A student scores the following marks in five tests 45, 54, 41, 57, 43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is

Options

A. `(10)/(3)`
B. `(10)/(sqrt(3))`
C. `(100)/(sqrt(3))`
D. `(100)/(3)`

Answer

B

Explanation

Let the marks in sixth tests is 'x', so
mean `=(41+45+43+54+57+x)/(6)=48 `(given)
`rArr (240+x)/(6)=48rArr 40+(x)/(6)=48`
`rArr (x)/(6)=8rArr x =48`
Now, standard deviation of these marks
`=sqrt((41^(2)+45^(2)+43^(2)+54^(2)+57^(2)+48^(2))/(6)-48^(2))`
` [because "standard deviation (SD)"=sqrt((Sigma x_(i)^(2))/(6)-(bar(x))^(2))]`
`=sqrt((((41^(2)-48^(2))+(45^(2)-48^(2))+(43^(2)-48^(2))),(+(54^(2)-48^(2))+(57^(2)-48^(2))))/(6))`
`=sqrt((((-7xx89)+(-3xx93)+(-5xx91)),(+(6xx102)+(9xx105)))/(6))`
`=sqrt((945+612-455-279-623)/(6))`
`=sqrt((1557-1357)/(6))=sqrt((200)/(6))=sqrt((100)/(3))=(10)/(sqrt(3))`