Descriptive Statistical Measures of the Binomial Distribution
A measure of central location and a measure of dispersion can be calculated for any random variable that follows a binomial distribution using the following formulae:
Mean:µ =np Standard deviation:σ
=
√
_________
np
(
1
–
p)
5.2 5.2
For Example 5.2 (insurance policy surrender study), where p = 0.20 andn = 10:
the mean (average) number of surrendered policies is (10)(0.2) = 2 policies, on average, out of 10 policies.
the standard deviation would be
√
_____________(10)(0.2)(0.8) = 1.27 policies.
Section 5.9 shows how to use Excel to compute binomial probabilities.
5.5
5.5 Poisson Poisson Probability Probability Distribut Distribution ion
A Poisson process is also a discrete process.
APoisson processPoisson process measures thenumber of occurrences of a particular outcome of a discrete random variable in a predetermined time
,space or volume interval
for which an average number of occurrences of the outcome is known or can be determined.
These are examples of a Poisson process:
the number of breakdowns of a machine in an eight-hou r shift
the number of cars arriving at a parking garage in a one-hour time interval the number of sales made by a telesales person in a week
the number of problems identified at the end of a construction project the number of particles of chorine in one litre of pool water
the number of typing errors on a page of a newspaper.
Applied Business Statistics
In each case, the number of occurrences of a given outcome of the random variable,x, can take on any integer value from 0, 1, 2, 3, … up to infinity (in theory).
The Poisson Question
‘What is the probability of x occurrences of a given outcome being observed in a predetermined time, space or volume interval?’
The Poisson question can be answered by applying thePoisson probability distributionPoisson probability distribution formula:
P(
x) =
____ e – λxλ!x forx = 0, 1, 2, 3 … 5.35.3 Where:λ = the mean number of occurrences of a given outcome of the random variablefor a predetermined time, space or volume interval e = a mathematical constant, approximately equal to 2.71828
x = number of occurrences of a given outcome for which a probability is required (x is the domain, which can be any discrete value from 0 to infinity)
Example 5.3
Example 5.3 Web-based Web-based Marketing StudyMarketing Study
A web-based travel agency uses its website to market its travel products (holiday packages). The agency receives an average of five web-based enquiries per day for its different travel products.
(a) What is the probability that, on a given day, the agency will receive only three web-based enquiries for its travel products?
(b) What is the probability that, on a given day, the travel agency will receiveat most two web-based enquiries for travel packages?
(c) What is the probability that the travel agency will receive more than four web-based enquiries for travel packages on a given day?
(d) What is the probability that the travel agency will receivemore than four web-based enquiries for travel packages in anytwo-day period?
Solution Solution
(a) The random variablex = number of web-based enquiries received per day ‘fits’ the Poisson process for the following reasons:
The random variable is discrete. The agency can receive {0, 1, 2, 3, 4, …} web-based enquiries per day. Any number of enquiries can be received per day.
The random variable is observed in a predetermined time interval (i.e. one day).
The average number of web-based enquiries per day, a, is known. Hereλ = 5.
Find P( x = 3) whenλ (average number of web-based enquiries per day) = 5.
P(x = 3) =e ____ –5.53 3!
= 0.006738 × 20.833 = 0.1404
Thus there is only a 14.04% chance that the travel agency will receive only three web-based enquiries on a given day when the average number of web-based enquiries per day is five.
Chapter 5 – Probability Distributions
(b) In terms of the Poisson question, ‘at most two web-based enquiries’ implies either 0 or 1 or 2 enquiries on a given day.
These possible outcomes (0, 1, 2) are mutually exclusive, and the combined probability can be found using the addition rule of probability for mutually exclusive events. Thus:
P(x ≤ 2) = P(x = 0) + P(x = 1) + P(x = 2)
Each probability is calculated separately using Formula 5.3.
P(x = 0 web-based enquiries):
P(x = 0) = ____ e –550
0! = 0.006738(1) = 0.00674 (Recall 0! = 1.)
P(x = 1 web-based enquiry):
P(x = 1) = ____ e –551
1! = 0.006738(5) = 0.0337 P(x = 2 web-based enquiries):
P(x = 2) = ____ e –552
2! = 0.006738(12.5) = 0.0842
Then P(x ≤ 2) = 0.00674 + 0.0337 + 0.0842 = 0.12464.
Thus there is only a 12.5% chance that at most two web-based enquiries for travel packages will be received by this travel agency on a given day, when the average number of web-based enquiries received per day is five.
(c) The Poisson question requires us to find P(x > 4) whenλ = 5.
Sincex is a discrete random variable, the first integer value ofx above four isx = 5.
Hence the problem beco mes one of finding:
P(x ≥ 5) = P(x = 5) + P(x = 6) + P(x = 7) + …
To solve this problem, the complementary rule of probability must be used. The complement ofx ≥ 5 is allx ≤ 4. Thus:
P(x ≥ 5) = 1 − P(x ≤ 4)
= 1 − [P( x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)]
= 1 − [0.0067 + 0.0337 + 0.0842 + 0.1404 + 0.1755] (Formula 5.3)
= 1 – 0.4405 = 0.5595
Thus there is a 55.95% chance that the travel agency will receive more than four web-based enquiries for their travel packages on a given day, when the average number of web-based enquiries received per day is five.
(d) Notice that the time interval over which the web-based enquiries are received has changed from one day totwo days. Thusλ = 5 per day, on average, must be adjusted toλ = 10 pertwo days, before the Poisson formula can be applied.
Then find P( x > 4) whereλ = 10.
P(x > 4) = P(x ≥ 5) = [P(x = 5) + P(x = 6) + P(x = 7) + … + P(x = ∞)]
Applied Business Statistics
This can be solved using the complementary rule as follows:
P(x > 4) = 1 − P(x ≤ 4)
= 1 − [P( x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)]
Each Poisson probability is calculated separately using Formula 5.3 withλ = 10.
P(x> 4) = 1 − [0.0000454 + 0.000454 + 0.00227 + 0.00757 + 0.01892]
= 1 – 0.02925 = 0.97075
Thus there is a 97.1% chance that the travel agency will receivemore than four web-based enquiries for their travel packages in anytwo-day period when the average number of web-based enquiries received is 10 per two-day period.
A Word of Caution
Always check that the interval for the average rate of occurrence is the same as the predetermined time, space or volume interval of the Poisson question. If not, always adjust the average rate to coincide with the predetermined interval in the question.