MPC-1a Random Events - Counting Statistics


OBJECTIVES:

To study and measure the occurrence of random events. This in preparation for study of the absorption of radiation by matter.

INTRODUCTION:

There are many type of events that are controlled by pure random chance: the number of babies born on a particular day of the week, the number of fender benders in Madison a year (assuming that the number of cars in Madison does not change), the number of people (out of one thousand) that feel better after taking a particular kind of pain reliever or the number of murders in a year in New York City.

In all these examples one discovers that, if one repeats the observation, the new number will differ from the preceding one. More importantly the number found can not be predicted using the knowledge of a preceding observation nor can it be used to predict the following one. For example the number of babies born on a particular Monday will usually be different from the number born on the succeeding Monday. Similarly, the number of patients (out of a different one thousand) that feel better after taking a particular kind of pain reliever, or the number of murders in the following year in New York City will be unpredictably different.

EXAMPLE I: AN IMAGINARY EXPERIMENT

$\textstyle \parbox{0.27\linewidth}{
The average number of births per day is $\o...
...from the average $\overline N$\ one can expect a single
observation to lie.
}$ $\textstyle \parbox{0.70\linewidth}{
\begin{center}
Table I: Number of babies b...
...ibution: $\sigma = \sqrt{12.8}= 3.58$\ } \\ \hline
\end{tabular}\end{center} }$

$\textstyle \parbox{0.54\linewidth}{
Making a \lq histogram' of the above data gives$\Rightarrow$
}$$\textstyle \parbox{0.46\linewidth}{
\includegraphics[height=4.8cm]{figs/l104/fnc1a-1.eps} \\
Fig.~1: Distribution of Monday births
}$

What does the figure tell you?

  1. It shows the `distribution' of the number of births, i.e. it shows the `frequency' of a certain number of births: there were two days with fifteen births, and two days with sixteen births, there were no days with less than ten or more than twenty-four births.
  2. It shows that days with a number of births close to the average occur more frequently than days in which the number of births is very different from the average.

EXAMPLE II

$\textstyle \parbox{0.25\linewidth}{
\begin{tabular}{p{1.75in}}
You are in the s...
...surprised to
find some squares with $30$\ or
$50$\ drop marks.
\end{tabular}}$$\textstyle \parbox{0.75\linewidth}{
\begin{center}
Table~II \\ \smallskip
\begi...
...2$} &
$\sigma = \sqrt{476.4/15} = 5.6$\ \\ \hline
\end{tabular}
\end{center}}$

\includegraphics[height=6.cm]{figs/l104/fnc1a-2.eps}
Fig. 2: The rain drop data

Again the value of $ \sigma$ calculated above tells how far from the average $ \overline{N}$ one can expect a single observation to be.

You may argue that the value of $ \sigma$ in the two examples was obtained using many observations; how can one know its value if one does only one observation?

In other words, if I make only one observation, for instance if I find that the number of drops in a single square is N = 35, what can I say about how far from the unknown but true value $ \overline{N}$ it is likely to be?

REQUIRED KNOWLEDGE:

Statistical theory predicts that if one records the counts of m different experiments: N1N2N3N4, ..... Nm and one calculates the average of these $ \overline{{N}}$ = ($ \Sigma_{1}^{m}$Ni)/m the counts will distribute themselves around the average value. Counts close to the average will occur frequently, counts very different from the average will occur infrequently.

If one makes a histogram of the number of times a certain count appears one finds a bell shaped curve called a ``Gaussian distribution'' centered on the average.
The histogram below looks a lot better than the ones you have seen above; this is so because it shows the distribution of many more observations.

\includegraphics[height=6.5cm]{figs/l104/fnc1a-3.eps}
Fig 3: A Gaussian distribution

The histogram shows that counts of about 100 are most frequent, and that counts of 70, or 130 are much less likely to occur; we can think of the ``width'' of the curve i.e. the range of the counts that occur most frequently is about 20.

THE STANDARD DEVIATION

Statistical theory predicts that the width of the curve, the ``standard deviation'' of the distribution is defined as:

$\displaystyle \sigma$ = $\displaystyle \sqrt{{\overline N }}$

The standard deviation $ \sigma$ is a measure of how wide the curve is; about 1/3 of the counts will lie outside the interval $ \overline{N}$ - $ \sigma$ to $ \overline{N}$ + $ \sigma$.
Only about 1/20 of the counts will lie outside the interval $ \overline{N}$ -2$ \sigma$ to $ \overline{N}$ +2$ \sigma$. The value of $ \sigma$ is calculated automatically by your computer, so you do not really have to worry about it. However here goes the formula, you may skip this if you wish.

$\displaystyle \sigma^{2}_{}$ = $\displaystyle \left(\vphantom{\sum_{i=1}^m (N_i - \overline N)^2}\right.$$\displaystyle \sum_{{i=1}}^{m}$(Ni - $\displaystyle \overline{N}$)2$\displaystyle \left.\vphantom{\sum_{i=1}^m (N_i - \overline N)^2}\right)$/m

EXPERIMENT

Instead of collecting data from the hospital, from the department of transportation or the NYPD, of the kind shown in the examples above, we will make our own homemade random distribution using the random decay of long lived radioactive nuclei. You will study the statistics of counting. You will find how the precision with which one can measure a decay rate R depends on the total number of counts, which in this case is proportional to the time interval during which you observe the radiation.
The rate R of the disintegration of the Radioactive sample is obtained by allowing the G-M counter to detect the radiation emitted for a length of time t, and dividing the number N of observed counts by the time: R = N/t.
If you were to repeat the experiment, for the same time interval t, the number of counts would almost certainly not be the same because of the random nature of radioactive decay. How accurately would you then know the real value of the rate? Intuition tells you that a rate measured with a long time interval t is going to be more reliable than one taken with a short time interval. But even so just how reliable would either of these be? A better way to ask this question is:
If I repeat the experiment, and obtain a different new rate, how big, on the average do I expect the difference between the two rates to be?


