Question

For a group of 100 candidates the mean and standard deviation of their marks were found to be 60 and 15 respectively. Later on it was found that the scores 45 and 72 were wrongly entered as 40 and 27. Find the correct mean and standard deviation.​

Answer 1

Step-by-step explanation:

Correct option is

D

none of these

We know that xˉ=n1∑i=1nxi

⇒∑xi=nxˉ=200×40=8000

⇒σ2=n1∑xi2−xˉ2

⇒∑xi2=n(σ2+xˉ2)=200(225+1600)=365000

Now corrected xi=8000−50+40=7990

Corrected Mean=2007990=39.95

⇒Corrected ∑xi2=365000−(50)2+(40)2

⇒Corrected ∑xi2=364100

⇒Corrected σ2=