You have been instructed to quantify and improve the failure rates associated wi

Question

You have been instructed to quantify and improve the failure rates associated with components made in a car assembly plant that’s just coming online (still working out a lot of bugs). Below are the incident rates of faults (flaw or problem, but still functions) and failures for selected systems and subsystems that are known to have the highest rates at the plant. You can assume the failures occur independent of each other (e.g., A wiring failures does not cause an alternator failure).

show your R and R studio code

2. Car Fault Assignment (Software) You have been instructed to quantify and improve the failure rates associated with components made in a car assembly plant that's just coming on (still working out a lot of bugs). Below are the incident rates of faults (flaw or problem, but still functions) and failures for selected systems and subsystems that are known to have the highest rates at the plant. You can assume the failures occur independent of each other (e.g., A wiring failures does not cause an alternator failure Subsystem P(fault) P(fault and failure) System Braking 1500 1/100000 brake pads 11000 12000 calipers 1100 brake line 1/20000 1/1500 master cylinder 1/1500 Drive train Engine 1/400000 head gasket 1/2000 engine block 1/2500 1/800000 Transmission 1/1500 each gear 1/7000 1500 Rear Differential 1/3000 Front Differential 1/550 1/3000 Electrical 160 160 wiring 1100 1150 alternator Note: To input these values into an R dataframe you can use the following syntax car.df-data frame (subsystem c brake pads calipers "brake line" master cylinder head cover engine bloc k" rear diff front .diff gea iring

Explanation / Answer

A. Probability of all the gears having faults = (1/1500)^6
B. Probability of 1 "gear" failure (one of the 2 gear fails) = 2 * (1/7000) * (1 - 1/7000) = 0.0002856735
Probability of 1 "gear" failure (one of the 6 gear fails) = 6 * (1/7000) * (1 - 1/7000)^5 = 0.0008565308
Probability of no "gear" failure (none of the 6 gear fails) = (1/7000)^0 * (1 - 1/7000)^6 = 0.9991432
Probability of any failure = 1 - 0.9991432 = 0.0008568
P(1 failure)/P(any failure) = 0.0008565308/0.0008568 = 0.9996858
So, 99.96% of failures could be avoided.

C. The function is
distribution <- function(size,prob) {
prob1 <- dbinom(1,6,prob)
prob2 <- dbinom(2,6,prob)
prob3 <- dbinom(3,6,prob)
prob4 <- dbinom(4,6,prob)
prob5 <- dbinom(5,6,prob)
prob6 <- dbinom(6,6,prob)
return(size*c(prob1,prob2,prob3,prob4,prob5,prob6))
}

prob.d <- distribution(100000,1/30)
It gives
> prob.d
[1] 1.688160e+04 1.455311e+03 6.691084e+01 1.730453e+00 2.386831e-02 1.371742e-04

prob.d <- distribution(100000,1/30)

D. P(Gear failing | Gear fault) = P(Gear failing and fault)/P(fault) = (1/7000)/(1/1500) = 15/70 = 3/14

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