The options given don't correspond to the values in the question. The options are from another question where mean is 5.3 and the probability being looked for is for exactly 3 occurrences and 5 occurrences
Answer:
A) 0.6511
B) Poisson distribution
Step-by-step explanation:
We are given;
Mean: μ = 5.4 minutes
Now, we want to find the probability that there are more than 5 occurrences in ten minutes.
This will be written as;
P(x > 5) = 1 - P(x < 5)
P(x < 5) = P(x = 4) + P(x = 3) + P(x = 2) + P(x = 1) + P(x = 0)
We will use Poisson distribution for this. The formula is;
P(x) = [(e^(-μ)) × μ^(x)]/x!
P(0) = [(e^(-5.4)) × 5.4^(0)]/0! = 0.004517
P(1) = [(e^(-5.4)) × 5.4^(1)]/1! = 0.02439
P(2) = [(e^(-5.4)) × 5.4^(2)]/2! = 0.06585
P(3) = [(e^(-5.4)) × 5.4^(3)]/3! = 0.11853
P(4) = [(e^(-5.4)) × 5.4^(4)]/4! = 0.16
Thus;
P(x < 5) = 0.16 + 0.11853 + 0.06585 + 0.004517 = 0.3489
Thus;
P(x > 5) = 1 - P(x < 5) = 1 - 0.3489 = 0.6511
B) like seen from solution A above, we used Poisson distribution formula.
Thus, random variable x satisfies Poisson distribution
of both numbers.</h2>