For the sheet-metal stamping machine in a certain factory, the time between failures, X1 has a mean between the failures, of 56 hours and a variance of 16 hours. The prepair time X2, has a mean time to repair of 5 hours and a variance of 4 hours.
a) If X1 and X2 are independent , find
the expected value and the variance
of Y=X1+X2, which represents one
operation-repair cycle.
b)Would you expect an operation-repair
cycle to last more than 75 hours?Why?
Work
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a) E(Y)= E(X1+X2)= EX1+EX2= 56+5=61
V(Y)= ... = 20
b) P(Y>75) = P(X1+X2 >75) =
I'm not sure where to go from here, i thought of using complement but i just got stuck.
ans: no, P(Y>75)<=0.1653
To find the expected value and variance of Y = X1 + X2, where X1 represents the time between failures and X2 represents the repair time, you have already correctly calculated the expected value as E(Y) = 61 hours.
To calculate the variance of Y, you can use the property that the variance of a sum of independent random variables is equal to the sum of their individual variances. Since X1 and X2 are independent, you can compute V(Y) as follows:
V(Y) = V(X1) + V(X2)
From the given information, V(X1) = 16 hours and V(X2) = 4 hours. Therefore,
V(Y) = 16 + 4 = 20 hours
Now, let's move on to the second question.
To determine whether an operation-repair cycle would last more than 75 hours, you need to calculate the probability of Y being greater than 75 hours, i.e., P(Y > 75).
One way to approach this is by using the complement rule:
P(Y > 75) = 1 - P(Y ≤ 75)
To calculate P(Y ≤ 75), you can use the fact that Y follows a normal distribution since it is the sum of two independent normal random variables (due to the central limit theorem). However, you haven't provided the distribution information for X1 and X2 (e.g., whether they follow a normal or different distribution), so it's a bit difficult to proceed with an exact calculation.
However, if we assume that X1 and X2 follow approximately normal distributions due to the large sample sizes in real-life scenarios, we can make an approximation.
Using the mean and variance you provided for Y (E(Y) = 61 and V(Y) = 20), you can calculate the z-score for the value 75:
z = (75 - E(Y)) / √V(Y)
Substituting the values, you get:
z = (75 - 61) / √20
simplifying,
z ≈ 2.5
Using a standard normal distribution table or calculator, you can find that P(Z > 2.5) ≈ 0.00621, where Z is a standard normal random variable.
Therefore,
P(Y > 75) = 1 - P(Y ≤ 75) ≈ 1 - 0.00621 ≈ 0.99379
So, the probability of an operation-repair cycle lasting more than 75 hours is approximately 0.99379.
In conclusion, it is unlikely for an operation-repair cycle to last more than 75 hours based on the given distribution assumptions and calculations.