Question 1030191: A call centre receives 20 call per hour on an average. Should the centre be employing more people if they expect more than 25 calls in an hour, when the centre cannot take more than 15% of the risk?
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website!
Question:
A call centre receives 20 call per hour on an average. Should the centre be employing more people if they expect more than 25 calls in an hour, when the centre cannot take more than 15% of the risk?
Solution:
In a probability problem, the first step is to identify the underlying probability distribution and then do the calculations accordingly.
The given problem provides a constant average number of calls (20/hour) which the call centre can handle normally. Exceptionally, the call centre "expects" 25+ calls per hour and wishes to know the risk (probability) of that happening.
I interpret that the management would like to know the if probability of 25 or less calls is more than 85%, i.e. ≤15% is maximum risk. (instead of "expecting" 25+ calls, in which case definitely more staff is needed).
On this basis, the problem satisfies the Poisson distribution, which has a probability given by:
P(x)=e^(-μ)*μ^(x)/x!
where
e=base to natural log
μ=mean number of calls per hour = 20
x=number of calls for which probability is to be calculated = 26,....∞
Since it is impossible to sum directly to infinity, we can make use of the fact that area under a distribution curve integrates/sums to 1.
So
P(x>25)=1-P(x≤25)
and calculators and software are available to do the summation.
So
P(x>25)=1-P(x≤25) with parameters μ=20 is
=1-0.887815
=0.112185
< 15%
so the manager could take a chance not to hire more staff for the given period, given instructions that risk should be less than 15%.
Details:
P(0)=2.06115*10^-9
P(1)=4.1223*10^-8
P(2)=4.1223*10^-7
P(3)=2.7482*10^-6
P(4)=1.3741*10^-5
P(5)=5.4964*10^-5
P(6)=1.83213*10^-4
P(7)=5.23467*10^-4
P(8)=0.00130866
P(9)=0.00290815
P(10)=0.0058163
P(11)=0.0105751
P(12)=0.0176251
P(13)=0.0271156
P(14)=0.0387366
P(15)=0.0516488
P(16)=0.064561
P(17)=0.0759541
P(18)=0.0843935
P(19)=0.0888353
P(20)=0.0888353
P(21)=0.084605
P(22)=0.0769136
P(23)=0.0668814
P(24)=0.0557345
P(25)=0.0445876
sum=0.0.887815
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