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The number of accidents that occureach day at a certain intersection alongwith the corresponding probabilities areshown.Accidents01234Probability.935.03.02.01.005Find the expected number of accidentseach day.

The Number Of Accidents That Occureach Day At A Certain Intersection Alongwith The Corresponding Probabilities AreshownAccidents01234Probability935030201005Find class=

Sagot :

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.

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