The following table contains the probability distribution for the number of traffic accidents (X) daily in a small town:
No of accidents (X)
0
1
2
3
4
5
P(X)
0.15
0.2
B
0.15
0.15
0.15
Determine:
B in the table
P(0 < X < 3)
P(X 3)
P(1 X < 4)
E[X]
Var(X)
To determine the missing probability in the table, we can use the fact that the sum of all probabilities must equal 1.
Sum of all probabilities:
0.15 + 0.2 + B + 0.15 + 0.15 + 0.15 = 1
B = 1 - (0.15 + 0.2 + 0.15 + 0.15 + 0.15)
B = 0.3
Now, let's calculate the probabilities for the given events:
P(0 < X < 3) = P(1 accident) + P(2 accidents) = 0.2 + 0.3 = 0.5
P(X ≥ 3) = P(3 accidents) + P(4 accidents) + P(5 accidents) = 0.15 + 0.15 + 0.15 = 0.45
P(1 ≤ X < 4) = P(1 accident) + P(2 accidents) + P(3 accidents) = 0.2 + 0.3 + 0.15 = 0.65
Expected value (E[X]):
E[X] = Σ(x * P(x)) = (0*0.15) + (1*0.2) + (2*0.3) + (3*0.15) + (4*0.15) + (5*0.15) = 0 + 0.2 + 0.6 + 0.45 + 0.6 + 0.75 = 2.6
Variance (Var(X)):
Var(X) = Σ((x - E[X])^2 * P(x)) = ((0-2.6)^2 * 0.15) + ((1-2.6)^2 * 0.2) + ((2-2.6)^2 * 0.3) + ((3-2.6)^2 * 0.15) + ((4-2.6)^2 * 0.15) + ((5-2.6)^2 * 0.15) = 6.76
Therefore, B = 0.3, P(0 < X < 3) = 0.5, P(X ≥ 3) = 0.45, P(1 ≤ X < 4) = 0.65, E[X] = 2.6, and Var(X) = 6.76.