1. Copyright © Cengage Learning. All rights reserved. 6 Normal Probability Distributions

2. Copyright © Cengage Learning. All rights reserved. 6.5 Normal Approximation of the Binomial

3. 3 As we know that the binomial distribution is a probability distribution of the discrete random variable x, the number of successes observed in n repeated independent trials. We will now see how binomial probabilities—that is, probabilities associated with a binomial distribution—can be reasonably approximated by using the normal probability distribution. Let’s look first at a few specific binomial distributions.

4. 4 Normal Approximation of the Binomial Figure 6.13 shows the probabilities of x for 0 to n in three situations: n = 4, n = 8, and n = 24. Figure 6.13 Binomial Distributions (a) Distribution for n = 4, p = 0.5(b) Distribution for n = 8, p = 0.5 (c) Distribution for n = 24, p = 0.5

5. 5 Normal Approximation of the Binomial For each of these distributions, the probability of success for one trial is 0.5. Notice that as n becomes larger, the distribution appears more and more like the normal distribution. To make the desired approximation, we need to take into account one major difference between the binomial and the normal probability distribution. The binomial random variable is discrete, whereas the normal random variable is continuous. Let’s look at the distribution of the binomial variable x, when n = 14 and p = 0.5.

6. 6 Normal Approximation of the Binomial The probabilities for each x value can be obtained from Table 2 in Appendix B. This distribution of x is shown in Figure 6.14. The Distribution of x when n = 14, p = 0.5 Figure 6.14

7. 7 Normal Approximation of the Binomial We see the very same distribution in Figure 6.15 in histogram form. Histogram for the Distribution of x when n = 14, p = 0.5 Figure 6.15

8. 8 Normal Approximation of the Binomial Let’s examine P (x = 4) for n = 14 and p = 0.5 to study the approximation technique. P (x = 4) is equal to 0.061 (see Table 2 in Appendix B), the area of the bar (rectangle) above x = 4 in Figure 6.16. Figure 6.16 Area of Bar above x = 4 is 0.061, for B(n = 14, p = 0.5)

9. 9 Normal Approximation of the Binomial The area of a rectangle is the product of its width and height. In this case the height is 0.061 and the width is 1.0, so the area is 0.061. Let’s take a closer look at the width. For x = 4, the bar starts at 3.5 and ends at 4.5, so we are looking at an area bounded by x = 3.5 and x = 4.5. The addition and subtraction of 0.5 to the x value is commonly called the continuity correction factor. It is our method of converting a discrete variable into a continuous variable.

10. 10 Normal Approximation of the Binomial Now let’s look at the normal distribution related to this situation. We will first need a normal distribution with a mean and a standard deviation equal to those of the binomial distribution we are discussing. Formulas (5.7) and (5.8) give us these values:  = np = (14)(0.5) = 7.0  = 1.87

11. 11 Normal Approximation of the Binomial The probability that x = 4 is approximated by the area under the normal curve between x = 3.5 and x = 4.5 is shown in Figure 6.17. Figure 6.17 Probability that x = 4 is Approximated by Shaded Area

12. 12 Normal Approximation of the Binomial Figure 6.18 shows the entire distribution of the binomial variable x with a normal distribution of the same mean and standard deviation superimposed. Notice that the bars and the interval areas under the curve cover nearly the same area. Figure 6.18 Normal Distribution Superimposed over Distribution for Binomial Variable x

13. 13 Normal Approximation of the Binomial The probability that x is between 3.5 and 4.5 under this normal curve is found by using formula (6.3),Table 3, and the methods outlined in Section 6.3: Z P (3.5 < x < 4.5) = = P(–1.87 < z < –1.34) = 0.0901 – 0.0307 = 0.0594

14. 14 Normal Approximation of the Binomial Since the binomial probability of 0.061 and the normal probability of 0.0594 are reasonably close, the normal probability distribution seems to be a reasonable approximation of the binomial distribution. The normal approximation of the binomial distribution is also useful for values of p that are not close to 0.5.

15. 15 Normal Approximation of the Binomial The binomial probability distributions shown in Figures 6.19 and 6.20 suggest that binomial probabilities can be approximated using the normal distribution. Binomial Distributions Figure 6.19 (a) Distribution for n = 4, p = 0.3 (b) Distribution for n = 8, p = 0.3(c) Distribution for n = 24, p = 0.3

16. 16 Normal Approximation of the Binomial Figure 6.20 Binomial Distributions (a) Distribution for n = 4, p = 0.1 (b) Distribution for n = 8, p = 0.1 (c) Distribution for n = 50, p = 0.1

17. 17 Normal Approximation of the Binomial Notice that as n increases, the binomial distribution begins to look like the normal distribution. As the value of p moves away from 0.5, a larger n is needed in order for the normal approximation to be reasonable.

18. 18 Normal Approximation of the Binomial The following rule of thumb is generally used as a guideline: Rule The normal distribution provides a reasonable approximation to a binomial probability distribution whenever the values of np and n(1 – p) both equal or exceed 5. By now you may be thinking, “So what? I will just use the binomial table and find the probabilities directly and avoid all the extra work.” But consider for a moment the situation presented in Example 21.

19. 19 Example 21 – Solving a Binomial Probability Problem with the Normal Distribution An unnoticed mechanical failure has caused of a machine shop’s production of 5000 rifle firing pins to be defective. What is the probability that an inspector will find no more than 3 defective firing pins in a random sample of 25? Solution: In this example of a binomial experiment, x is the number of defectives found in the sample, n = 25, and p = P (defective) =.

20. 20 Example 21 – Solution To answer the question using the binomial distribution, we will need to use the binomial probability function: for x = 0, 1, 2, …, 25 We must calculate the values for P(0), P(1), P(2), and P(3), because they do not appear in Table 2. This is a very tedious job because of the size of the exponent. In situations such as this, we can use the normal approximation method. cont’d

21. 21 Example 21 – Solution Now let’s find P (x  3) by using the normal approximation method. We first need to find the mean and standard deviation of x, using the formulas:  = np = (25) = 8.333  cont’d

22. 22 Example 21 – Solution = 2.357 These values are shown in the figure. cont’d

23. 23 Example 21 – Solution The area of the shaded region (x  3.5) represents the probability of x = 0, 1, 2, or 3. Remember that x = 3, the discrete binomial variable, covers the continuous interval from 2.5 to 3.5. P (x is no more than 3) = P (x  3) (for a discrete variable x) = P (x  3.5) (for a continuous variable x) cont’d

24. 24 P(x  3.5) = P (z  –2.05) = 0.0202 Thus, P (no more than three defectives) is approximately 0.02. Example 21 – Solution cont’d