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      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

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

      2 P(Z≥3.08)=1−Φ(3.08)=1−0.9990=0.0010

      3 P(−1.21≤Z≤2.28)=Φ(1.81)−Φ(−1.21)=0.9887−0.1131=0.8756

      4 P(0.67≤Z≤6.28)=Φ(6.28)−Φ(0.67)=1−0.7486=0.2514

      5 P(−4.21≤Z≤0.84)=Φ(0.84)−0=0.7995

      The pth percentile of the standard normal can be found by looking up the cumulative probability p100 inside of the standard normal tables given in Appendix A and then working backward to find the value of Z. Unfortunately, in some cases the value of p100 will not be listed inside the table, and in this case, the value closest to p100 inside of the table should be used for finding the approximate value of the pth percentile.

       Example 2.37

      Determine the 90th percentile of a standard normal distribution.

Solutions To find the 90th percentile, the first step is to find 90100=0.90, or the value closest to 0.90, in the cumulative standard normal table. The row of the standard normal table containing the value 0.90 is given in Table 2.11. Since 0.8997 is the value closest to 0.90 that occurs in the row labeled 1.2 and the column labeled 0.08, the 90th percentile of a standard normal distribution is roughly z=1.2+0.08=1.28.

Table 2.11 An Excerpted Section of the Cumulative Normal Table

Φ(z)=P(Z≤z)
z	0.00	0.01	0.02	0.03	0.04	0.05	0.06	0.07	0.08	0.09
⋮
1.2	0.8849	0.8869	0.8888	Скачать книгу В начало < 35 36 37 38 39 40 41 42 43 44 > В конец