Answer:
0.12
Explanation:
The expected value of a probability distribution can be obtained using the formula:
[tex]\sum ^n_{i\mathop=1}x_i\cdot P(x_i)[/tex]
Therefore, the expected number of accidents will be:
[tex]\begin{gathered} E(X)=(0\times0.935)+(1\times0.03)+(2\times0.02)+(3\times0.01)+(4\times0.005) \\ =0+0.03+0.04+0.03+0.02 \\ =0.12 \end{gathered}[/tex]
The expected number of accidents each day is 0.12.
