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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

      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

      Figure 2.26 P(Z≤1.65).

       Example 2.35

      Figure 2.27 P(−1.35≤Z≤1.51).

      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,

upper P left-parenthesis negative 1.35 less-than-or-equal-to upper Z less-than-or-equal-to 1.51 right-parenthesis equals normal upper Phi left-parenthesis 1.51 right-parenthesis minus normal upper Phi left-parenthesis negative 1.35 right-parenthesis equals 0.9345 minus 0.0885 equals 0.8640

       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)

      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.

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

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