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Chapter 6:. CONTINUOUS RANDOM VARIABLES AND THE NORMAL DISTRIBUTION. CONTINUOUS PROBABILITY DISTRIBUTION. Table 6.1 Frequency and Relative Frequency Distributions of Heights of Female Students. Figure 6.1 Histogram and polygon for Table 6.1.
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Chapter 6: CONTINUOUS RANDOM VARIABLES AND THE NORMAL DISTRIBUTION
CONTINUOUS PROBABILITY DISTRIBUTION Table 6.1Frequency and Relative Frequency Distributions of Heights of Female Students
CONTINUOUS PROBABILITY DISTRIBUTION Two characteristics • The probability that x assumes a value in any interval lies in the range 0 to 1 • The total probability of all the (mutually exclusive) intervals within which x can assume a value of 1.0
Figure 6.3 Area under a curve between two points. Shaded area is between 0 and 1 x = a x = b x
Figure 6.4 Total area under a probability distribution curve. Shaded area is 1.0 or 100% x
Figure 6.5 Area under the curve as probability. Shaded area gives the probability P (a ≤x ≤b) a b x
Figure 6.6 Probability that x lies in the interval 65 to 68 inches.
Figure 6.8 Probability “from 65 to 68” and “between 65 and 68”.
THE NORMAL DISTRIBUTION • Normal Probability Distribution • A normal probability distribution , when plotted, gives a bell-shaped curve such that • The total area under the curve is 1.0. • The curve is symmetric about the mean. • The two tails of the curve extend indefinitely.
Figure 6.11 Normal distribution with mean μand standard deviation σ. Standard deviation = σ Mean =μ x
Figure 6.12 Total area under a normal curve. The shaded area is 1.0 or 100% μ x
Figure 6.13 A normal curve is symmetric about the mean. Each of the two shaded areas is .5 or 50% .5 .5 μ x
Figure 6.14 Areas of the normal curve beyond μ± 3σ. Each of the two shaded areas is very close to zero μ μ–3σ μ + 3σ x
Figure 6.15 Three normal distribution curves with the same mean but different standard deviations. σ = 5 σ = 10 σ = 16 x μ = 50
Figure 6.16 Three normal distribution curves with different means but the same standard deviation. σ = 5 σ = 5 σ = 5 µ = 20 µ = 30 µ = 40 x
THE STANDARD NORMAL DISTRIBTUION • Definition • The normal distribution with μ = 0 and σ = 1 is called the standard normal distribution.
Figure 6.17 The standard normal distribution curve. σ = 1 µ = 0 z -3 -2 -1 0 1 2 3
THE STANDARD NORMAL DISTRIBTUION • Definition • The units marked on the horizontal axis of the standard normal curve are denoted by z and are called the z values or z scores. A specific value of z gives the distance between the mean and the point represented by z in terms of the standard deviation.
Figure 6.18 Area under the standard normal curve. Each of these two areas is .5 . 5 . 5 -3 -2 -1 0 1 2 3 z
Example 6-1 • Find the area under the standard normal curve between z = 0 and z = 1.95.
Table 6.2 Area Under the Standard Normal Curve from z = 0 to z = 1.95 Required area
Figure 6.19 Area between z = 0 to z = 1.95. Shaded area is .4744 0 1.95 z
Example 6-2 • Find the area under the standard normal curve from z = -2.17 to z = 0.
Solution 6-2 • Because the normal distribution is symmetric about the mean, the area from z = -2.17 to z = 0 is the same as the area from z = 0 to z = 2.17. • Area from -2.17 to 0 = P(-2.17≤ z ≤ 0) = .4850
Figure 6.20 Area from z = 0 to z = 2.17 equals area from z = -2.17 to z = 0. Because the symmetry the shaded areas are equal -2.17 0 z 0 2.17 z
Table 6.3 Area Under the Standard Normal Curve from z = 0 to z = 2.17 Required area
Figure 6.21 Area from z = -2.17 to z = 0. Shaded area is .4850 -2.17 0 z
Example 6-3 • Find the following areas under the standard normal curve. • Area to the right of z = 2.32 • Area to the left of z = -1.54
Solution 6-3 • Area to the right of 2.32 = P (z > 2.32) = .5 - .4898 = .0102 as shown in Figure 6.22
Figure 6.22 Area to the right of z = 2.32. This area is .4898 from Table VII Shaded area is .5 - .4898 - .0102 .4898 0 2.32 z
Solution 6-3 • Area to the left of -1.54 = P (z < -1.54) = .5 - .4382 = .0618 as shown in Figure 6.23
Figure 6.23 Area to the left of z = -1.54. This area is .4382 from Table VII Shaded area is .5 - .4382 = .0618 .4382 0 z -1.54
Example 6-4 • Find the following probabilities for the standard normal curve. • P (1.19 < z < 2.12) • P (-1.56 < z < 2.31) • P (z > -.75)
Solution 6-4 • P (1.19 < z < 2.12) = Area between 1.19 and 2.12 = .4830 - .3830 = .1000 as shown in Figure 6.24
Figure 6.24 Finding P (1.19 <z < 2.12). Shaded area = .4830 - .3830 = .1000 z 0 1.19 2.12
Solution 6-4 • P (-1.56 < z < 2.31) = Area between -1.56 and 2.31 = .4406 + .4896 = .9302 as shown in Figure 6.25
Figure 6.25 Finding P (-1.56 < z < 2.31). Total shaded area = .4406 + .4896 = .9302 .4406 .4896 z -1.56 0 2.31
Solution 6-4 • P (z > -.75) = Area to the right of -.75 = .2734 + .5 = .7734 as shown in Figure 6.26
Figure 6.26 Finding P (z > -.75). Total shaded area = .2734 + .5 = .7734 .2734 .5 -.75 0 z
Figure 6.27 Area within one standard deviation of the mean. Total shaded area is .3413 + .3413 = .6826 or 68.26% .3413 .3413 -1.0 0 1.0 z
Figure 6.28 Area within two standard deviations of the mean. Total shaded area is .4772 + .4772 = .9544 or 95.44% .4772 .4772 -2.0 0 2.0 z
Figure 6.29 Area within three standard deviations of the mean. Total shaded area is .4987 + .4987 = .9974 or 99.74% .4987 .4987 z -3.0 0 3.0
Example 6-5 • Find the following probabilities for the standard normal curve. • P (0 < z < 5.67) • P (z < -5.35)
Solution 6-5 • P (0 < z < 5.67) = Area between 0 and 5.67 = .5 approximately as shown in Figure 6.30
Figure 6.30 Area between z = 0 and z = 5.67. Shaded area is approximately .5 0 z 5.67
Solution 6-5 • P (z < -5.35) = Area to the left of -5.35 = .5 - .5 = .00 approximately as shown in Figure 6.31
Figure 6.31 Area to the left of z = -5.35. This area is approximately .5 Shaded area is approximately .00 -5.35 0 z
