Probability Query

[Polgi-Wan]

A cell center can process (i.e. deliver within 10 seconds) up to 1800 text messages per minute. If actual average text traffic is 28 messages per second, what is the probability that in a given second, 1 of every 20 text messages will be delayed (i.e. not delivered in 10 sec)?

Could someone pls help me out on this question? If okay, kindly display solution.

Thanks and g00d day! : )

[alkatran]

so 30 msg per sec top (1800/60 duh)
and 28 msg per sec currently

so if it was 40 msg per sec and 30 was top..
40/30
simplify 4/3
reverse 3/4 will be recived
(3/4) / 3 = 1/4 will be probably be delayed

so 28/30
14/15
14/15 / 15 = 0.0622222222
this is 'probably' the answer.
(vb gaveme this as 6.22222222E-02 i might have translated number wrong too)

[galway]

Actually with the given data it is impossible to calculate. According to your numbers the call center can process 30 messages a second. You also give the average rate as 28 messages a second, but all that tells us is that during any given second there is greater then a 50% chance a message will be delivered on time.

In order to determine what you are asking the stadard deviation must be known. Otherwise we really have no idea what the distribution of the data looks like.

[Guv]

I am not sure how to solve this problem, but I can provide some useful information. I hope the following gives you some useful ideas about how to attack your problem.

A Poisson distribution is applicable to this problem. Poisson probabilities are calculated from the following formula.

Probability( k ) = (e^ - Mean) * (Mean^k) / k!

The Mean need not be an integer, but k must be. Not sure if the Gamma function can be used for non integer k.

The above gives the probability of k occurrences, given the Mean number expected.

For the problem proposed here, 28 messages are expected every second (Mean = 28). In this case the Poisson probabilities are as follows.

Probability( k ) = (e^ - 28) * (28^k) / k!

For example.

Probability(35 messages in a given second) = (e^ - 28) * (28^35) / k!

For your problem, you can expect service to be okay if less than 30 messages per second occur, but service will be marginal if exactly 30 messages occur in some second, while more than 30 messages per second should cause a problem.

A program written for my HP calculator gives the following probabilities for your problem.

P( > 30 messages) = .309 651
P( = 30 messages) = .067 738
P( < 30 messages) = .622 610 (No problem about 62% of the time).

I do not know how to determine the specific probability that you requested, namely the following.

. . . what is the probability that in a given second, 1 of every 20 text messages will be delayed (i.e. not delivered in 10 sec)?

Given some reasonable assumptions, the following seems valid.

If 31 messages occur in a given second, one will not be delivered in 10 seconds (1 in 30 fail)
If 32 messages occur in a given second, two will not be delivered in 10 seconds. (1 in 16 fail)
If 33 messages occur in a given second, three will not be delivered in 10 seconds.(1 in 11 fail)

The probabilities associated with the above are as follows.

P( 31 ) = .061 183
P( 32 ) = .053 535
P( 33 ) = .045 424

I do not know how to analyze the situation further. It seem obvious that the system has a chance of catching up after a second in which too many messages arrive, but might get hopelessly far behind if it has 5-10 bad seconds in a row.

A common situation for which Poisson probabilities are used relates to analysis of disease clusters. For example, suppose that leukemia is known to strike 5 people out of every 30,000, and there is a town with a population of 30,000 which has 8 cases. The following are the Poisson probabilities.

P( > 7 ) = .133 372
P( = 7 ) = .104 445
P( < 7 ) = .762 183

The above would suggest that perhaps the town is unlucky. In the absence of some other reason to suspect a causative factor, the medical community would probably ignore this town. However, suppose there were 11 cases, with the following Poisson probabilities?

P( > 10 ) = .013 695
P( = 10 ) = .018 133
P( < 10 ) = .968 172

The above might result in the medical people doing some serious investigation of the town to hunt for a causative factor. Not sure how they would react if no evidence of a causative factor were found. Note that about 1 in 73 towns are expected to have luck as bad as the above.