Column 1: the class intervals

Find the minimum and the maximum of the data. The difference between these gives the range of the data. Choose a number of class intervals (usually up to about 20) so that the class range can be Chosen as some multiple of 10, 100, 1000, etc. (like the class range of 100 above). The data range divided by the class range and then rounded up to the integer above gives the number of classes. The class intervals are chosen so that the lowest class minimum is less than the minimum data value. In our example, the lowest class minimum has been chosen as 900. Add the class range of 100 to find the class interval, for example, 900–1000. Carryon adding the class range to find the next class interval until you pass the maximum value.

Column 2: the class midpoint

The class midpoint is found by the maximum value in the class interval – the minimum value in the class interval divided by 2. For the interval 1300–1400, the class midpoint is (1400 – 1300)/2 = 1350.

The class midpoint is taken as a representative value for the class. That is, for the sake of simplicity we treat the data in the class 1300–1400 as

though there were 47 values at 1350. This is an approximation that is not too serious if the classes are not too wide.

Column 3: class frequency and the total number in the sample

The class frequency is found by counting the number of data values that fall within that class range. For instance, there are 47 values between 1300 and 1400. You have to decide whether to include values that fall exactly on the class boundary in the class above or the class below. It does not usually matter which you choose as long as you are consistent within the whole data set.

The total number of values in the sample is given by the sum of the frequencies for each class interval.

Column 4: f_ix_i and the mean of the sample, $\overline{x}$

This is the product of the frequency and the class midpoint. If the class midpoint is taken as a representative value for the class then f_ix_i gives the sum of all the values in that class. For instance, in the eighth class, with class midpoint 1650, there are 190 values. If we say that each of the values is approximately 1650 then the total of all the values in that class is 1650 × 190 = 313 500. Summing this column gives the approximate total value for the whole sample= 1 682 700 h. If we only used one light bulb at a time and changed it when it failed then the 1000 light bulbs would last approximately 1 682 700 h. The mean of the sample is this total divided by the number in the sample, 1000. Giving 1 682 700/1000 = 1682.7 h. The mean is a measure of the central tendency. That is, if you wanted a simple number to sum up the lifetime of these light bulbs you would say that the average life was 1683 h.

Column 5: the cumulative frequency

The cumulative frequency gives the number that falls into the current class interval or any class interval that comes before it. You could think of it as ‘the number so far’ function. To find the cumulative frequency for a class, take the number in the current class and add on the previous cumulative frequency for the class below, for example, for 1900–2000 we have a frequency of 92. The cumulative frequency for 1800–1900 is 859. Add 859 + 92 to get the cumulative frequency of 951. That is, 951 light bulbs in the sample have a lifetime below 2000 h. Notice that the cumulative frequency of the final class must equal the total number in the sample. This is because the final class must include the maximum value in the sample and all the others have lifetimes less than this.

Column 6: (X_i – $\overline{x}$)², the squared deviation

This column is used in the process of calculating the standard deviation (see column 7). (x_i– $\overline{x}$) represents the amount that the class midpoint differs from the mean of the sample. If we want to measure how spread out around the mean the data are, then this would seem like a useful number. However, adding up (x_i–$\overline{x}$) for the whole sample will just give zero, as the positive and negative values will cancel each other. Hence, we take the square, so that all the numbers are positive. For the sixth class interval the class midpoint is 1450; subtracting the mean, 1682.7, gives – 232.7, which when squared is 54149.2. This value is the square of the deviation from the mean, or just the squared deviation.

Column 7: f_i(x_i − $\overline{x}$)², the variance and the standard deviation

For each class we multiply the frequency by the squared deviation, calculated in column 6. This gives an approximation to the total squared deviation for that class. For the sixth class we multiply the squared deviation of 54 149 by the frequency 103 to get 5 577 347. The sum of this column gives the total squared deviation from the mean for the whole sample. Dividing by the number of sample points gives an idea of the average squared deviation. This is called the variance. It is found by summing column 7 and dividing by 1000, the number in the sample, giving a variance of 39 120. A better measure of the spread of the data is given by the square root of this number, called the standard deviation and usually represented by σ. Here $\sigma =\sqrt{39?120}\approx ?198$.

Column 8: the relative frequency

If instead of 1000 data values in the sample we had 2000, 10 000, or 500, we might expect that the proportion falling into each class interval would remain roughly the same. This proportion of the total number is called the relative frequency and is found by dividing the frequency by the total number in the sample. Hence, for the third class, 1100–1200, we find 5/1000 = 0.005.

Column 9: the cumulative relative frequency

By the same argument as above we would expect the proportion with a lifetime of less than 1900 h to be roughly the same whatever the sample size. This cumulative relative frequency can be found by dividing the cumulative frequency by the number in the sample. Notice that the cumulative relative frequency for the final class interval is 1. That is, the whole of the sample has a lifetime of less than 2000 h.

The data can be represented in a histogram as in Figure 21.3, which gives a simple pictorial representation of the data. The right-hand side has a scale equal to the left-hand scale divided by the number in the sample. These readings, therefore, give the relative frequencies and the cumulative relative frequency as given in Table 21.1.

We may want to sum up our findings with a few simple statistics. To do this we use a measure of the central tendency and the dispersion of the data, which are calculated as already described above. These are summarized below.

