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Basic Terms In Statistics

April 01, 2012 | Comments: 0 | Views: 187


Before I start with describing some basic terms in statistics, I would stress that I am not actually a statistician, nevertheless, anticipate that I have provided enough guidance here to throw light on specific terminologies. This article summarizes a couple of simple statistics terminologies and tries to expand these with a little word on the best way they might be made use of in practice. The treatment of statistics in risk management is very large and it is not the intention below to present true examples in this area but to ensure that you acquire some idea of their fundamental meaning.

Mean (average or expected value):

We can select any kind of event and determine values that summarize that event. If we were to measure twelve values, they might be:

Values: 3, 5, 5, 6, 7, 9, 10, 11, 12, 12, 15, 16

The total of these values is: 111

The number of values is: 12

Thus, clearly the mean will be 111 divided by 12 = 9.25


(3 + 5 + 5 + 6 + 7 + 9 + 10 + 11 + 12 + 12 + 15 + 16) / 12 = 111/12 = 9.25

The value of 9.25 is the 'expected value' which you could anticipate from assessing a one off value for this event, but of course, in practice there is a spectrum of values. The 'expected values' are accumulative.

If we had 4 separate events and we took quite similar measurements for each one we could finish up with 'expected values' for each event of:

5.3, 8.4, 6.3 and 11.3

If we next considered the result of determining the outcome of the 4 events we might just anticipate a total value of:

(5.3 + 8.4 + 6.3 + 11.3) = 31.3

Once again, the genuine value will be in a range.


We referred earlier to an event wherein we obtained the following values:

Values: 3, 5, 5, 6, 7, 9, 10, 11, 12, 12, 15, 16

If we were to predict an activity with these values we would be assuming that they are all just as likely for calculation of the mean or the 'expected value'. Simply put if there is little possibility of the initial 6 values being measured, then simply the final 6 matter, and so the 'expected value' will be:

(10 + 11 + 12 + 12 + 15 + 16) / 6 = 76/6 = 12.7

Therefore, in practice not only is there a distribution of values, their likelihood of occurring may equally be different.

We might take a simple look at a task in a project. We may wish to understand by how much time it might be delayed before it begins. A project manager could call upon the specialist for his opinion and he might advocate 16 working weeks. The project manager could make use of this detail with regard to his planning. Nevertheless, this estimate is going to be predicated upon distinct assumptions which the project manager ought to challenge.

If the expert is 100 % confident that there will definitely be 16 working weeks hold up that's fine, however, it is not often the case. What we do know is that there will be a setback. The likelihood of this happening is 1, i.e, it will occur. We could, now, take into account other probable instances.

Let us assume that we have a selection of delays (in weeks) each one with a particular likelihood. We might have:


6.........................0.3......................6 x 0.3 = 1.8

16.......................0.5.....................16 x 0.5 = 8.0

20.......................0.2.....................20 x 0.2 = 4.0

The 'contribution' reflects a 'weighted' value. Note that the sum of each of the likelihoods adds up to 1, which will be the chance of the hold up actually occurring.

The expected value this time becomes:

(1.8 + 8.0 + 4.0) = 13.8 working weeks

This is actually the more likely value that the project manager could apply in his plans instead of 16 weeks.

If we considered a series of 4 similar activities we may end up with a complete delay of 13.8 x 4 = 55.2 weeks (if executed in series and not in parallel). If we had used an isolated assessment of 16 weeks the possible complete delay to the project would have been 64 working weeks (approximately 16 % extra).

Whilst assessing a single activity the influence is not much of a problem, but if you are assessing many events the differences can easily accumulate.

The value of 16 working weeks hold up becomes the most likely (highest probability) and is 2.2 working weeks more than the 13.8 working weeks above.

This is because the spread of the values is 'skewed' somewhat. Had there existed a balanced distribution the 'expected value' would have calculated as being the same as the initial estimate, that is, 16 working weeks.


12......................0.25......................12 x 0.25 = 3.0

16......................0.5........................16 x 0.5 = 8.0

20......................0.25.......................20 x 0.2 = 5.0

Above the contributions are:

(3.0 + 8.0 + 5.0) = 16 working weeks

With any luck, the above information has given a little understanding into one of the basic terms in statistics, the 'mean' of a series of values'.

Look out for more terms in other articles soon.

At the same time as writing articles about items like basic terms in statistics, for example, we generate direct jargon free strategies covering a wide range for business and personal use. If you would like more help and advice why not go to Prince2 2009

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