In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume that the prob

Question

In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume that the probability that a wafer passes the test is 0.9 and that wafers are independent. Determine the cumulative distribution function of the number of wafers from a lot that pass the test at specified values. Round your answers to three decimal places (e.g. 98.765).

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Kaylee 2 weeks 2021-11-25T15:33:22+00:00 1 Answer 0 views 0

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    2021-11-25T15:34:58+00:00

    Answer:

    P(X=0)= 0.001

    P(X=1)=  0.027

    P(X=2)= 0.243

    P(X=3)= 0.729

    Step-by-step explanation:

    Hello!

    If X represents the variable “number of wafers that pass the test” and three wafers of a lot were tested, X can take the following values:

    0 wafers out of 3 passed the test.

    1 wafer out of 3 passed the test.

    2 wafers out of 3 passed the test.

    3 wafers passed the test.

    If each wafer has a probability of 0.9 of passing the test, that means that there is a 1 – 0.9= 0.1 probability of the wafer not passing the test then you can calculate the probabilities as:

    > P(X=0)= 0.1*0.1*0.1= 0.1³= 0.001

    (“The first didn’t pass the test”, “the second didn’t pass the test” and “the third didn’t pass the test”)

    Since the three wafers are independent you can apply that P(A∩B)=P(A)*P(B), and because the three of them didn’t pass the test, having or not an order of testing of the wafers does not change the fact that you have only one combination of outcomes possible.

    For the next probabilities, we don’t have a given order in the wafers, so to calculate the corresponding probabilities, you have to count all possible combinations:

    >P(X=1)= (0.9*0.1*0.1)+(0.1*0.9*0.1)+(0.1*0.1*0.9)= 3*(0.9*0.1²)= 0.027

    (“The first passed the test”, “the second didn’t pass the test” and “the third didn’t pass the test”) or (“The first didn’t pass the test”, “the second passed the test” and “the third didn’t pass the test”) or (“The first didn’t pass the test”, “the second didn’t pass the test” and “the third passed the test”)

    >P(X=2)= (0.9*0.9*0.1)+(0.9*0.1*0.9)+(0.1*0.9*0.9)= 3*(0.1*0.9²)= 0.243

    (“The first passed the test”, “the second passed the test” and “the third didn’t pass the test”) or (“The first passed the test”, “the second didn’t pass the test” and “the third passed the test”) or (“The first didn’t pass the test”, “the second passed the test” and “the third passed the test”)

    Now for the last observation, it is the same as the first one, since the three passed the test, there is only one possible combination of outcomes:

    >P(X=3)= 0.9*0.9*0.9= 0.9³= 0.729

    (“The first passed the test”, “the second passed the test” and “the third passed the test”)

    I hope it helps!

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