Central tendency – the mean

The most commonly used measure of the central tendency is the mean, $\overline{x}$

$\overline{x}=\frac{1}{n}\sum _i{f}_i{x}_i=\sum _i\frac{f_i{x}_i}{n}$

where x_iis a representative value for the class and f_iis the class frequency. The summation is over all classes.

If the data have not been grouped into classes then the mean is found by summing all the individual data values and dividing by the number in the sample:

$\overline{x}=\frac{1}{n}\sum _{i}xi$

where the summation is now over all sample values

Dispersion – the standard deviation

The variance is the mean square deviation given by

$\upsilon ={\sigma }^2=\frac{1}{n}\sum _{i}fi{\left(x_i-\overline{x}\right)}^2=\sum _i\frac{fi}{n}{\left(x_i-\overline{x}\right)}^2$

where $\overline{x}$ is the sample mean, x_iis a representative value for the class, f_iis the class frequency, and the summation is over all classes. This value is the same as

$\upsilon ={\sigma }^2=\frac{1}{n}\sum _i{f}_i{x}_i^2-{\overline{x}}^2$

which can some times be quicker to calculate.

For data not grouped into classes we have

$\upsilon ={\sigma }^2=\frac{1}{n}\sum _i\left(x_i-{\overline{x}}^2\right)$

or

$\upsilon ={\sigma }^2=\frac{1}{n}\sum _i{x}_i^2-{\overline{x}}^2$

where the summation is now over all sample values.

The standard deviation is given by the square root of the variance:

$\sigma =\sqrt{\frac{1}{n}\sum _i{f}_i{\left(x_i-\overline{x}\right)}^2}.$

We have already calculated the average lifetime and the standard deviation of the lifetimes of the sample of light bulbs. We repeat the calculation below. The mean is given by

$\begin{array}{l}\text{class}?\text{midpoint}\hfill \\ ?????????\downarrow \hfill \\ (2\times 950+0\times 1050+5\times 1150+23\times 1250+47\times 1350+103\times 1450\hfill \\ ??\uparrow \hfill \\ \text{class}?\text{frequency}\hfill \\ +160\times 1550+190\times 1650+165\times 1750+164\times 1850+92\times 1950\hfill \\ +49\times 2050)/1000=\mathrm{1682.7.}\hfill \end{array}$

The variance is given by

$\begin{array}{l}\text{class midpoint}??????\text{sample}?\text{mean}\hfill \\ ?????????????????\downarrow ??????????????????????\downarrow \hfill \\ (2?\times {\left(950-1682.7\right)}^2+0\times {\left(1050-1682.7\right)}^2+5\times {\left(1150-1687.7\right)}^2\hfill \\ ??\uparrow \hfill \\ \text{class}?\text{frequency}\hfill \\ +23\times {\left(1250-1682.7\right)}^2+47\times {\left(1350-1682.7\right)}^2\hfill \\ +103\times {\left(1450-1682.7\right)}^2+160\times {\left(1550-1682.7\right)}^2\hfill \\ +190\times {\left(1650-1682.7\right)}^2+165\times {\left(1750-1682.7\right)}^2\hfill \\ +164\times {\left(1850-1682.7\right)}^2+92\times {\left(1950-1682.7\right)}^2\hfill \\ +49\times {\left(2050-1682.7\right)}^2/1000=39?120420/1000\approx \mathrm{39120.4.}\hfill \end{array}$

Hence, the standard deviation is

$\sqrt{39120.4}\approx 198.$

Can we agree with the manufacturer's claim that the light bulbs have a lifetime of 1500 h? The number with a lifetime of less than 1500 h is 180 (the cumulative frequency up to 1500 h). This represents only 0.18, less than 20% of the sample. However, this is probably an unreasonable number to justify the manufacturers claim as nearly 20% of the customers will get light bulbs that are not as good as advertised. We might accept a small number, say 5%, failing to live up to a manufacturer's promise. Hence, it would be better to claim a lifetime of around 1350 h with a relative cumulative frequency between 0.03 and 0.077. Approximately 0.053 or just over 5% would fail to live up to the manufacturer's amended claim.

Example 21.1

A sample of 2000-Ω resistors were tested and their resistances were found as below:

$\begin{array}{lllllllllllll}1997\hfill & 1998\hfill & 2004\hfill & 2002\hfill & 1999\hfill & 2000\hfill & 2001\hfill & 2002\hfill & 1999\hfill & 1998\hfill & 1997\hfill & 1999\hfill & 2001\hfill \\ 2003\hfill & 2005\hfill & 1996\hfill & 2000\hfill & 2000\hfill & 1998\hfill & 1999\hfill & 1998\hfill & 2001\hfill & 2003\hfill & 2002\hfill & 1996\hfill & 2002\hfill \\ 1995\hfill & 2000\hfill & 2000\hfill & 1999\hfill & 1997\hfill & 2004\hfill & 2001\hfill & 1999\hfill & 1999\hfill & 1998\hfill & 2002\hfill & 2003\hfill & 2002\hfill \\ 2001\hfill & 1999\hfill & 1998\hfill & 1997\hfill & 2002\hfill & 2001\hfill & 2000\hfill & 1999\hfill & 2001\hfill & 1999\hfill & 1998\hfill & 2000\hfill & 1999\hfill \\ 1998\hfill & 2000\hfill & 2002\hfill & 1999\hfill & 2001\hfill & 2000\hfill & 2001\hfill & 2002\hfill & 2004\hfill & 1996\hfill & 2000\hfill & 2002\hfill & 2000\hfill \\ 1998\hfill & 2000\hfill & 2005\hfill & 2000\hfill & 1999\hfill & 2000\hfill & 2000\hfill & 2001\hfill & 1999\hfill & 2001\hfill & 1997\hfill & 2002\hfill & 2001\hfill \\ 2004\hfill & 2000\hfill & 2000\hfill & 1999\hfill & 1998\hfill & 2001\hfill & 1998\hfill & 2000\hfill & 2002\hfill & 1998\hfill & 1998\hfill & 1999\hfill & 2002\hfill \\ 1999\hfill & 2001\hfill & 2002\hfill & 2006\hfill & 2000\hfill & 2001\hfill & 2001\hfill & 2002\hfill & 2003\hfill & \hfill & \hfill & \hfill & \hfill \end{array}$

