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The number of pits in a corroded steel coupon follows a Poisson distribution with a mean of 6 pits per

Posted on October 23, 2021 By Natishtaylor1p8dirz 3 Comments on The number of pits in a corroded steel coupon follows a Poisson distribution with a mean of 6 pits per

The number of pits in a corroded steel coupon follows a Poisson distribution with a mean of 6 pits per cm2. Let X represent the number of pits in a 1 cm2 area. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.

Find P(X = 2).

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Comments (3) on “The number of pits in a corroded steel coupon follows a Poisson distribution with a mean of 6 pits per”

  1. Expert says:
    October 23, 2021 at 7:52 pm

    i use my brain for prombles like this maybe you could try it

    step-by-step explanation:

    Reply
  2. Expert says:
    October 23, 2021 at 11:13 pm

    i will with my problems and all the other things that need to   be on by the way

    step-by-step explanation:

    Reply
  3. Trackg8101 says:
    October 24, 2021 at 6:44 am

    P(X=2)=0.0446

    Step-by-step explanation:

    If X follows a Poisson distribution, the probability p to have x pits in 1 cm2 is calculated as:

    [tex]p(x)=\frac{e^{-m}*m^x}{x!}[/tex]

    Where m is the mean of pits per cm2, so, replacing m by 6 pits per cm2, we get that the probability is equal to:

    [tex]p(x)=\frac{e^{-6}*6^x}{x!}[/tex]

    Now, the probability P(x=2) that there are 2 pits in a 1 cm2 is calculated as:

    [tex]p(x=2)=\frac{e^{-6}*6^2}{2!}\\p(x=2)=0.0446[/tex]

    Reply

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