328 Chapter 7 The Normal Distribution SUMMARY Following are the areas under the normal curve to use when the continuity correction is applied. P(a ≤ X ≤ b) Find the area between a − 0.5 and b + 0.5. a − 0.5 b + 0.5 P(X ≤ b) Find the area to the left of b + 0.5. b + 0.5 P(X ≥ a) Find the area to the right of a − 0.5. a − 0.5 P(X = a) Find the area between a − 0.5 and a + 0.5. a − 0.5 a + 0.5 Use the following steps to compute a binomial probability with the normal approximation. Explain It Again We will use Table A.2 when using the normal approximation to the binomial: The normal approximation is useful when computing by hand. If technology is to be used, the exact probability can be calculated, so the normal approximation is less useful. Procedure for Computing Binomial Probabilities with the Normal Approximation Step 1: Check to see that the conditions np ≥ 10 and n(1 − p) ≥ 10 are both met. If so, it is appropriate to use the normal approximation. If not, the probability must be calculated with the binomial distribution (see Section 6.2). Step 2: Compute the mean ��X and the standard deviation ��X. Step 3: For each endpoint, determine whether to add 0.5 or subtract 0.5. Step 4: Sketch a normal curve, label the endpoints, and shade in the area to be found. Step 5: Find the area using Table A.2 or technology. Note, however, that if you are using technology, you may be able to compute the probability exactly without using the normal approximation. EXAMPLE 7.27 Using the continuity correction to compute a probability The Statistical Abstract of the United States reported that 66% of students who graduated from high school in 2012 enrolled in college. One hundred high school graduates are sampled. Let X be the number who enrolled in college. Find P(X ≤ 75). 66 75.5 z = 2.01 Figure 7.35 Solution Step 1: Check the assumptions. The number of trials is n = 100 and the success probability is p = 0.66. Therefore np = (100)(0.66) = 66 ≥ 10 and n(1 − p) = (100)(1 − 0.66) = 34 ≥ 10. We can use the normal approximation. Step 2: We compute the mean and standard deviation of X: √ ��X = np = (100)(0.66) = 66 ��X = np(1 − p) = √ (100)(0.66)(1 − 0.66) = 4.73709 Step 3: Since the probability is P(X ≤ 75), we compute the area to the left of 75.5. Step 4: We sketch a normal curve, and label the mean of 66 and the point 75.5. Step 5: We use Table A.2 to find the area. The z-score for 75.5 is z = 75.5 − 66 4.73709 = 2.01 From Table A.2 we find that the probability is 0.9778. See Figure 7.35.
navidi_monk_elementary_statistics_2e_ch7-9
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