) = 9.5, \(\frac{1}{n}\) (Σ [(xi)^2 – (X)^2] = 6.25, Σ xi = 95 and \(\frac{1}{10}\) Σ[(X)^2] = 96.5 Corrected MEAN = (95 + 15)/11 = 10 Corrected Variance = [(1/11) x (965 + 225)] – 100 = 90/11 ⇒ Standard deviation = √Variance = √(90/11) = 2.86. |
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