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. Figure 6.8 (p. 231) The standard normal curve. Figure 6.9 (p. 232) Equal normal tail probabilities.

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

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

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

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

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

Converting a normal probability to a standard normal probability

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

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.

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.

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