Question

6. The mean number of typing errors in a document is known to follow a Poisson distribution w...

6. The mean number of typing errors in a document is known to follow a Poisson distribution with mean 1.5 per page. Find the probability that on a page chosen at random there are

 a) no mistakes, (0.2231) 

 b) more than 2 mistakes (0.1912) 


7. Consider a computer system with Poisson job-arrival stream at an average of 2 per minute. Determine the probability that in any one-minute interval there will be 

a) 0 jobs; (135) 

b) exactly 2 jobs; (18) 

c) at most 3 arrivals (0.8571)

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Answer #1

6)

a)

Here, λ = 1.5 and x = 0
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X = 0)
P(X = 0) = 1.5^0 * e^-1.5/0!
P(X = 0) = 0.2231
Ans: 0.2231

b)

Here, λ = 1.5 and x = 2
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X > 2) = 1 - P(X <= 2).
P(X > 2) = 1 - (1.5^0 * e^-1.5/0!) + (1.5^1 * e^-1.5/1!) + (1.5^2 * e^-1.5/2!)
P(X > 2) = 1 - (0.2231 + 0.3347 + 0.251)
P(X > 2) = 1 - 0.8088 = 0.1912


7)

a)

Here, λ = 2 and x = 0
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X = 0)
P(X = 0) = 2^0 * e^-2/0!
P(X = 0) = 0.135
Ans: 0.135

b)

Here, λ = 1 and x = 2
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X = 2)
P(X = 2) = 1^2 * e^-1/2!
P(X = 2) = 0.18
Ans: 0.18


c)

Here, λ = 2 and x = 3
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X <= 3).
P(X <= 3) = (2^0 * e^-2/0!) + (2^1 * e^-2/1!) + (2^2 * e^-2/2!) + (2^3 * e^-2/3!)
P(X <= 3) = 0.1353 + 0.2707 + 0.2707 + 0.1804
P(X <= 3) = 0.8571

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