Posted on Categories Uncategorized   Leave a comment on The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12 . If μ and σ^2 denote the mean and variance of the correct observations respectively, then 15(μ+μ^2+σ^2 ) is equal to

The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12 . If μ and σ^2 denote the mean and variance of the correct observations respectively, then 15(μ+μ^2+σ^2 ) is equal to

Ans. (2521) Sol. Let the incorrect mean be μ’ and standard deviation be σ’ We have μ’=(Σxi)/15=12⇒Σxi=180 As per given information correct Σxi=180-10+12 ⇒μ( correct mean )=182/15 Also σ’=√((Σxi^2)/15-144)=3⇒Σxi ^2=2295 Correct Σxi ^2=2295-100+144=2339 σ^2 ( correct variance )=2339/15-(182×182)/(15×15) Required value =15(μ+μ^2+σ^2 ) =15(182/15+(182×182)/(15×15)+2339/15-(182×182)/(15×15)) =15(182/15+2339/15) =2521

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