Question #58982

potholes requiring repairing occur at an average rate of 3.2 potholes per kilometer. what is the probability that no potholes require repair in 5km road and what ia the probability that at most 3 potholes require a 200meter repair
1

Expert's answer

2016-04-08T10:16:04-0400

Answer on Question #58982 – Math – Statistics and Probability

Question

Potholes requiring repairing occur at an average rate of 3.2 potholes per kilometer.

a) What is the probability that no potholes require repair in 5km road and

b) what is the probability that at most 3 potholes require a 200meter repair?

Solution

To find the probability of a specific number of successes in the given number of trials, one should use Poisson distribution.


P(k)=λkeλk!.P(k) = \frac{\lambda^k e^{-\lambda}}{k!}.


a) λ=3.2×5=16\lambda = 3.2 \times 5 = 16;

k=0k = 0;

P(0)=160e160!=1.13×107P(0) = \frac{16^0 e^{-16}}{0!} = 1.13 \times 10^{-7}

b) λ=3.2×0.2=0.64\lambda = 3.2 \times 0.2 = 0.64;

P(k3)=PkiP(k \leq 3) = \sum P_{k_i}

k=0,1,2,3k = 0, 1, 2, 3

P(0)=0.640e0.640!=0.527;P(0) = \frac{0.64^0 e^{-0.64}}{0!} = 0.527;

P(1)=0.641e0.641!=0.337;P(1) = \frac{0.64^1 e^{-0.64}}{1!} = 0.337;

P(2)=0.642e0.642!=0.108;P(2) = \frac{0.64^2 e^{-0.64}}{2!} = 0.108;

P(3)=0.643e0.643!=0.023;P(3) = \frac{0.64^3 e^{-0.64}}{3!} = 0.023;

P(k3)=0.527+0.337+0.108+0.023=0.995.P(k \leq 3) = 0.527 + 0.337 + 0.108 + 0.023 = 0.995.

Answer: a) 1.13×1071.13 \times 10^{-7}; b) 0.995.

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