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

The mean and standard deviation  of 20 observation  are found to be   10 and 2 respectively .On rechecking  it was found that  an observation  8 was incorrect.Calculate the incorrect  mean and standard deviation  (a)If the wrong iterm was  ommited  (b) If it is replaced by 12

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

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

x^− =((x_1 +x_2 +...+8+...+x_(20) )/(20))=10   Sum=x_1 +x_2 +...+8+...+x_(20) =200  When 8 is ommited  Sum=200−8=192  Number of observation=20−1=19  x^− =((Σx_i )/n)=((192)/(19))=10(2/(19))  When 8 is replaced by 12:  Σx_i =200−8+12=204  x^− =((Σx_i )/n)=((204)/(20))=10(1/5)

$$\overset{−} {{x}}=\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +...+\mathrm{8}+...+{x}_{\mathrm{20}} }{\mathrm{20}}=\mathrm{10} \\ $$$$\:{Sum}={x}_{\mathrm{1}} +{x}_{\mathrm{2}} +...+\mathrm{8}+...+{x}_{\mathrm{20}} =\mathrm{200} \\ $$$${When}\:\mathrm{8}\:{is}\:{ommited} \\ $$$${Sum}=\mathrm{200}−\mathrm{8}=\mathrm{192} \\ $$$${Number}\:{of}\:{observation}=\mathrm{20}−\mathrm{1}=\mathrm{19} \\ $$$$\overset{−} {{x}}=\frac{\Sigma{x}_{{i}} }{{n}}=\frac{\mathrm{192}}{\mathrm{19}}=\mathrm{10}\frac{\mathrm{2}}{\mathrm{19}} \\ $$$${When}\:\mathrm{8}\:{is}\:{replaced}\:{by}\:\mathrm{12}: \\ $$$$\Sigma{x}_{{i}} =\mathrm{200}−\mathrm{8}+\mathrm{12}=\mathrm{204} \\ $$$$\overset{−} {{x}}=\frac{\Sigma{x}_{{i}} }{{n}}=\frac{\mathrm{204}}{\mathrm{20}}=\mathrm{10}\frac{\mathrm{1}}{\mathrm{5}} \\ $$

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

great sir .thanks

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

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