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Question Number 155574 by peter frank last updated on 02/Oct/21

$$\mathrm{The}\mathrm{mean}\mathrm{and}\mathrm{standard}\mathrm{deviation}$$$$\mathrm{of}20\mathrm{observation}\mathrm{are}\mathrm{found}\mathrm{to}\mathrm{be}$$$$10\mathrm{and}2\mathrm{respectively}.\mathrm{On}\mathrm{rechecking}$$$$\mathrm{it}\mathrm{was}\mathrm{found}\mathrm{that}\mathrm{an}\mathrm{observation}$$$$8\mathrm{was}\mathrm{incorrect}.\mathrm{Calculate}\mathrm{the}\mathrm{incorrect}$$$$\mathrm{mean}\mathrm{and}\mathrm{standard}\mathrm{deviation}$$$$\left(\mathbf{a}\right)\mathbf{I}\mathrm{f}\mathbf{the}\mathbf{wrong}\mathbf{iterm}\mathbf{was}\mathbf{ommited}$$$$\left(\mathbf{b}\right)\mathbf{If}\mathbf{it}\mathbf{is}\mathbf{replaced}\mathbf{by}12$$

Answered by Rasheed.Sindhi last updated on 02/Oct/21

$$\stackrel{-}{x}=\frac{x_1+x_2+...+8+...+x_{20}}{20}=10$$$$Sum=x_1+x_2+...+8+...+x_{20}=200$$$$When8isommited$$$$Sum=200-8=192$$$$Numberofobservation=20-1=19$$$$\stackrel{-}{x}=\frac{\mathrm{\Sigma }{x}_i}{n}=\frac{192}{19}=10\frac{2}{19}$$$$When8isreplacedby12:$$$$\mathrm{\Sigma }{x}_i=200-8+12=204$$$$\stackrel{-}{x}=\frac{\mathrm{\Sigma }{x}_i}{n}=\frac{204}{20}=10\frac{1}{5}$$

Commented by peter frank last updated on 02/Oct/21

$$\mathrm{great}\mathrm{sir}.\mathrm{thanks}$$

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