E.1 Suppose a machine breaks down occasionally as a result of a particular part

Question

E.1 Suppose a machine breaks down occasionally as a result of a particular part that wears out, and supposed these breakdowns occur randomly and independently. The average number of breakdowns per 8-hour day is four, arid the distribution of the breakdowns does not change. a. Find the probability that no breakdowns will occur during a given 8-hour day. b. Find the probability that at most. two breakdowns will occur during the first hour of the day. c. What is the minimum number of spare parts that management should have on hand on a given 8-hour day if it wants to be at least 90% sure that the machine will not be idle at any time during the day because of a lack of parts? E.2 A boat broker in Florida receives an average of 26 orders per year for an exotic model of cruiser. Assuming that the demand for this model remains the same throughout the year, what is the probability that the broker will receive the following? (Assume there are 5 weeks in a year.) a. exactly 1 order for this model iii a given week. b. exactly 2 orders for this model over a given two-week period. c. exactly 4 orders for this model over a given two-week period.

Explanation / Answer

This is poission distribution,

mean for 8 four is m8 = 4

mean for one hour is m1 = 1/2

No breakdown in 8-hour shift,

P(x=r) = e^(-m8) * (m8)^r / r!

P(x=0) = e^(-4) * 4^0/0! = 0.0183

b.

at most two breakdown in first one hour,

P(x<=2) = P(x=0) + P(x=1) + P(x=2)

P(x<=2) = e^(-1/2) * (1/2)^0/0! + e^(-1/2) * (1/2)^1/1! + e^(-1/2)* (1/2)^2/2! = 0.9856

So for 90% confidence,

manager have to keep seven sparepart.

2.

a.

mean = 26/52 =1/2

P(x=1) = e^(-1/2) * (1/2)^1/1! = 0.3032

b.

mean for 2 week = 26/26 = 1

P(x=2) = e^(-1) * (1)^2 /2! = 0.1839

c.

mean for 2 week =1

P(x=4) = e^(-1) * (1)^4 /4! = 0.01532

mean= 4 x P(x) Cumulative 0 0.018316 0.01831564 1 0.073263 0.09157819 2 0.146525 0.23810331 3 0.195367 0.43347012 4 0.195367 0.62883694 5 0.156293 0.78513039 6 0.104196 0.88932602 7 0.05954 0.94886638 8 0.02977 0.97863657 9 0.013231 0.99186776 10 0.005292 0.99716023

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