Page 155 - Applied statistics and probability for engineers

Section 4-8/Exponential Distribution 133

4-108. A set of 200 independent patients take antiacid medi- (b) Approximate the probability that more than 65 cabs pass
cation at the start of symptoms, and 80 experience moderate within a 10-hour day.
to substantial relief within 90 minutes. Historically, 30% of (c) Approximate the probability that between 50 and 65 cabs
patients experience relief within 90 minutes with no medi- pass in a 10-hour day.
cation. If the medication has no effect, approximate the (d) Determine the mean hourly rate so that the probability
probability that 80 or more patients experience relief of symp- is approximately 0.95 that 100 or more cabs pass in a
toms. What can you conclude about the effectiveness of this 10-hour data.
medication? 4-111. The number of (large) inclusions in cast iron follows
a Poisson distribution with a mean of 2.5 per cubic millimeter.
4-109. Among homeowners in a metropolitan area, 75% recy-
Approximate the following probabilities:
cle plastic bottles each week. A waste management company
(a) Determine the mean and standard deviation of the number
services 1500 homeowners (assumed independent). Approxi-
of inclusions in a cubic centimeter (cc).
mate the following probabilities:
(b) Approximate the probability that fewer than 2600 inclu-
(a) At least 1150 recycle plastic bottles in a week
sions occur in a cc.
(b) Between 1075 and 1175 recycle plastic bottles in a week
(c) Approximate the probability that more than 2400 inclu-
4-110. Cabs pass your workplace according to a Poisson pro- sions occur in a cc.
cess with a mean of ive cabs per hour. (d) Determine the mean number of inclusions per cubic mil-
(a) Determine the mean and standard deviation of the number limeter such that the probability is approximately 0.9 that
of cabs per 10-hour day. 500 or fewer inclusions occur in a cc.
4-8 Exponential Distribution

The discussion of the Poisson distribution deined a random variable to be the number of laws
along a length of copper wire. The distance between laws 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 law is detected. As you might expect, the distribution of X can be obtained from knowledge of
the distribution of the number of laws. The key to the relationship is the following concept. The
distance to the irst law exceeds three millimeters if and only if there are no laws within a length
of three millimeters—simple but suficient for an analysis of the distribution of X.
In general, let the random variable N denote the number of laws in x millimeters of wire.
If the mean number of laws is λ per millimeter, N has a Poisson distribution with mean λx.
We assume that the wire is longer than the value of x. Now
(
P X > x) = ( 0 e − x λ x λ ( ) 0 = e − x λ
P N = ) =
Therefore, ! 0
F x ( ) = ( x) = − e −λ x , x ≥ 0
P X ≤
1
is the cumulative distribution function of X. By differentiating F x ( ), the probability density
function of X is calculated to be
f x ( ) = λ e −λ x , x ≥ 0
The derivation of the distribution of X depends only on the assumption that the laws in the wire
follow a Poisson process. Also, the starting point for measuring X does not matter because the
probability of the number of laws in an interval of a Poisson process depends only on the length
of the interval, not on the location. For any Poisson process, the following general result applies.
Exponential
Distribution The random variable X that equals the distance between successive events from a
Poisson process with mean number of events λ > 0 per unit interval is an exponential
random variable with parameter λ. The probability density function of X is
f x ( ) = λ e − x λ for 0 ≤ x < ∞ (4-14)

The exponential distribution obtains its name from the exponential function in the prob-
ability density function. See plots of the exponential distribution for selected values of λ in