While calculating the mean and variance of 10 redings,a student wrongly used the reading 52 for the correct reading 25.He obtained the mean and variance as 45 and 16 respectively .Find the correct mean and the variance .

Answer

N/a

Explanation

Given, `n=10,barx=45 and sigma^(2)=16`
,`therefore barx=45rArr(Sigmax_(i))/n=45`
`rArr (Sigmax_(i))/(10)=45rArrSigmax_(i)=450`
Corrected `Sigmax_(i)=450-52+25=423 `
`therefore barx=(423)/(10)=42.3`
`rArr sigma^(2)=(Sigmax_(i)^(2))/(n)-((Sigmax_(i))/n)^(2)`
`rArr 16=(Sigmax_(i)^(2))/10-(45)^(2)`
`rArr Sigmax^(2)=10(2025+16)`
`rArr Sigmax_(i)^(2)=20410`
`therefore "Corrected" Sigmax_(i)^(2)=20410-(52)^(2)+(25)^(2)=18331`
and Corrected `sigma^(2)=(18331)/(10)-(42.3)^(2)=43.81`