Group the data in class intervals and represent it using a table and a histogram. Calculate the mean resistance and the standard deviation.

Solution We follow the method of building up the table as described previously.

Column 1: class intervals. To decide on class intervals look at the range of values presented. The minimum value is 1995 and the maximum is 2006. A reasonable number of class intervals would be around 10, and in this case as the data are presented to the nearest whole number, the smallest possible class range we could choose would be 1 Ω. A 1-Ω class range would give 12 classes, which seems a reasonable choice. If we choose the class midpoints to be the integer values then we get class intervals of 1994.5–1995.5, 1995.5–1996.5, etc. We can now produce Table 21.2.

Table 21.2

Frequency distribution of resistances of a sample of resistors. N represents the number of classes, n the number in the sample

(1) Resistance (Ω)	(2) Class mid-point xi	(3) Class frequency fi	(4) fi (x)	(5) Cumulative frequency Fi	(6) (xi - $\overline{x}$)2	(7) fi(xi −x)2	(8) Relative frequency fi /n	(9) Relative cumulative frequency Fi/n
1994.5–1995.5	1995	1	1995	1	26.52	26.52	0.01	0.01
1995.5–1996.5	1996	3	5988	4	17.22	51.66	0.03	0.04
1996.5–1997.5	1997	5	9985	9	9.92	49.6	0.05	0.09
1997.5–1998.5	1998	13	25974	22	4.62	60.06	0.13	0.22
1998.5–1999.5	1999	17	33983	39	1.32	22.44	0.17	0.39
1999.5–2000.5	2000	19	38000	58	0.02	0.38	0.19	0.58
2000.5–2001.5	2001	16	32016	74	0.72	11.52	0.16	0.74
2001.5–2002.5	2002	15	30030	89	3.42	51.3	0.15	0.89
2002.5–2003.5	2003	4	8012	93	8.12	32.48	0.04	0.93
2003.5–2004.5	2004	4	8016	97	14.82	59.28	0.04	0.97
2004.5–2005.5	2005	2	4010	99	24.52	47.04	0.02	0.99
2005.5–2006.5	2006	1	2006	100	34.22	34.22	0.01	1
$\begin{array}{c}\frac{100??????200015}{{\sum }_i{f}_i=n????\overline{x}\frac{200015}{100}}\\ ?????\overline{x}\frac{1}{n}{\sum }_i{f}_i{x}_i\\ \overline{x}=2000.15\end{array}?????F_N=n$						$\begin{array}{c}\frac{446.5??????1}{{\sigma }^2=\frac{1}{n}{\sum }_i{f}_i{\left(x_i-\overline{x}\right)}^2????{\sum }_i\frac{f_i}{n}=1}\\ {\sigma }^2=\frac{446.5}{100}\\ \sigma =2.11\end{array}?????\frac{F_N}{n}=1$

The other columns are filled in as follows:

Column 2. To find the frequency, count the number of data values in each class interval. The sum of the frequencies gives the total number in the sample; in this case 100.

Column 3. The cumulative frequency is given by adding up the frequencies so far. The number of resistances up to 1995.5 is 1. As there are three in the interval 1995.5–1996.5, add 1 + 3 = 4 to get the number up to a resistance of 1996.5. The number up to 1997.6 is given by 4 + 5 = 9, etc.

Column's 4, 6, and 7 help us calculate the mean and standard deviation. To find column 4 multiply the frequencies (column 3) by the class midpoints (column 2) to get f_ix_i.

The sum of this column is an estimate of the total if we added all the values in the sample together. Therefore, dividing by the number of items in the sample gives the mean:

$\begin{array}{lll}\text{Mean}\hfill & =\hfill & (1\times 1995+3\times 1996+5\times 1997+13\times 1998\hfill \\ \hfill & \hfill & +17\times 1999+19\times 2000+16\times 2001+15\times 2002\hfill \\ \hfill & \hfill & +4\times 2003+4\times 2004+2\times 2005+1\times 2006)/100\hfill \\ \hfill & =\hfill & 200?015/100=\mathrm{2000.15.}\hfill \end{array}$

To find the standard deviation we calculate the variance first. The variance is the mean squared deviation. Find the difference between the mean and each of the class intervals and square it. This gives column 6. Multiply by the class frequency to get column 7. Finally, add them all up and divide by the total number. This gives the variance:

${\sigma }^2=\frac{446.5}{100}=4.465$

and the standard deviation

$\sigma =\sqrt{446.5}\approx \mathrm{2.11.}$

The relative frequency (given in column 8) is the fraction in each class interval; that is, the frequency divided by the total number. For example, the relative frequency in the class interval 2000.5–2001.5 is 16/100 = 0.16

The cumulative relative frequency is the cumulative frequency divided by the total number n. The cumulative frequency up to 2000.5 is 58. Divided by the total number gives 0.58 for the relative cumulative frequency.

