A shopkeeper wants to check the average number of cars sold per call. Past record of sales is shown below :
| Sales of cars (Units) | 0 | 1 | 2 | 3 |
| Probabilities | $\frac{1}{6}$ | $\frac{1}{2}$ | $\frac{3}{10}$ | $\frac{1}{30}$ |
The expected number of cars sold is :
Answer & explanation
Correct answer: option 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 $1.2$).