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Normal Distribution

  1. Mar 8, 2006 #1
    Hi all,
    I need help with a problem.

    The lifetimes of interactive computer chips are normally distributed with mean u = 1.4 * 10^6 hours and sigma = 3 * 10^5 hours. What is the approximate probability that a batch of 100 chips will contain at least 20 whose lifetimes are less than 1.8 * 10^6.

    Here is what I did:

    Let X be the lifetime of a chip, and Y be the number of chips whose lifetimes are less than 1.8 * 10^6.

    X ~ N(1.4 * 10^6, (3 * 10^5)^2)
    P(X < 1.8 * 10^6) = P(Z < ((1.8 * 10^6 - 1.4 * 10^6) / 3 * 10^5)) = P(Z < 1.33) = 0.9082

    np = 100 * 0.9082 = 90.82
    np(1-p) = 100 * 0.9082 * 0.0918 = 8.337
    Y ~ N(90.82, 8.337)
    Using normal distribution to approximate binomial distribution,
    P(Y >= 20) = P(Y >= 19.5) (continuity correction) = P(Z >= ((19.5 - 90.82) / sqrt(8.337)) = P(Z >= -24.7)

    At this point, I'm stuck, because I can't find a value for P(Z >= -24.7).

    What have I done wrong?

    Any help is appreciated. Thank you.

    Regards,
    Rayne
     
  2. jcsd
  3. Mar 8, 2006 #2
    Why even use a binomial approximation?
    Let Y denote the number of computer chips whose lifetimes are less than 1.8*10^6. Then Y~Bin(100, 0.9082).
     
  4. Mar 10, 2006 #3
    Because then I would have to calculate 20 terms of (100 C r)*(0.9082)^r * 0.0918^100-r --- from r = 0 to r = 19. I thought using an approximation would make the calculations much less tedious.
     
    Last edited: Mar 10, 2006
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