We can sum up this sample by saying that the mean resistance is 2000.15 with a standard deviation of 2.11. These simple figures can be used to sum up how closely the claim that they are resistors of 2000Ω can be justified. We can also picture the frequency distribution using a histogram as in Figure 21.4

Returning to the lifetimes of the light bulbs we might want to ask what average lifetime and standard deviation might lead to less than 5% of the light bulbs having a lifetime less than 1500 h?

To answer this sort of problem we need to build up a theory of statistical models and use probability theory.

Relative frequency and probability

The probability of an event is related to its relative frequency. If I chose a light bulb from the sample in Section 21.2, then the probability of its lifetime being between 1400 and 1500 h is 0.103. This is the same as the relative frequency for that class interval, that is, the number in the class interval divided by the total number of light bulbs in the sample. Here, we have assumed that we are no more likely to pick any one light bulb than any other, that is, each outcome is equally likely. The histogram of the relative frequencies (Figure 21.3 using the right-hand scale) gives the probability distribution function (or simply the probability function) for the lifetimes of the sample of light bulbs.

As an introduction to probability, examples are often quoted involving throwing dice or dealing cards from a pack of playing cards. These are used because the probabilities of events are easy to justify and not because of any particular predilection on the part of mathematicians to a gambling vocation!

Example 21.2

A die has the numbers one to six marked on its sides. Draw a graph of the probability distribution function for the outcome from one throw of the die, assuming it is fair.

Solution If we throw the die 10 000 times we would expect the number of times each number appeared face up to be roughly the same. Here, we have assumed that the die is fair; that is, any one number is as likely to be thrown as any other. The relative frequencies would be approximately 1/6. The probability distribution function, therefore, is a flat function with a value of 1/6 for each of the possible outcomes of 1,2,3,4,5,6. This is shown in Figure 21.5

Notice two important things about the probabilities in the probability distribution for the die:

1. each probability is less than 1;

2. the sum of all the probabilities is 1:

$1/6+1/6+1/6+1/6+1/6+1/6=1.$

Example 21.3

A pack of cards consists of four suits, hearts, diamonds, spades, and clubs. Each suit has 13 cards; that is, cards for the numbers 1 (the ace) to 10 and a Jack, Queen, and King. Draw a graph of the probability distribution function for the outcome when dealing one card from the pack, where only the suit is recorded. Assume the card is replaced each time and the pack is perfectly shuffled.

Solution If a card is selected, the suit recorded hearts, spades, clubs, or diamonds, the card is placed back in the pack and the pack is shuffled. If this is repeated (say 10 000 times) then we might expect that each suit will occur as often as any other, that is, 1/4 of the time. The probability distribution function again is a flat distribution, and has the value of 1/4 for each of the possible four outcomes. The probability distribution for suits is given in Figure 21.6.

Notice again that in Example 21.3 each probability is less than 1 and that the sum of all the probabilities is 1:

$\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=1.$

We have seen that a probability distribution can be represented using a graph. Each item along the x-axis has an associated probability. Probability is a function defined on some set. The set is called the sample space, S, and contains all possible outcomes of the random system. The probability distribution function is often abbreviated to p.d.f.

Some definitions

A trial is a single observation on the random system, for example, one throw of a die, one measurement of a resistance in the example in Section 21.2.

The sample space is the set of all possible outcomes, for example, for the die it is the set {1, 2, 3, 4, 5, 6}, and for the resistance problem it is the set of all possible measured resistances. This set may be discrete or continuous. An event is a set of outcomes. For instance, A is the event of throwing less than 4 and B is the event of throwing a number greater than or equal to 5.

An event is a subset of the sample space S.

Notice that in the case of a continuous sample space an event is also a continuous set, represented by an interval of values. For example, C is the event that the resistance lies in the interval 2000 ± 1.5.

Probability

The way that probability is defined is slightly different for the case of a discrete sample space or a continuous sample space.

Discrete sample space

The outcome of any trial is uncertain; however, in a large number of trials the proportion showing a particular outcome approaches a certain number. We call this the probability of that outcome. The probability distribution function, or simply probability function, gives the value of the probability that is associated with each outcome. The probability function obeys two important conditions:

1. all probabilities are less than, or equal to 1, that is, $0\le p\left(x\right)\le 1,$ where x is any outcome in the sample space;

2. the probabilities of the individual outcomes sum to 1, that is, ∑ p(x) = 1.

Continuous sample space

Considering a continuous sample space. We assign probabilities to intervals. We find the probability of, for instance, the resistance being in the interval 2000 ± 0.5. Hence, we assign probabilities to events and not to individual outcomes. In this case, the function that gives the probabilities is called the probability density function and we can find the probability of some event by integrating the probability density function over the interval. For instance, if we have a probability density function f (x), where x can take values from a continuous sample space, then the probability of x being in the interval a to b is given by, for example,

$p\left(a<x<b\right)={\int }_a^{b}f\left(x\right)\text{d}x.$

The probability density function obeys the condition:

${\int }_{-\infty }^{\infty }f\left(x\right)\text{d}x=1$

that is, the total area under the graph of the probability density function must be 1.

