Question

Assume a Poisson random variable has a mean of 5 successes over a 125-minute period.

a. Find the mean of the random variable, defined by the time between successes
b. What is the rate parameter of the appropriate exponential distribution? (Round your answer to 2 decimal places.)
Rate parameter
c. Find the probability that the time to success will be more than 63 minutes. (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)

Answer 1

a

`\mu  =  25`

b

`\lambda  =0.04`

c

`P(X > 65 ) =0.0742`

Explanation:

From the question we are told that

The number of success is n = 5

The time duration is N = 125 minutes

Generally the mean of the random variable is mathematically represented as

`\mu  =  (N)/(n)`

=>
`\mu  =  (125)/(5)`

=>
`\mu  =  25`

Generally the rate parameter is mathematically represented as

`\lambda  =  (1)/(\mu)`

=>
`\lambda  =  (1)/(25)`

=>
`\lambda  =0.04`

Generally the cumulative distribution function for exponential distribution function is

`F(x) = \left \{0 \ \ \ \ \ \  x \le  0}} \atop {1 -e^(-\lambda t )\   x>0}} \right.`

Generally the probability that the time to success will be more than 63 minutes is mathematically represented as

`P(X > 65 ) =  1 - P(X \le 65)`

Here

`P(X \le 65) =  1 -e^(-65 *  \lambda )`

=>
`P(X \le 65) =  1 -e^(-65 *  0.04 )`

=>
`P(X \le 65) = 0.92573`

So

`P(X > 65 ) =  1 - 0.92573`

`P(X > 65 ) =0.0742`