Page 144 - Applied Statistics And Probability For Engineers

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122 CHAPTER 4 CONTINUOUS RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS

4-63. The manufacturing of semiconductor chips produces ment, approximate the probability that at least 1 of the pages
2% defective chips. Assume the chips are independent and in error are in the sample.
that a lot contains 1000 chips. 4-68. Hits to a high-volume Web site are assumed to follow
(a) Approximate the probability that more than 25 chips are a Poisson distribution with a mean of 10,000 per day.
defective. Approximate each of the following:
(b) Approximate the probability that between 20 and 30 chips (a) The probability of more than 20,000 hits in a day
are defective. (b) The probability of less than 9900 hits in a day
4-64. A supplier ships a lot of 1000 electrical connectors. A (c) The value such that the probability that the number of hits
sample of 25 is selected at random, without replacement. in a day exceed the value is 0.01
Assume the lot contains 100 defective connectors. 4-69. Continuation of Exercise 4-68.
(a) Using a binomial approximation, what is the probability (a) Approximate the expected number of days in a year (365
that there are no defective connectors in the sample? days) that exceed 10,200 hits.
(b) Use the normal approximation to answer the result in part (b) Approximate the probability that over a year (365 days)
(a). Is the approximation satisfactory? more than 15 days each have more than 10,200 hits.
(c) Redo parts (a) and (b) assuming the lot size is 500. Is the nor- 4-70. The percentage of people exposed to a bacteria who
mal approximation to the probability that there are no defec- become ill is 20%. Assume that people are independent. Assume
tive connectors in the sample satisfactory in this case?
that 1000 people are exposed to the bacteria. Approximate each
4-65. An electronic office product contains 5000 elec- of the following:
tronic components. Assume that the probability that each (a) The probability that more than 225 become ill
component operates without failure during the useful life of (b) The probability that between 175 and 225 become ill
the product is 0.999, and assume that the components fail (c) The value such that the probability that the number of peo-
independently. Approximate the probability that 10 or more ple that become ill exceeds the value is 0.01
of the original 5000 components fail during the useful life of 4-71. A high-volume printer produces minor print-quality
the product.
errors on a test pattern of 1000 pages of text according to a
4-66. Suppose that the number of asbestos particles in a sam- Poisson distribution with a mean of 0.4 per page.
ple of 1 squared centimeter of dust is a Poisson random variable (a) Why are the number of errors on each page independent
with a mean of 1000. What is the probability that 10 squared cen- random variables?
timeters of dust contains more than 10,000 particles? (b) What is the mean number of pages with errors (one or more)?
4-67. A corporate Web site contains errors on 50 of 1000 (c) Approximate the probability that more than 350 pages
pages. If 100 pages are sampled randomly, without replace- contain errors (one or more).

4-8 CONTINUITY CORRECTION TO IMPROVE
THE APPROXIMATION (CD ONLY)

4-9 EXPONENTIAL DISTRIBUTION

The discussion of the Poisson distribution defined a random variable to be the number of
flaws along a length of copper wire. The distance between flaws is another random variable
that is often of interest. Let the random variable X denote the length from any starting point on
the wire until a flaw is detected.
As you might expect, the distribution of X can be obtained from knowledge of the
distribution of the number of flaws. The key to the relationship is the following concept. The
distance to the first flaw exceeds 3 millimeters if and only if there are no flaws within a length
of 3 millimeters—simple, but sufficient for an analysis of the distribution of X.
In general, let the random variable N denote the number of flaws in x millimeters of wire.
If the mean number of flaws is per millimeter, N has a Poisson distribution with mean x .
We assume that the wire is longer than the value of x. Now,

e x 1 x2 0
P1X x2 P1N 02 e x
0!