Equally likely events

If all outcomes are equally likely then the probability of an event E from a discrete sample set is given by

$\frac{\text{The}?\text{number}?\text{of}?\text{outcomes}?\text{in}?\text{E}}{\text{The}?\text{total}?\text{number}?\text{of}?\text{outcomes}?\text{in}?\text{S}}.$

That is, the probability of E is equal to the proportion of the whole sample space that is in E, when each of the outcomes are equally likely.

Example 21.4

What is the probability on one throw of a die getting a number less than 3?

Solution Give this event the name A. Then A = { 1, 2} and S = {1,2,3,4,5, 6}. The number in A is 2 and the number in S is 6. Therefore, as each outcomes is equally likely, p(A), the probability of A is 2/6 = 1/3.

This particular result leads to another way of representing probability – by using area.

If we use a rectangle to represent the set of all possible outcomes, S, then an event is a subset of S, that is, one section of the rectangle and if all outcomes are equally likely then its probability can be pictured by the proportion of S that is in A. We will put a dotted line around the picture representing the set A to indicate the number in A or the area of A (see Figure 21.7).

This way of picturing the probability of an event can help in remembering some of the probabilities of combined events.

Disjoint events are events with no outcomes in common. They cannot happen simultaneously.

Example 21.5

A is the event that a card chosen from a playing pack is under 6 (counting ace as low, i.e. = 1) and B is the event of choosing a picture card (Jack, Queen, or King). Find the probability that a card chosen from the pack is under 6 or is a picture card.

Solution

A and B are disjoint. As each outcome is considered to be equally likely the probabilities are easy to find:

$p\left(\text{A}\right)=\frac{\text{Number}?\text{in}?\text{A}}{\text{Number}?\text{in}?\text{S}}.$

There are 52 cards; therefore, 52 possible outcomes in S, so

$p\left(\text{A}\right)=\frac{20}{52}=\frac{5}{13}.$

Also

$p\left(\text{B}\right)=\frac{\text{Number}?\text{in}?\text{B}}{\text{Number}?\text{in}?\text{S}}=\frac{12}{52}=\frac{3}{13}.$

We want to find the probability of A or B happening; that is, the probability that the card is either under 6 or is a picture card:

We can see that:

$\begin{array}{l}p\left(\text{A}\cup \text{B}\right)=\frac{\text{Number}?\text{in}?\text{A}?\cup ?\text{B}}{\text{Number}?\text{in}?\text{S}}=\frac{32}{52}=\frac{8}{13}\hfill \\ \frac{\text{Number}?\text{in}?\text{A}?+?\text{number}?\text{in}?\text{B}}{\text{Number}?\text{in}?\text{S}}=\frac{\text{Number}?\text{in}?\text{A}}{\text{Number}?\text{in}?\text{S}}+\frac{\text{Number}?\text{in}?\text{B}}{\text{Number}?\text{in}?\text{S}}\hfill \\ p\left(\text{A}?\cup ?\text{B}\right)=p\left(\text{A}\right)+p\left(\text{B}\right)\hfill \\ \frac{8}{13}=\frac{5}{13}+\frac{3}{13}.\hfill \end{array}$

We can also see this by using the idea of area to picture it, as in Figure 21.8.

We can also consider the probability of A not happening; that is, the probability of the complement of A, which we represent by A′. As A and A′ are disjoint.

$\begin{array}{l}p(A\prime )+p\left(\text{A}\right)=1\hfill \\ p(A\prime )=1-\left(\text{A}\right).\hfill \end{array}$

This is shown in Figure 21.9.

The simplest sort of trial to consider is one with only two outcomes; for instance, tossing a coin. Each trial has the outcome of head or tail and each is equally likely.

We can picture repeatedly tossing the coin by drawing a probability tree. The tree works by drawing all the outcomes and writing the probabilities for the trial on the branches. Let us consider tossing a coin three times. The probability tree for this is shown in Figure 21.11.

There are various rules and properties we can notice.

Rules of probability trees and repeated trials

1. The probabilities of outcomes associated with each of the vertices can be found by multiplying all the probabilities along the branches leading to it from the top of the tree. HTH has the probability

$\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}$

2. At each level of the tree, the sum of all the probabilities on the vertices must be 1.

3. The probabilities along branches out of a single vertex must sum to 1. For example, after getting a head on the first trial we have a probability of 1/2 of getting a head on the second trial and a probability of 1/2of getting a tail. Together they sum to 1.
The fourth property is one that is only true when the various repeated trials are independent, that is, the result of the first trial has no effect on the possible result of the second trial, etc.

4. For independent trials, the tree keeps repeating the same structure with the same probabilities associated with the branches. Here, the event of getting a head on the third toss is independent of the event of getting a head on the first or second toss.

Using the tree we can find various probabilities, as we see in Example 21.7.

Picking balls from a bag without replacement

We have 20 balls in a bag. Ten are red and ten are black. A ball is picked out of the bag, its colour recorded and then it is not replaced into the bag. There are two possible outcomes of each trial, red (R) or black (B). Let us consider three trials and their associated outcomes in a probability tree, as shown in Figure 21.12.

To find the probabilities we consider how many balls remain in each case. If the first ball chosen is red then there are only 19 balls left of which 9 are red and 10 are black; therefore, the probability of picking a red ball on the second trial is only 9/19 and the probability of picking black is 10/19.

The first three rules given in the last example apply.

