(H) – scored 1, or tail (T) – scored 0, are mutually exclusive and these are the only events which can happen. If we let X be a random variable that is the number of heads shown on a single toss and is therefore either 1 or 0, then the probability distribution, for X is: probability (H) = ½; probability (T) = ½ and is shown graphically in Figure 4.3a.
Figure 4.3 Examples of probability distributions. (a) Probability distribution for the number of heads (X) shown in one toss of a coin. (b) Probability distribution for the number of heads (Y) shown in two tosses of a coin. (c) Probability distribution for the number of heads in four tosses of a coin. (d) Probability distribution for the number of heads in ten tosses of a coin.
What happens if we toss two coins at once? We now have four possible events: HH, HT, TH, and TT. There are all equally likely and each has probability ¼. If we let Y be the number of heads then Y has three possible values 0, 1, and 2. Y = 0 only when we get TT and has probability ¼. Similarly, Y = 2 only when we get HH, so has probability ¼. However, Y = 1 either when we get HT or TH and so has probability ¼ + ¼ = ½. The probability distribution for Y is shown in Figure 4.3b. The distribution for the number of heads becomes more symmetrical as the number of coin tosses increases (Figures 4.3c and d).
In general, we can think of the tosses of the coin as trials, each of which can have an outcome of success (H) or failure (T). These distributions are all examples of what is known as the Binomial distribution. In addition, we will discuss two more distributions that are the backbone of medical statistics: the Poisson and the Normal. Each of the distributions is known as the probability distribution function, which gives the probability of observing an event. The corresponding formulas are given in Section 4.7. These formulas contain certain constants, known as parameters, which identify the particular distribution, and from which various characteristics of the distribution, such as its mean and standard deviation, can be calculated.
4.2 The Binomial Distribution
If a group of patients is given a new treatment such as acupuncture for the relief of a particular condition, such as tension type headache, then the proportion p being successfully treated can be regarded as estimating the population treatment success rate π (here, π denotes a population value and has no connection at all with the mathematical constant 3.14159). The sample proportion p is analogous to the sample mean
For a fixed sample size n the shape of the Binomial distribution depends only on π. Suppose n = 5 patients are to be treated, and it is known that on average 0.25 will respond to this particular treatment. The number of responses actually observed can only take integer values between 0 (no responses) and 5 (all respond). The Binomial distribution for this case is illustrated in Figure 4.4a. The distribution is not symmetric; it has a maximum at one response and the height of the blocks corresponds to the probability of obtaining the particular number of responses from the five patients yet to be treated.
Figure 4.4 Binomial distribution for π = 0.25 and various values of n. The horizontal scale in each diagram shows the value of r the number of successes.
Figure 4.4 illustrates the shape of the Binomial distribution for various n and π = 0.25. When n is small (here 5 and 10), as in Figure 4.4a and b, the distribution is skewed to the right. The distribution becomes more symmetrical as the sample size increases (here 20 and 50) as in Figure 4.4c and d. We also note that the width of the bars decreases as n increases since the total probability of unity is divided amongst more and more possibilities.
If π were set equal to 0.5, then all the distributions corresponding to those of Figure 4.4 would be symmetrical whatever the size of n. On the other hand, if π = 0.75 then the distributions would be skewed to the left.
We can use the properties of the Binomial distribution when making inferences about proportions, as we shall see in subsequent chapters.
Example – Probability of Corn Resolving
Farndon et al. (2013) give the successful response rate to scalpel treatment in 94 patients with foot corns as 21%. From their data we have p = 20/94 = 0.21. Suppose a podiatrist treated four patients with corns with scalpel debridement. What is the probability that at most one patient responds to treatment? This implies that either 0 or 1 respond. We can use formula 4.1, with r = 0 to give 1 × 0.210 × (1–0.21)4 = 0.3895 and with r = 1 to give 4 × 0.211 × (1–0.21)3 = 0.4142. Summing these two probabilities gives P = 0.3895 + 0.4142 = 0.8037.
4.3 The Poisson Distribution
The Poisson distribution is used to describe discrete quantitative data such as counts that occur independently and randomly in time or space at some average rate. For example, the number of deaths in a town from a particular disease per day, or the number of admissions to a particular hospital in a day typically follows a Poisson distribution.
The Poisson random variable is the count of the number of events that occur independently and randomly in time at some rate, λ. The formula for a Poisson distribution is given as Eq. (4.2) in Section 4.9.
We can use our knowledge of the Poisson distribution to calculate the anticipated number of hospital admissions on any particular day or the number of deaths from lung cancer in a year in a town.
