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The probability density function of the time to failure of an electronic component in a copier (in hours) is f(x) = (e-x/1000 )/1000 for x > 0. Determine the the number of hours at which 12% of all components have failed. Please enter the answer to 2 decimal places.

Question
The probability density function of the time to failure of an electronic component in a copier (in hours) is f(x) = (e-x/1000 )/1000 for x > 0. Determine the the number of hours at which 12% of all components have failed. Please enter the answer to 2 decimal places.
 
check_circleAnswer
Step 1

The time to failure of an electronic component in copier follows Exponential Distribution with pdf f(x) = (e-x/1000)/1000 for x > 0.

We need to find the value of x ­(number of hours) at which 12% of all component have failed.

Step 2

We have the pdf of exponential distribution but as per the question we need to find the probability that in x hours all components will fail that means we need a cumulative probability till point x.

So, we ...

F(x)1-e
-Ax
Now
F(x)1-e
F(x)1e00Putting the value of lambda)
As per the question we have the given value of F(x) which is equal to 0.12
Therefore
0.12=1-e1000
0.88 e1000
Taking log both sides
In (e)
In (0.88)= 1000
-x
1000x In (0.88) -x
-127.83 -x
x =127.83
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F(x)1-e -Ax Now F(x)1-e F(x)1e00Putting the value of lambda) As per the question we have the given value of F(x) which is equal to 0.12 Therefore 0.12=1-e1000 0.88 e1000 Taking log both sides In (e) In (0.88)= 1000 -x 1000x In (0.88) -x -127.83 -x x =127.83

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