Question

Network breakdowns are unexpected rare events that occur every 3 weeks, on the average. Compute the probability of more than 4 breakdowns during a 21-week period

Answer 1

Answer:

0.827

Step-by-step explanation:

Data provided in the question:

Probability of breakdown, p = once in 3 weeks i.e $\frac{1}{3}$

number of weeks n = 21

now,

mean, λ = np

=  $\frac{1}{3}\times21$

= 7

P(X > 4) = 1 - ( P(X ≤ 4))

using Poisson distribution

P(X = x) = $\frac{e^{-\lambda}\lambda^x}{x!}$

Thus,

P(X = 0) = $\frac{e^{-7}7^0}{0!}$

= 0.00091

P(X = 1) = $\frac{e^{-7}7^1}{1!}$

= 0.00638

P(X = 2) = $\frac{e^{-7}7^2}{2!}$

= 0.02234

P(X = 3) = $\frac{e^{-7}7^3}{3!}$

= 0.05213

P(X = 4) = $\frac{e^{-7}7^4}{4!}$

= 0.09123

Thus,

P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= 0.00091 + 0.00638 + 0.02234 + 0.05213 + 0.09123

= 0.17299

Therefore,

P(X > 4) = 1 - 0.17299

= 0.827