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Chapter 6 The Normal Distribution. In this handout: The standard normal distribution Probability calculations with normal distributions The normal approximation to the binomial. Figure 6.8 (p. 231) The standard normal curve. It is customary to denote the standard normal variable by Z.

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Chapter 6 The Normal Distribution


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    1. Chapter 6The Normal Distribution In this handout: The standard normal distribution Probability calculations with normal distributions The normal approximation to the binomial

    2. Figure 6.8 (p. 231)The standard normal curve. It is customary to denote the standard normal variable by Z. Figure 6.9 (p. 232)Equal normal tail probabilities.

    3. From the table, P[Z ≤ 1.37] = .9147 An upper tail normal probability P[Z > 1.37] = 1 - P[Z ≤ 1.37] = 1 - .9147 = .0853

    4. Figure 6.11 (p. 233)Normal probability of an interval. P[-.155 < Z < 1.6] = P[Z < 1.6] - P[Z < -.155] = .9452 - .4384 = .5068

    5. Figure 6.12 (p. 233)Normal probabilities for Example 3. P[Z < -1.9 or Z > 2.1] = P[Z < -1.9] + P[Z > 2.1] = .0287 + .0179 = .0466

    6. Determining an upper percentile of the standard normal distribution • Problem: Locate the value of z that satisfies P[Z > z] = 0.025 • Solution: P[Z < z] = 1- P[Z> z] = 1- 0.025 = 0.975 From the table, 0.975 = P[Z < 1.96]. Thus, z = 1.96

    7. Determining z for given equal tail areas • Problem: Obtain the value of z for which P[-z < Z < z] = 0.9 • Solution: From the symmetry of the curve, P[Z < -z] = P[Z > z] = .05 From the table, P[Z < -1.645] = 0.05. Thus z = 1.645

    8. Converting a normal probability to a standard normal probability

    9. Converting a normal probability to a standard normal probability (example) • Problem: Given X is N(60,4), find P[55 < X < 63] • Solution: The standardized variable is Z = (X – 60)/4 x=55 gives z=(55-60)/4 = -1.25 x=63 gives z=(63-60)/4 = .75 Thus, P[55<X<63] = P[-1.25 < Z < .75] = P[Z < .75] - P[Z < -1.25] = .7734 - .1056 = .6678

    10. When the success probability p of is not too near 0 or 1 and the number of trials is large, the normal distribution serves as a good approximation to the binomial probabilities. Figure 6.16 (p. 241)The binomial distributions for p = .4 and n = 5, 12, 25.

    11. How to approximate the binomial probability by a normal? The normal probability assigned to a single value x is zero. However, the probability assigned to the interval x-0.5 to x+0.5 is the appropriate comparison (see figure). The addition and subtraction of 0.5 is called the continuity correction.

    12. Example: Suppose n=15 and p=.4. Compute P[X = 7]. Mean = np = 15 * .4 = 6 Variance = np(1-p) = 6 * .6 = 3.6 sd = 1.897 P[6.5 < X < 7.5] = P[(6.5 -6)/1.897 < Z < (7.5 - 6)/1.897] = P[.264 < Z < .791] = .7855 - .6041 = .1814