1. The probabilities of outcomes associated with each of the vertices can be found by multiplying all the probabilities along the branches leading to it from the top of the tree. BRB has the probability:

$\frac{10}{20}\times \frac{10}{19}\times \frac{9}{18}=\frac{900}{6840}=\frac{1}{76}.$

2. At each level of the tree the sum of all the probabilities on the vertices must be 1; for example, at the second level we have

$\frac{90}{380}+\frac{100}{380}+\frac{100}{380}+\frac{90}{380}=\frac{9+10+10+9}{38}=\frac{38}{38}=1.$

3. The probabilities along branches out of a single vertex must sum to 1; for example, after picking red on the first trial we have a probability of 9/19 of picking red and 10/19 of picking black. Together they sum to 1.

The fourth property is no longer true as the various repeated trials are not independent. The result of the first trial has an effect on the possible result of the second trial, etc.

Using the tree we can find the probability of various events as in Example 21.8.

Multiplication law of probability

$p\left(\text{A}?\cap ?\text{B}\right)=p\left(\text{A}\right)p\left(\text{B}|\text{A}\right).$

This law applies for any two events A and B. It is the law used in finding the probabilities of the vertices of the probability trees.

Example 21.10

It is known that 10% of a selection of 100 electrical components are faulty. What is the probability that the first two components selected are faulty if the selection is made without replacement?

Solution The probability we are looking for is:

p(first faulty ∩ second faulty)

= p(first faulty) p(second faulty | first faulty)

Here there are only two possibilities at each stage, faulty or not faulty. The probability tree is as shown in Figure 21.15.

Notice that their are only 99 components left after the first trial and whether the first was faulty or not changed the probability that the second is faulty or otherwise. Each branch of the tree, after the first layer, represents a conditional probability.

The answer to our problem is therefore

$\begin{array}{l}p\left(\text{first}?\text{faulty}?\cap ?\text{second}?\text{faulty}\right)\hfill \\ ??=p\left(\text{first}?\text{faulty}\right)p\left(\text{second}?\text{faulty}?|?\text{first}?\text{faulty}\right)\hfill \\ ??=\frac{10}{100}\times \frac{9}{99}=\frac{1}{110}.\hfill \end{array}$

Condition of independence

If two events A and B are independent then

$p\left(\text{B}|\text{A}\right)=p\left(\text{B}\right).$

Notice that the multiplication law changes in this case, as described below.

Multiplication law of probability for independent events

$p\left(\text{A}\cap \text{B}\right)=p\left(\text{A}\right)p\left(\text{B}\right).$

There is one other much-quoted law of probability that completes all the basic laws from which probabilities can be worked out. That is Bayes's theorem and it comes from the multiplication law.

Bayes's theorem

As p(A ∩ B) = p(A) p(B|A) then as A ∩ B = B ∩ A we can also write p(B ∩ A) = p(B) p(A|B) and putting the two together gives p(A) p(B|A) = p(B) p(A|B) or

$p\left(\text{B}|\text{A}\right)=\frac{p\left(\text{A}|\text{B}\right)p\left(\text{B}\right)}{p\left(\text{A}\right)}.$

Bayes's theorem is important because it gives a way of swapping conditional probabilities that may be useful in diagnostic situations where not all of the conditional probabilities can be found directly.

We denote ‘The probability that A does not fail’ as p(A). Then the probability that S fails is p(S’) = 1 – p(S).

As they are in series (see Figure 21.16), S will function if both A and B function: S = A ∩ B. Therefore

$p\left(\text{S}\right)=p\left(\text{A}\cap \text{B}\right)$

as A and B are independent

$\begin{array}{l}p\left(\text{S}\right)=p\left(\text{A}\right)p\left(\text{B}\right)\hfill \\ p(S\prime )=1-p\left(\text{A}\right)p\left(\text{B}\right).\hfill \end{array}$

Example 21.12

An electrical circuit has three components in series (see Figure 21.17). One has a probability of failure within the time of operation of the system of 1/5; the other has been found to function on 99% of occasions and the third component has proved to be very unreliable with a failure once for every three successful runs. What is the probability that the system will fail on a single operation run.

Solution Using the method of reasoning above we call the components A, B, and C. The information we have is

$\begin{array}{c}p(A\prime )=\frac{1}{5}\\ p\left(\text{B}\right)=0.99\\ p(C\prime ):p\left(\text{C}\right)=1:3\end{array}$

The probability we would like to find is p(S’), the probability of failure of the system and we know that the system is in series so S = A ∩ B ∩ C; that is, all the components must function for the system to function. We can find the probability that the system will function and subtract it from 1 to find the probability of failure:

$p(S\prime )=1-p\left(\text{S}\right).$

As all the components are independent,

$p\left(\text{A}\cap \text{B}\cap \text{C}\right)=p\left(\text{A}\right)p\left(\text{B}\right)p\left(\text{C}\right).$

We can find p(A), p(B), and p(C) from the information we are given: p(A) = 1 – p(A’) = 1 −1/5 = 0.8 and p(B) is given as 0.99.

Given that p(C') : p(C) = 1 : 3, C fails once in every 1 + 3 occasions, so p(C') = 1/4 = 0.25 and p(C) = 1 – p(C') = 0.75.

We can now find p(A)p(B)p(C) = 0.8 × 0.99 × 0.75 = 0.594. This is the probability that the system will function, so the probability it will fail is given by 1 – 0.594 = 0.406.

