0% of those then = 0 and o,=Sy. who bought a competitor's multiplexers had also been on the same It is often necessary to draw conclusions from the order in which training course. During that year the company had 20% of the items are ranked. For example, two customers may rank the styling multiplexer market share. The company wishes to know what per- of a telephone and we need to know if there is any correlation centage of buyers actually went on its training course, in order to between their rankings. This may be done by using the Rank discover the effectiveness of this course. correlation coefficient (R) given by Equation 1.22, where d is the If B denotes that a person bought the company's product and T difference between the two ranks for each item and n is the number that he went on the training course then the problem is to find of items. P(B|T). From the data P(B) = 0.2, P(B) = 0.8. Then from Equation 1.29 expression 1.30 is obtained. (1.22) (1.30) Probability distributions 1/7 Probability distributions 0.001 m = 2 There are several mathematical formulae with well defined charac- m - 4 m = 1 teristics and these are known as probability distributions. If a prob- lem can be made to fit one of these distributions then its solution is simplified. Distributions can be discrete, when the characteristic can only take certain specific values, such as 0, 1, 2, etc., or they can be continuous, where the characteristic can take any value. 0.999 1.9.1 Binomial distribution Expectation The binomial probability distribution is given by Equation 1.31. Figure 1.7 Poisson probability paper +"C,p"q" + (1.31) an event will occur at least m times when the mean (or expected) value np is known. p is the probability of an event occurring, q (=1 - p) is the probability of an event not occurring and n is the number of selec- 1.9.3 Normal distribution tions. The probability of an event occurring m successive times is given The normal distribution represents continuous events and is shown by the binomial distribution as in Equation 1