heart attacks, the probability of survival will be influenced by many factors including severity of heart attack, delay in treatment, age, and ability to change diet and lifestyle following a heart attack. Because each individual is different, the probability of survival is not going to be constant over the n individuals in the study, and hence, the binomial probability model does not apply.
2.4.2 The Normal Probability Model
The choice of a probability model for continuous variables is generally based on historical data rather than a particular set of conditions. Just as there are many discrete probability models, there are also many different probability models that can be used to model the distribution of a continuous variable. The most commonly used continuous probability model in statistics is the normal probability model.
The normal probability model is often used to model distributions that are expected to be unimodal and symmetric, and the normal probability model forms the foundation for many of the classical statistical methods used in biostatistics. Moreover, the distribution of many natural phenomena can be modeled very well with the normal distribution. For example, the weights, heights, and IQs of adults are often modeled with normal distributions.
Several properties of a normal distribution are listed below.
PROPERTIES OF A NORMAL DISTRIBUTION
A normal distribution
is a bell- or mound-shaped distribution.
is completely characterized by its mean and standard deviation. The mean determines the center of the distribution and the standard deviation determines the spread about the mean.
has probabilities and percentiles that are determined by the mean and standard deviation.
is symmetric about the mean.
has mean, median, and mode that are equal (i.e., μ=μ~=M).
has probability density function given by
Example 2.33
The intelligence quotient (IQ) is based on a test of aptitude and is often used as a measure of an individual’s intelligence. The distribution of IQ scores is approximately normally distributed with mean 100 and standard deviation 15. The normal probability model for IQ scores is given in Figure 2.25.
Figure 2.25 The approximate distribution of IQ scores with µ = 100 and σ = 15.
The standard normal, which will be denoted by Z, is a normal distribution having mean 0 and standard deviation 1. The standard normal is used as the reference distribution from which the probabilities and percentiles associated with any normal distribution will be determined. The cumulative probabilities for a standard normal are given in Tables A.1 and A.2; because 99.95% of the standard normal distribution lies between the values −3.49 and 3.49, the standard normal values are only tabulated for z values between −3.49 and 3.49. Thus, when the value of a standard normal, say z, is between −3.49 and 3.49, the tabled value for z represents the cumulative probability of z, which is P(Z≤z) and will be denoted by Φ(z). For values of z below −3.50, Φ(z) will be taken to be 0 and for values of z above 3.50, Φ(z) will be taken to be 1. Tables A.1 and A.2 can be used to compute all of the probabilities associated with a standard normal.
The values of z are referenced in Tables A.1 and A.2 by writing z=a.bc as z=a.b+0.0c. To locate a value of z in Table A.1 and A.2, first look up the value a.b in the left-most column of the table and then locate 0.0 c in the first row of the table. The value cross-referenced by a.b and 0.c in Tables A.1 and A.2 is Φ(z)=P(Z≤z). The rules for computing the probabilities for a standard normal are given below.
COMPUTING STANDARD NORMAL PROBABILITIES
1 For values of z between −3.49 and 3.49, the probability that Z≤z is read directly from the table. That is,
2 For z≤−3.50 the probability that Z≤z is 0, and for z≥3.50 the probability that Z≤z is 1.
3 For values of z between −3.49 and 3.49, the probability that Z≥z is
4 For values of z between −3.49 and 3.49, the probability that a≤Z≤b is
Example 2.34
To determine the probability that a standard normal random variable is less than 1.65, which is the area shown in Figure 2.26, look up 1.6 in the left-most column of Table A.2 and 0.05 in the top row of this table, which yields P(Z≤1.65)=0.9505.
Figure 2.26 P(Z≤1.65).
Example 2.35
Determine the probability that a standard normal random variable lies between −1.35 and 1.51 (see Figure 2.27).
Figure 2.27 P(−1.35≤Z≤1.51).
Solutions First, look up the cumulative probabilities for both −1.35 and 1.51 in Tables A.1 and A.2. The probability that Z≤−1.35 and the probability that Z≤1.51 are shown in Figure 2.28.
Figure 2.28 The areas representing P(Z≤1.51) and P(Z≤−1.35).
Subtracting these probabilities yields P(−1.35≤Z≤1.51), and thus,
Example 2.36
Using the standard normal tables given in Tables A.1 and A.2, determine the following probabilities for a standard normal distribution:
1 P(Z≤−2.28)
2 P(Z≤3.08)
3 P(−1.21≤Z≤2.28)
4 P(1.21≤Z≤6.28)
5 P(−4.21≤Z≤0.84)
Solutions Using the normal table in the appendix
1 P(Z≤−2.28)=Φ(−2.28)=0.0113
