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:
6.1. B in the table
6.2. P(0 < X < 3)
6.3. P(X 3)
6.4. P(1 X < 4)
6.5. E[X]
6.6. Var(X)
6.1. To find B in the table, we need to sum up the probabilities for each number of accidents and subtract from 1 since the total probability should equal 1:
B = 1 - (0.15 + 0.2 + 0.15 + 0.15 + 0.15) = 1 - 0.8 = 0.2
So, B = 0.2
6.2. P(0 < X < 3) = P(X=1) + P(X=2)
P(0 < X < 3) = 0.2 + 0.15 = 0.35
6.3. P(X ≥ 3) = P(X=3) + P(X=4) + P(X=5)
P(X ≥ 3) = 0.15 + 0.15 + 0.15 = 0.45
6.4. P(1 ≤ X < 4) = P(X=1) + P(X=2) + P(X=3)
P(1 ≤ X < 4) = 0.2 + 0.15 + 0.15 = 0.5
6.5. The expected value E[X] is calculated by multiplying each number of accidents by its probability and summing up the results:
E[X] = (0*0.15) + (1*0.2) + (2*B) + (3*0.15) + (4*0.15) + (5*0.15)
E[X] = 0 + 0.2 + 2B + 0.45 + 0.6 + 0.75
E[X] = 0.2 + 0.45 + 0.6 + 0.75
E[X] = 2B + 2
6.6. The variance Var(X) is calculated as the sum of the squared differences between each number of accidents and the mean, multiplied by their probabilities:
Var(X) = (0-E[X])^2 * P(X=0) + (1-E[X])^2 * P(X=1) + (2-E[X])^2 * B + (3-E[X])^2 * P(X=3) + (4-E[X])^2 * P(X=4) + (5-E[X])^2 * P(X=5)
Substitute the value of E[X] calculated in 6.5 to find Var(X).