Answer:
a) 0.30327
b) 0.07582
Step-by-step explanation:
This is a Poisson distribution problem
Poisson distribution probability function is given as
P(X = x) = (e^-λ)(λˣ)/x!
where λ = mean = 3 power surges per hour
x = variable whose probability is required
a) If a single disturbance in 10 minutes will crash the computer, what is the probability of the computer crashing, that is probability of a single disturbance (power surge) in 10 minutes?
λ = 3 disturbances per hour = 1 disturbance per 20 minutes = 0.5 disturbance per 10 minutes.
P(X = x) = (e^-λ)(λˣ)/x!
λ = 0.5 disturbance per 10 minutes
x = 1 disturbance per 10 minutes
P(X = 1) = (e^-0.5)(0.5¹)/1! = (e⁻⁰•⁵)(0.5)/1!
= 0.30327
b) If two disturbances in 10 minutes will crash the computer, what is the probability of the computer crashing, that is probability of two disturbances (power surges) in 10 minutes?
λ = 3 disturbances per hour = 1 disturbance per 20 minutes = 0.5 disturbance per 10 minutes.
P(X = x) = (e^-λ)(λˣ)/x!
λ = 0.5 disturbance per 10 minutes
x = 2 disturbance per 10 minutes
P(X = 2) = (e^-0.5)(0.5²)/2! = (e⁻⁰•⁵)(0.25)/2!
= 0.07582
Hope this Helps!!!