EQUIPMENT

 $&bull#bullet;$

Geiger-Muller counter$\scriptstyle \dagger$ and stand. This device detects the ($ \beta$ or $ \gamma$) radiation from the radioactive sources.
$\scriptstyle \dagger$

The Geiger Muller tube was first developed by Hans Geiger(1882-1945) who collaborated with Sir Rutherford at Manchester in the early work on radioactivity which later led to the discovery of the atomic nucleus by Rutherford. The counter was then perfected by Muller in 1928 and is now known as the Geiger-Muller (GM) counter. The counter consists of a fine wire along the axis of a gas filled metal tube. The wire is made about 500V positive with respect to the tube. When ionizing radiation, such as $ \gamma$ rays or $ \beta$ particles, enters the counter it breaks up (ionizes) a few gas atoms releasing electrons which are rapidly accelerated toward the positive wire. These electrons collide with other gas molecules, releasing new electrons. Finally, an avalanche of many millions of electrons reaches the wire and produces a voltage pulse large enough to be detected and counted. The whole process can be initiated by a single $ \alpha$, $ \beta$ or $ \gamma$ ray entering the counter.

The Geiger counter is mounted vertically in a stand.
\includegraphics[height=5.2cm]{figs/l104/fnc1a-4.eps}

Figure 4: The Geiger-Muller counter and stand



PRECAUTIONS: The Geiger counter has a very thin window to permit the entry of $ \beta$ radiation. The window is protected by a plastic cap. In this laboratory exercise you will use $ \gamma$ emitting Cobalt sources, the cap should be left on.


PROCEDURE I:


In order to obtain statistics fairly rapidly for different time intervals you will be divided into three groups of students, each group will use different time intervals. You will then compare results at the end.

  1. Click on the Launch NC-1A icon below (web version) to initiate the PASCO DataStudio software window. The screen should show three data windows: a table window, a graph window and a histogram window.

  2. A total of 336 seconds have been allocated for collecting the data. If you wish to verify or change this: 1) CLICK on the Setup icon 2) In the Experimental Setup window CLICK on the Options Icon 3) In the Sampling Options window CLICK on the Automatic Stop tab.

  3. By default the time period between data collection points is set to 12 seconds. To change this: 1) CLICK on the Setup icon 2) In the Experimental Setup window DCLICK on the Geiger Counter icon 3) modify the ``Seconds between Samples" setting.
    $ \Rightarrow$Group #1 should set the count time period to 3 seconds, and will therefore observe the count in 112 3 second intervals.
    $ \Rightarrow$Group #2 should set the count time period to 12 seconds and will therefore observe the count in 28 12 second intervals.
    $ \Rightarrow$Group #3 should set the count time period to 48 seconds and will therefore observe the count in 7 48 second intervals.
    All groups will compare results at the end.

  4. Stack two Cobalt sources on top of each other in slot #2 (the second slit counting from the top).
    Prepare a table as shown below; you will enter your results, and those of other groups in this table.

  5. CLICK on the START icon and then wait until the computer has finished taking data. Enter the mean ( $ \overline{N}$), and the standard deviation in Table III.

  6. To modify the histogram properties (e.g., bin size) DCLICK in the histogram window.

TABLE III

$ \overline{N}$ $ \sigma$ $ \sqrt{{{\overline N}}}$ 100 x $ {{\sigma} \over {\overline N}}$ 100 x $ {{\sigma - \sqrt{{\overline N}}} \over {\sqrt{{\overline N}}}}$
mean std. dev.    rel. unc.
           
           
           
           
           

ANALYSIS OF THE DATA

  1. Enter the data obtained by the other groups in Table III

  2. Calculate the value of $ \sqrt{{\overline N}}$ and enter it in column 3 of the table.

  3. Calculate the value of 100 x $ {{\sigma} \over {\overline N}}$ and enter it in column 4 of the table.

  4. Plot a graph of the relative uncertainty in the determination of the disintegration rate 100 x $ \sigma$/$ \overline{N}$ versus $ \overline{N}$.

  5. If you have time calculate the value of 100 x $ {{\sigma - \sqrt{{\overline N}}} \over {\sqrt{{\overline N}}}}$ and enter it in column 5 of the table.

  6. Plot a graph of the percent error 100 x $ {{\sigma - \sqrt{{\overline N}}} \over {\sqrt{{\overline N}}}}$ versus $ \sqrt{{\overline N}}$.

QUESTIONS

Q1

Does the precision in the measurement of the decay rate depend on on the number of counts? How?
Q2

A pharmaceutical firm tests a new medication by comparing a group of patients who were given the medicine with a group that received a placebo. The results are the following:
of 1000 patients who were given the medicine 750 were cured.
of 1000 patients who were given the placebo 600 were cured.
Is the new medicine effective? Why?
How would your opinion change, if the numbers were 100, 75 and 60?


Michael Winokur 2007-09-07