Components in parallel

If a system S consists of two components A and B in parallel, as in Figure 21.18, then S fails if both A and B fail.

Again denote ‘The probability that A does not fail’ as p(A). Then the probability that S fails is

$p(S\prime )=1-p\left(\text{S}\right).$

As they are in parallel, S will function if either A or B functions: S = A ∪ B. Therefore

$p\left(\text{S}\right)=p\left(\text{A}\cup \text{B}\right).$

A and B are independent, but not disjoint. They are not disjoint because both A and B can function simultaneously. From the addition law of probabilities given in Section 21.4

$p\left(\text{A}\cup \text{B}\right)?p\left(\text{A}\right)+p\left(\text{B}\right)-p\left(\text{A}\cap \text{B}\right).$

As A and B are independent,

$p\left(\text{A}\cap \text{B}\right)=p\left(\text{A}\right)+p\left(\text{B}\right).$

Therefore

$\begin{array}{ll}p\left(\text{S}\right)\hfill & =p\left(\text{A}\cup \text{B}\right)=p\left(\text{A}\right)+p\left(\text{B}\right)-p\left(\text{A}\right)p\left(\text{B}\right)\hfill \\ p(S\prime )\hfill & =1-\left(p\left(\text{A}\right)+p\left(\text{B}\right)-p\left(\text{A}\right)p\left(\text{B}\right)\right)\hfill \\ \hfill & =1-p\left(\text{A}\right)-p\left(\text{B}\right)+p\left(\text{A}\right)p\left(\text{B}\right).\hfill \end{array}$

As this is a long-winded expression it may be more useful to look at the problem the other way round. S fails only if both A and B fail: S' = A' ∩B'. So

$p(S\prime )=p\left(A\prime \cap B\prime \right)=p(A\prime )p(B\prime )$

as A and B are independent. Finally, p(S′) = p(A’) p(B’).

This is a simpler form that we may use in preference to the previous expression we derived. It must be equivalent to our previous result, so we should just check that by substituting p(A') = 1 – p(A) and p(B') = 1 – p(B). Hence

$\begin{array}{l}p(S\prime )=p(A\prime )p(B\prime )\hfill \\ p(S\prime )=\left(1-p\left(\text{A}\right)\right)\left(1-p\left(\text{B}\right)\right)\hfill \\ p(S\prime )=1-p\left(\text{A}\right)-p\left(\text{B}\right)+p\left(\text{A}\right)p\left(\text{B}\right)\hfill \end{array}$

which is the same as we had before.

Let us try a mixed example with some components in series and others in parallel.

Example 21.13

An electrical circuit has three components, two in parallel, components A and B, and one in series, component C. They are arranged as in Figure 21.19. Component's A and B are identical components with a 2/3 probability of functioning and C has a probability of failure of 0.1%. Find the probability that the system fails.

Solution We denote ‘The probability that A functions’ as p(A). The information we have is

$\begin{array}{l}p\left(\text{A}\right)=\frac{2}{3}\hfill \\ p\left(\text{B}\right)=\frac{2}{3}\hfill \\ p(C\prime )=0.001?\Rightarrow ?p\left(\text{C}\right)=\mathrm{0.999.}\hfill \end{array}$

The probability we would like to find is p(S‘), the probability of failure of the system and we know that the system will function if (either A or B function) and C functions. So

$S=\left(\text{A}\cup \text{B}\right)\cap \text{C}\text{.}$

From the multiplication law for independent events we have:

$p=\left(\left(\text{A}\cap \text{B}\right)\cap \text{C}\right)=p\left(\text{A}\cup \text{B}\right)p\left(\text{C}\right)\text{.}$

From the addition law for non-disjoint events (A and B are not disjoint because both A and B can occur), we have:

$p\left(\text{A}\cup \text{B}\right)=p\left(\text{A}\right)p\left(\text{B}\right)-p\left(\text{A}\right)p\left(\text{B}\right)$

$p\left(\text{A}\cup \text{B}\right)=\frac{2}{3}+\frac{2}{3}-\frac{2}{3}\times \frac{2}{3}=\frac{8}{9}\approx \mathrm{0.8889.}$

So p(S) = p (A ∪ B) p(C) = 0.8889 × 0.999 = 0.888. Hence, the probability that the system functions is 0.888.

Finding probabilities from a continuous graph

Before we look at the normal distribution in more detail we need to find out how to relate the graph of a continuous function to our previous idea of probability. In Section 21.3, we identified the probability of a class with its relative frequency in a frequency distribution. That was all right when we had already divided the various sample values into classes. The problem with the normal distribution is that is has no such divisions along the x-axis and no individual class heights, just a nice smooth curve.

To overcome this problem we define the probability of the outcome lying in some interval of values, as the area under the graph of the probability function between those two values as shown in Figure 21.22.

As we found in Chapter 7, the area under a curve is given by the integral; therefore, for a continuous probability distribution, f (x), we define

$p\left(x?\text{lies}?\text{between}?a?\text{and}?b\right)={\int }_a^{b}f\left(x\right)\text{d}x.$

The cumulative distribution function gives us the probability of this value or any previous value (it is like the cumulative relative frequency). A continuous distribution thus becomes the ‘the area so far’ function and therefore becomes the integral from the lowest possible value that can occur in the distribution up to the current value.

