Normal Distribution

  • Thread starter wu_weidong
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  • #1
31
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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
 

Answers and Replies

  • #2
128
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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).
 
  • #3
31
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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:

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