A shopkeeper wants to check the average number of cars sold per call. Past record of sales is shown below :

The expected number of cars sold is :

1.2

Given distribution

$X:0,1,2,3$

$P(X):\frac{1}{6},\frac{1}{2},\frac{3}{10},\frac{1}{30}$

Expected value

$E(X)=\sum xP(X)$

$E(X)=0\cdot\frac{1}{6}+1\cdot\frac{1}{2}+2\cdot\frac{3}{10}+3\cdot\frac{1}{30}$

$=\frac{1}{2}+\frac{6}{10}+\frac{3}{30}$

$=\frac{1}{2}+\frac{3}{5}+\frac{1}{10}$

Convert to denominator $10$

$=\frac{5}{10}+\frac{6}{10}+\frac{1}{10}$

$=\frac{12}{10}=\frac{6}{5}$

$=1.2$

The expected number of cars sold is $\frac{6}{5}$ (or $1.2$).