Normal distribution as an approximation to the Poisson distribution

Figure 4.4indicates that when the mean,μ, equals 5, the shape of the Poisson distribution is similar to that of the normal distribution. When μ equals or

Table 4.4. Data from X-ray counting experiment.

1 1 1 1 1 0 0 6 1 2

0 4 1 1 2 3 1 0 3 3

4 1 1 0 0 2 4 1 3 5

6 0 1 1 4 6 0 0 0 1

1 1 2 2 1 1 0 2 3 1

1 4 0 2 0 0 3 3 2 4

0 2 2 1 1 2 2 0 0 1

1 1 3 2 2 0 0 2 0 1

1 1 2 2 1 0 1 4 0 1

1 1 0 1 0 2 2 0 0 3

4.3 THE POISSON DISTRIBUTION 163

exceeds 5, the normal distribution is preferred to the Poisson distribution when the calculation of probabilities is required. This is due to the fact that summing a large series is tedious, unless computational aids are available. Even with aids like Excel, some summations cannot be determined. When r,μ^rand r! are large, the result of a calculation usingequation 4.9can exceed the numerical range of the computer causing an‘overflow’ error to occur.

Example 5

In an X-ray counting experiment, the mean number of counts in a period of 1 minute is 200. Use the Poisson and normal distributions to calculate the probability that, in a 1 minute period, the number of counts occurring would be exactly 200.

ANSWER

To find the probability using the Poisson distribution, we use equation 4.9 with μ = r = 200, i.e.

PðrÞ ¼μ^re^μ r! ; so that

Pð200Þ ¼200²⁰⁰e^200 200! :

This is where we must stop, as few calculators or computer programs can cope with numbers as large as 200! or 200²⁰⁰.

Using the normal distribution we use the approximation that P(r = 200)≅ P(199.5 ≤ x ≤ 200.5), where x is a continuous random variable.

Take the population standard deviation of the normal distribution to be ffiffiffiμ

p ¼ ffiffiffiffiffiffiffiffip200

¼ 14:142.

Now

Pð199:5 x 200:5Þ ¼ Pð1 x 200:5Þ  Pð1 x 199:5Þ: (4:12) The two terms on the right hand side ofequation 4.12may be determined by using the NORM.DIST() function in Excel.

Entering =NORM.DIST(200.5,200,14.142,TRUE) into a cell in Excel returns the number 0.5141 and entering =NORM.DIST(199.5,200,14.142,TRUE) into another cell returns the number 0.4859.

It follows that P(r = 200) = 0.5141− 0.4859 = 0.0282.

Exercise H

Using the data inexample 5, determine the probability that in any 1 minute time period, the number of counts lies between the inclusive limits of 180 and 230.

4.4 Review

When random processes, such as coin tossing, radioactive decay, or the scat-tering of particles, lead to outcomes which may be counted, we turn to discrete probability distributions to assist in determining the probability that certain outcomes will occur. The starting point for discussing discrete distributions is to consider the binomial distribution. In particular, we found that the binomial distribution is useful for determining probabilities when there are only two possible outcomes from a single trial and the probability of those outcomes is fixed from trial to trial. This requires that we know (or are able to determine experimentally) the probability of a successful outcome occurring in a single trial. In addition, the binomial distribution is helpful in deriving another important distribution, the Poisson distribution. This distribution is appropri-ate when the frequency of occurrence of‘rare’ events is required, such as the decay of a radioactive nucleus. Both the binomial and the Poisson distributions become more‘normal-like’ as the number of trials increases. Owing to the comparative ease with which calculations may be performed using the normal distribution, it is used extensively as an approximation to the binomial and Poisson distributions when determining probabilities or frequencies.

Whether an experiment involves discrete or continuous quantities, the outcome of a measurement or a trial cannot be predicted with complete certainty. We must therefore acknowledge that ‘uncertainty’ is an inherent and very important feature of all experimental work. With the results obtained inchapters 3 and 4, we are in a position to discuss the important topic of uncertainty in measurement.

End of chapter problems

(1) Determine

C_10;5; C15;2; C42;24; C580;290

(suggestion: use the COMBIN() function in Excel).

(2) The screen for a laptop computer is controlled by 480000 transistors. If the probability of a faulty transistor occurring during manufacture of the screen is 2 × 10^–7, determine the probability that a screen will have

(i) one faulty transistor;

(ii) more than one faulty transistor.

If 5000 screens are supplied to a computer manufacturer, how many screens would you expect to have one or more faulty transistors?

END OF CHAPTER PROBLEMS 165

(3) Ten percent of ammeters in a laboratory have blown fuses and so are unable to measure electric currents. An electronics experiment requires the use of 3 ammeters. If 25 students are each given 3 ammeters, how many students would you expect to have:

(i) 3 functioning ammeters;

(ii) 2 functioning ammeters;

(iii) less than 2 functioning ammeters?

(4) An electroencephalograph (EEG) is an instrument which uses small electro-des pressed against the scalp to detect the electrical activity of the human brain.

When an electrode is pressed against the scalp there is a probability of 0.89 that it will make good enough electrical contact to the skin to allow faithful record-ing of brain activity. In a study, 24 electrodes are used and of those, at least 20 must make good electrical contact with the scalp for an acceptable assessment of brain activity to be made.

(i) What is the probability that at least 20 electrodes will make good contact to the scalp?

A new type of electrode is trialled which has a probability of 0.72 of making good contact with the scalp.

(ii) How many electrodes must be pressed against the scalp so that the proba-bility that at least 20 of them will make good contact is the same as in part (i)?

(5) When a specimen is bombarded with high energy electrons (such as occurs in an electron microscope) the specimen emits X-rays. In an experiment, the number of 14.4 keV X-rays emerging from a specimen was counted in 10 successive 100 s time intervals. The data obtained are shown inTable 4.5.

Use the data to estimate:

(i) the population mean for the number of counts in 100 s;

(ii) the population standard deviation.

(6) A thin tape of superconductor is inspected forflaws that would affect its capacity to carry an electrical current. Sixty strips of tape, each of length 1 m, are examined and the number offlaws in each strip is recorded. These are shown in table 4.6.

(i) Calculate the mean number offlaws per metre.

(ii) Assuming the Poisson distribution to be valid, determine the expected frequency for 0, 1, 2, 3, 4, and 5flaws (hint: first calculate P(0), P(1), etc., usingequation 4.9).

(7) An environmental scanning electron microscope (ESEM) is used to image specimens in a‘poor’ vacuum environment. Owing to the high concentration of air molecules in the microscope, electrons emitted from the electron gun in an

ESEM are scattered by the molecules as they travel towards a specimen. An acceptable image of the specimen is produced as long as 20% or more of the electrons reach the specimen unscattered.

Assuming that the Poisson distribution may be used to describe the prob-ability of electron scattering in an ESEM and that 20% of the electrons are unscattered, determine:

(i) the number of times, on average, that an electron is scattered as it travels from electron gun to specimen;

(ii) the probability that an electron will undergo more than 4 scattering‘events’;

(iii) the probability that an electron will undergo two or less scattering‘events’.

(8) The mean number of cyclones to strike a region in Western Australia per year is 0.2. What is the probability that over a period of twenty years, one or more cyclones will strike that region?

Table 4.6. Number offlaws in superconducting tapes.

Number of flaws, N

Number of 1 m strips of tape with N flaws

0 18

1 19

2 12

3 9

4 1

5 1

Table 4.5. Number of X-rays counted in 10 periods each of duration 100 s.

118 131 136 119 131

122 119 129 124 98

END OF CHAPTER PROBLEMS 167