Copyright © Cengage Learning. All rights reserved. 8 PROBABILITY DISTRIBUTIONS AND STATISTICS

Copyright © Cengage Learning. All rights reserved. 8.5 The Normal Distribution

3 Computations of Probabilities Associated with Normal Distributions

4 Areas under the standard normal curve have been extensively computed and tabulated. Table 2, Appendix D, gives the areas of the regions under the standard normal curve to the left of the number z; these areas correspond, of course, to probabilities of the form P(Z < z) or P(Z  z). The next example illustrates the use of this table in computations involving the probabilities associated with the standard normal variable.

5 Example 1 Let Z be the standard normal variable. Make a sketch of the appropriate region under the standard normal curve, and then find the values of a. P(Z 0.5) c. P(0.24 < Z < 1.48) d. P(–1.65 < Z < 2.02)

6 Example 1(a) – Solution The region under the standard normal curve associated with the probability P(Z < 1.24) is shown in Figure 18. Figure 18 P(Z < 1.24)

7 Example 1(a) – Solution To find the area of the required region using Table 2, Appendix D, we first locate the number 1.2 in the column and the number 0.04 in the row, both headed by z, and read off the number appearing in the body of the table. Thus, P(Z < 1.24) =.8925 cont’d

8 Example 1(b) – Solution The region under the standard normal curve associated with the probability P(Z > 0.5) is shown in Figure 19a. Figure 19(a) P(Z > 0.5) cont’d

9 Example 1(b) – Solution Observe, however, that the required area is, by virtue of the symmetry of the standard normal curve, equal to the shaded area shown in Figure 19b. Thus, P(Z > 0.5) = P(Z < –0.5) =.3085 Figure 19(b) P(Z < – 0.5) cont’d

10 Example 1(c) – Solution The probability P(0.24 < Z < 1.48) is equal to the shaded area shown in Figure 20. This area is obtained by subtracting the area under the curve to the left of z = 0.24 from the area under the curve to the left of z = 1.48; that is, P(0.24 < Z < 1.48) = P(Z < 1.48) – P(Z < 0.24) =.9306 –.5948 =.3358 Figure 20 P(0.24 < Z < 1.48) cont’d

11 Example 1(d) – Solution The probability P(–1.65 < Z < 2.02) is given by the shaded area shown in Figure 21. Figure 21 P(– 1.65 < Z < 2.02) cont’d

12 Example 1(d) – Solution We have P(–1.65 < Z < 2.02) = P(Z < 2.02) – P(Z < –1.65) =.9783 –.0495 =.9288 cont’d

13 Computations of Probabilities Associated with Normal Distributions We now turn our attention to the computation of probabilities associated with normal distributions whose means and standard deviations are not necessarily equal to 0 and 1, respectively. As was mentioned earlier, any normal curve may be transformed into the standard normal curve.

14 Computations of Probabilities Associated with Normal Distributions In particular, it may be shown that if X is a normal random variable with mean  and standard deviation, then it can be transformed into the standard normal random variable Z by means of the substitution The area of the region under the normal curve (with random variable X) between x = a and x = b is equal to the area of the region under the standard normal curve between z = (a –  )/  and z = (b –  )/ .

15 Computations of Probabilities Associated with Normal Distributions In terms of probabilities associated with these distributions, we have (Figure 28). (16) Figure 28

16 Computations of Probabilities Associated with Normal Distributions Similarly, we have Thus, with the help of Equations (16)–(18), computations of probabilities associated with any normal distribution may be reduced to the computations of areas of regions under the standard normal curve. (17) (18)

17 Example 3 Suppose X is a normal random variable with  = 100 and  = 20. Find the values of: a. P(X < 120) b. P(X > 70) c. P(75 < X < 110)

18 Example 3(a) – Solution Using Equation (17) with  = 100,  = 20, and b = 120, we have = P(Z < 1) =.8413 Use the table of values of Z.

19 Example 3(b) – Solution Using Equation (18) with  = 100,  = 20, and a = 70, we have P(X > 70) = P(Z > –1.5) = P (Z < 1.52) =.9332 Use the table of values of Z. cont’d

20 Example 3(c) – Solution Using Equation (16) with  = 100,  = 20, a = 75, and b = 110, we have P(75 < X < 110) = P(– 1.25 < Z < 0.5) cont’d

21 Example 3(c) – Solution P(Z < 0.5) – P(Z < 1.25) =.6915 –.1056 =.5859 See Figure 29.. Use the table of value of Z. Figure 29 cont’d

22 Practice p. 482 Self-Check Exercises #1 & 2