The cumulative distribution up to a value a is represented by

$F\left(a\right)={\int }_{-\infty }^{a}f\left(x\right)\text{d}x$

and it is the total area under to graph of the probability function up to a; this is shown in Figure 21.23.

We can also use the cumulative distribution function to represent probabilities of a certain interval. The area between two values can be found by subtracting two values of the cumulative distribution function as in Figure 21.24.

However, there is a problem with the normal distribution function in that is not easy to integrate! The probability density function for x, where x is N(μ σ²) is given by

$f\left(x\right)=\frac{1}{\sqrt{2\pi \sigma }}{\text{e}}^{-{\left(x-\mu \right)}^2/2{\sigma }^2}.$

It is only integrated by using numerical methods. Hence, the values of the integrals can only be tabulated. The values that we have tabulated are the areas in the tail of the standardized normal distribution; that is

${\int }_u^{\infty }f\left(z\right)\text{d}z$

where f (z) is the probability distribution with 0 mean (μ = 0) and standard deviation of 1 (σ = 1). This is shown in Figure 21.25 and tabulated in Table 21.3. In order to use these values we need to use ideas of transformation of graphs from Chapter 2 to transform any normal distribution into its standardized form.

The standardized normal curve

The standardized normal curve is obtained from the normal curve by the substitution z = (x – μ) /σ and it converts the original distribution into one with zero mean and standard deviation 1. This is useful because we can use a table of values for z given in Table 21.3 to perform calculations.

Finding the probability that x lies between a given range of values

Supposing we have decided that a sample of resistors have a mean of 10.02 and a standard deviation of 0.06, then what percentage lie inside an acceptable tolerance of 10 ± 0.1?

We want to find the area under the normal curve N(10.02,0.06²) between x = 9.9 and x = 10.1, that is, the shaded area in Figure 21.26.

First, convert the x values to z values, by using z = (x – μ) / σ:

$\begin{array}{lll}x=9.9\hfill & \Rightarrow \hfill & z=\frac{9.9-10.02}{0.06}=-2\hfill \\ x=10.1\hfill & \Rightarrow \hfill & z=\frac{10.1-10.02}{0.06}=1.3333\hfill \end{array}$

So we now want to find the shaded area for z values (which will be the same area as above), shown in Figure 23.27.

In order to use Table 21.3, we need to express the problem in terms of the proportion that lies outside of the tolerance limits, as in Figure 21.28.

We use the table of the standardized normal distribution to find the proportion less than z = – 2. As the curve is symmetric this will be the same as the proportion greater than z = 2. From the table this gives 0.02275.

The proportion greater than z = 1.33 from the table is 0.09176. Hence, the proportion that lies outside of our limits is

$0.02275+0.09176=\mathrm{0.11451.}$

As the total area is 1, the proportion within the limits is 1 – 0.11451 = 0.88549.

Mean and standard deviation of a continuous distribution

We can find the mean and standard deviation of a continuous distribution by using integration to replace the summation over all values. The mean is given by

$\mu =\int xf(x)\text{d}x$

where the integration is over all values in the sample space for x. For the exponential distribution this gives

$\mu =\int x\lambda {\text{e}}^{-\lambda x}\text{d}x.$

which can be found by using integration by parts (Chapter 7) to be $1/\lambda .$

The standard deviation is given by

$\sigma =\sqrt{\int {(x-\mu )}^{2}f(x)\text{d}x}$

where the integration is over all values of x. For the exponential distribution this gives

$\sigma =\sqrt{{\int }_0^{\infty }{(x-1/\lambda )}^2\lambda {\text{e}}^{-\lambda x}\text{d}x.}$

Again using integration by parts, we obtain $\sigma =1/\lambda .$

So we see that the mean is 1/λ, as is the standard deviation for the exponential distribution.

Mean and variance of a single trial

The mean of a discrete distribution can be found by using µ = Σxp (x) and the variance is

${\sigma }^2=\sum {(x-\mu )}^{2}p(x)$

where the summation is over the sample space.

We can use these to find the mean and variance of a single trial with only two outcomes, success or failure. The outcome of success has the value 1 and occurs with probability θ and the outcome of failure has the value 0 with probability 1 – θ. Then, the mean is given by 1 × θ + (1 – θ) × 0 = θ.

The variance of a single trial is given by

$\begin{array}{l}{(1-\theta )}^2\theta +{(0-\theta )}^2(1-\theta )=\theta -2{\theta }^2+{\theta }^3+{\theta }^2-{\theta }^3=\theta (1-\theta ).\\ \uparrow ??\uparrow ??\uparrow ??\uparrow ??\uparrow ??\uparrow \\ x??\mu ??p(1)??x??\mu ??p(0)\end{array}$

The standard deviation is the square root of the variance:

$\sigma =\sqrt{\theta \left(1-\theta \right).}$

The mean and standard deviation of the binomial distribution

The expressions involving a summation over the entire sample space can be used to find the mean and standard deviation of the binomial distribution but they take a bit of manipulation to find. Instead, we can take a short cut and use the fact that each trial is independent. The mean of the union of n trials is given by the sum of the means of the n trials. Similarly (for independent trials only), the variance of the union of the n trials is given by the sum of the variances.

Therefore, the mean of the binomial distribution for n trials is given by the number of trials × mean for a single trial = nθ. The variance is given by nθ (1 – θ) and therefore the standard deviation is

$\sigma =\sqrt{n\theta \left(1-\theta \right).}$