Correct mean and CV

For a group of 100 students, the mean and coefficient of variation of their marks were found to be 60 and 25, respectively. Later, it was found that the scores 45 and 70 were wrongly entered as 40 and 27. Find the corrected mean and coefficient of variation.

Final Answer:

m =  60.48
C =  2.4445

Step-by-step explanation:

$$\begin{gathered} m_1=60 \\ {C}_1=25 \\ n=100 \\ C=\frac{\sigma }{m} \\ m=\sigma /C \\ m=\frac{\Sigma a}{n} \\ m=\frac{n\cdot m_1-40-27+45+70}{n}=\frac{100\cdot 60-40-27+45+70}{100}=60.48 \end{gathered}$$

$$\begin{gathered} {\sigma }_1=m_1\cdot C_1=60\cdot 25=1500 \\ {r}_1={\sigma }_1^2=1500^2=2250000 \\ {r}_1=\Sigma (a-m_1{)}^2 \\ a=m_1+\sqrt{r_1}/n=60+\sqrt{2250000}/100=75 \\ {r}_2=r_1-(40-m_1{)}^2-(27-m_1{)}^2=2250000-(40-60{)}^2-(27-60{)}^2=2248511 \\ b=m_1+\sqrt{r_2}/(n-2)=60+\sqrt{2248511}/(100-2)\doteq 75.3011 \\ {r}_3=(n-2)\cdot (b-m{)}^2+(45-m{)}^2+(70-m{)}^2=(100-2)\cdot (75.3011-60.48{)}^2+(45-60.48{)}^2+(70-60.48{)}^2\doteq 21857.3064 \\ \sigma =\sqrt{r_3}=\sqrt{21857.3064}\doteq 147.8422 \\ C=\frac{\sigma }{m}=\frac{147.8422}{60.48}=2.4445 \end{gathered}$$

Try another example