Suppose that in the production of 60-ohm radio resistors, nondefective items are those that have a resistance between 58 and 62 ohms and the

Question

Suppose that in the production of 60-ohm radio resistors, nondefective items are those that have a resistance between 58 and 62 ohms and the probability of a resistor’s being defective is The resistors are sold in lots of 200, with the guarantee that all resistors are nondefective. What is the probability that a given lot will violate this guarantee? (Use the Poisson distribution.)

in progress 0
niczorrrr 6 months 2021-07-24T18:53:12+00:00 1 Answers 42 views 0

Answers ( )

    0
    2021-07-24T18:54:55+00:00

    Answer:

    The probability that a given lot will violate this guarantee is 1 - e^{-200x}, in which x is the probability of a resistor being defective.

    Step-by-step explanation:

    In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

    P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

    In which

    x is the number of sucesses

    e = 2.71828 is the Euler number

    \mu is the mean in the given interval.

    The probability of a resistor’s being defective is x:

    This means that \mu = nx, in which n is the number of resistors.

    The resistors are sold in lots of 200

    This means that n = 200, so \mu = 200x

    What is the probability that a given lot will violate this guarantee?

    This is:

    P(X \geq 1) = 1 - P(X = 0)

    In which

    P(X = 0) = \frac{e^{-200x}*(200x)^{0}}{(0)!} = e^{-200x}

    So

    P(X \geq 1) = 1 - e^{-200x}

    The probability that a given lot will violate this guarantee is 1 - e^{-200x}, in which x is the probability of a resistor being defective.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )