Handbook of Civil Engineering Calculations

2. Compute the probability that X < 3
   Use the relation P(X) = n^l\em(^\ where e = base of natural logarithms = 2.71828
   Thus, e
   m
   = e
   1
   
   
   • 6
     = 4.95303. Then P(O) = (1.6)°/(4.95303 x 1) = 0.2019; P(I) = (1.6)
     3
     /
     (4.95303 x i) = 0.3230; P(2) = (1.6)
     2
     /(4.95303 x 2) = 0.2584; P(3) = (1.6)
     3
     /(4.95303 x 6)
     = 0.1378. Sum these results to obtain P(X < 3) = 0.9211.
   
   
   
   


3. Compute the probability that X > 3
   P(X > 3) =1-0.9211=0.0789.

Related Calculations: The probabilities in step 2 also can be found by referring
to a table of Poisson probability. The foregoing discussion pertains to an event that occurs
in time, but analogous comments apply to an event that occurs in space. For example, as-
sume that a firm manufactures long rolls of tape. Defects in the tape occur randomly, but
the average number of defects in a 300-m length, as measured across long distances, is
constant. The number of defects in a given length of tape has a Poisson distribution. The
Poisson distribution is an extreme case of the binomial distribution. As the probability
that event E will occur on a single trial becomes infinitesimally small and the number of
trials becomes infinitely large, the binomial distribution approaches the Poisson distribu-
tion as a limit.

COMPOSITE EVENT WITH

POISSON DISTRIBUTION

With reference to the preceding calculation procedure, a counting device is installed to
determine the number of particles emitted. The probability that the device will actually
count an emission is 0.90. Find the probability that the number of emissions counted in a
20-s interval will be 3.

Calculation Procedure:

1. Compute the average number of emissions counted in 20 s
   In the present case, event E is that an emission is counted. This event is a composite of
   two basic events: a particle is emitted, and the device functions properly. Thus, m =
   20(0.08)(0.90) =1.44.


2. Compute the probability thatXs 3
   Use the equation given in the preceding calculation procedure, or P(3) = (1.44)3/
   [eL44(3!)] = 0.1179.

NORMAL DISTRIBUTION

A continuous random variable X has a normal probability distribution with an arithmetic
mean of 14 and a standard deviation of 2.5. Find the probability that on a given occasion
X will assume a value that (a) lies between 14 and 17; (b) lies between 12 and 16.2; (c) is
less than 10.