Question

Arithmetic mean of marks (out of $25$) scored by $10$ students was $15$. One of the student, named Karishma enquired the other $9$ students and find the deviations from her marks are noted as $-8,-6,-3,-1,0,2,3,4,6$. Find Karishma's marks.

Answer 1

We are given,

The Arithmetic mean of $10$ Data is $5$

i.e. $\overline{x}=15$ of ten students.

So, $x_1+x_2+x_3+......+x_9+x_{10}=150\to \left(1\right)$

Let Karishma marks is $x_1$

now, we are given the deviation

of marks from Karishma marks

So, as we know

$di=\overline{x}-xi$, So, $xi=\overline{x}-di$, where $i=2,3,,......,10$ So, we get $x_2=23,x_3=21,......,x_9=11$ &

$x_{10}=9$ So, from $\left(1\right)$

$x_1=150-(x_2+x_3+......+x_9+x_{10})$

$=150-(23+21+18+16+15+13+12+11+9)$

$x_1=150-\left(138\right)=12$

So, Karishna mark is $12$