1 the number of machine breakdown per day at a factory...

1. The number of machine breakdown, per day at a factory follows the Poisson distribution. Based on a random observation during 30 days, it was found that the number of machine breakdown was 2.5 machines / day. Please answer the following question: a) Please calculate the expected value of machine breakdown for the period of: i. One day ii. Two days iii. Three days. b) What is the probability of having 2 machine breakdowns in two days period? c) What is the probability of having more than 2 machine breakdowns in one day period? d) What is the probability of having 2 machine breakdowns in the first day and 1 machine breakdown in the second day, for two consecutive days? e) If the loss of each machine breakdown is IDR 1 million, what is the expected value of the loss from machine breakdown per day?

Transcript

00:01 All right, so our dear student, as per that, numerate policy, we can solve only first four subparts, okay? so let x be the number of machine breakdowns per day, okay? therefore, probability distribution of x is a poison of limda, let x is a poison of lambda 2 .5, okay? so therefore, the probability mass function will be equal to.

00:31 Probability of x is equal to e power minus limda limda power x divided by x pectoral where x will be starting bram 0 1 and up to so on okay so let this will be equal to e power minus 2 pi 2 .5 power x divided by x pectoral so this is so let here this will be x poison up person up lambda and e up x will be lambda and the variance of x will be lambda okay so let here in the first one day one day will be equal to 2 .5 which is lambda and second one is two days this will be equal to two times limba which is two times two point five this is is pipe and three days three day will be equal to three times limda which is three times two point five this is seven point five okay now in part b here let let the probability of two machine probability of two machine two machine breakdowns two machine breakdowns in two days in two days period period is nothing but probability of y is equal to two and this will be equal to e power pi power two derived by two pectoral so this is 0 .0 8422 okay now let here in part c probability of more than two more than two machine more than two machine breakdowns in one day period okay in one day period is equal to this will be equal to probability of x is greater than two okay so let me write it so this is one minus probability of x is less than equal to two which will be equal to one minus probability x is equal to zero plus probability of x is equal to 1 plus probability of x is equal to 2 okay so this will be equal to 1 minus 0 .08 208 5 plus 0 .205 plus 0 .252 plus 0 .2565 which is 1 minus 0 .5 4381 -131 this is 0 .456 -186 .9 okay and let here in part d, now we have to point that probability of two machines, probability of two machine breakdown in the past day, in first day and one machine, one machine breakdown, one machine breakdown, one machine breakdown in second day, okay? so this will be equal to probability of x is equal to 2...

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