The number of breakdowns of a machine In a manufacturing plant has a Poisson dis

Question

The number of breakdowns of a machine In a manufacturing plant has a Poisson distribution with mean of two breakdowns per month. a) Find the probability that there will be less than 3 breakdowns during the next three months. (keep at least 3 decimal places) b) Assume that the machine will not be used if it has more than 4 breakdowns in two months. What is the probability that 8 out of such 10 machines will not be used during the next two months? Assume machines work independently. (keep at least 3 decimal places)

Explanation / Answer

Q. 10 (a) Here = 2 breakdowns per month

(a) so average number of breakdowns in next 3 months t = 2 * 3 = 6 breakdowns

so Pr( x<3) = Pr(0) + Pr(1) + Pr(2) where x is the number of breakdowns

= (e-t) (tx) / x!

= e-6 * 60 / 0! + e-6 * 61 / 1! +  e-6 * 62/ 2! = 0.062

(b) we have to first calculate probability of the Machine having more than 4 breakdowns two months.

so average number of breakdowns in next 2 months t = 2 * 2 = 4 breakdowns

so Pr( x > 4) = by applying same formula or using poisson calculator

Pr(x>4) = 0.3712

so now we have to calculate that 8 out of 10 such machines will not be used during the next two months. we will use binomial distribution to calculate where n = 10 and p = 0.3712

so Pr( x = 8; 10 ; 0.3712) = BIN(8; 10; 0.3712) = 10C8 (0.3712)8 (1 - 0.3712)2 = 0.0064

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