The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. It was found that by mistake one data value was taken as 25 instead of 35 . If α and √β are the mean and standard deviation respectively for correct data, then (α, β) is :

Question

The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. It was found that by mistake one data value was taken as 25 instead of 35 . If α and √β are the mean and standard deviation respectively for correct data, then (α, β) is : (A) (11,26) (B) (10.5,25) (C) (11,25) (D) (10.5,26)

Tardigrade Admin · Accepted Answer

Given: Mean ( barx)=(∑ xi/20)=10 or Sigma x i =200 (incorrect) or 200-25+35=210= Sigma X i (Correct) Now correct overline x =(210/20)=10.5 again given S.D =2.5(σ) σ2=( Sigma xi2/20)-(10)2=(2.5)2 or Sigma x i 2=2125 (incorrect) or Sigma x i 2=2125-252+352 =2725 (Correct) ∴ correct σ2=(2725/20)-(10.5)2 underlineσ2=26 or σ=26 ∴ underlineα=10.5, β=26

Q. The mean and standard deviation of $20$ observations were calculated as $10$ and $2.5$ respectively. It was found that by mistake one data value was taken as $25$ instead of $35.$ If $α$ and $β$ are the mean and standard deviation respectively for correct data, then $(α, β)$ is :

$(11, 26)$

$(10.5, 25)$

$(11, 25)$

$(10.5, 26)$

Q. The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. It was found that by mistake one data value was taken as 25 instead of 35. If α and β​ are the mean and standard deviation respectively for correct data, then (α,β) is :

(11,26)

(10.5,25)

(11,25)

(10.5,26)

Q. The mean and standard deviation of $20$ observations were calculated as $10$ and $2.5$ respectively. It was found that by mistake one data value was taken as $25$ instead of $35.$ If $α$ and $β$ are the mean and standard deviation respectively for correct data, then $(α, β)$ is :

$(11, 26)$

$(10.5, 25)$

$(11, 25)$

$(10.5, 26)$