Answer:
a. P(x = 3) = 0.061313
b. The anticipated number of radio blackouts = 4
c. It's unclear when the next radio blackout will occur.
d. P(x = 4) = 0.19537
Step-by-step explanation:
Based on the provided details:
A Poisson process is utilized to model the occurrence of this event:
Minor radio blackouts occur, on average, two times annually.
Therefore:
For question (a)
time t = 1/2 (i.e., six months)
Let x represent the likelihood of 3 incidents during the remaining months of the year.
Then:
P(x = 3) = 0.061313
b).
The expected number of radio blackouts over two years is calculated as follows:
We start with:
t = 2 years
E(x) = λ × t
E(x) = 2 × 2
E(x) = 4
Thus, the predicted number of radio blackouts = 4
c).
The time we must wait until the probability of witnessing the next radio blackout reaches at least 0.5 is as follows:
In this case, time (t) =???
Thus:
P(x =1) = 0.5
Consequently, we are unable to ascertain the probability (50%) of the next radio blackout.
d)
We can compute the probability that the time until the fourth blackout is at most 2 years as follows:
Here;
x =4, t = 2
Thus:
P(x = 4) = 0.19537