The random variable x is the number of occurrences of an event over an interval of ten minutes. It can be assumed that the probability of an occurrence is the same in any two time periods of an equal length. It is known that the mean number of occurrences in ten minutes is 5.4. The probability that there are more than 5 occurrences in ten minutes is:____.a. .0241.b. .0771.c. .1126.d. .9107.The random variable x satisfies which of the following probability distributions?
A. Normal.B. Poisson.C. Binomial

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