Originally Posted by **wu_weidong**

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).