Normal Approximations to Binomial Distributions In 4.2 you learned how to find binomial probabilities. For instance, consider a surgical procedure that has an 85% chance of success. When a doctor performs this surgery on 10 patients, you can use the binomial formula to find the probability of exactly two successful surgeries. But what if the doctor performs the procedure on 150 patients and you want to find the probability of fewer than 100 successful surgeries. You would have to find the probability of 1 through 100 successes and then add them together. This is not a practical approach. So, we can use a normal distribution to approximate a binomial distribution. Normal Approximation to a Binomial Distribution If and, then the binomial random variable x is approximately normally distributed, with mean and standard deviation where n is the number of independent trials, p is the probability of success in a single trial, and q is the probability of failure in a single trial. 1
This is valid because, if you look at the graphs at the bottom of page 275, where p = 0.25, q = 1 0.25 = 0.75, and n = 4, n = 10, n = 25, and n = 50, the shape of the binomial distribution becomes more similar to a normal distribution. Example 1 Determine whether you can use a normal distribution to approximate the distribution of x. If you can, find the mean and standard deviation. If not, explain why. 1. In a survey of 8 to 18 year old heavy media users in the United States, 47% said they get fair or poor grades (C's or below). You randomly select forty five 8 to 18 year old heavy media users in the United States and ask them whether they get fair or poor grades. Both np and nq are and so we can use a normal distribution with 2. In a survey of 8 to 18 year old heavy media users in the United States, 23% said they get fair or poor grades (C's or below). You randomly select forty five twenty 8 to 18 year old heavy media users in the United States and ask them whether they get fair or poor grades. Since np is not greater than or equal to 5, we cannot use a normal distribution to approximate the distribution of x. 2
Continuity Correction When you use a continuous normal distribution to approximate a binomial probability, you need to move 0.5 unit to the left and right of the midpoint to include all possible x values in the interval (see graphs on page 277). When you do this, you are making acontinuity correction. Example 2 Use a continuity correction to convert each binomial probability to a normal distribution probability. 1. The probability of getting between 270 and 310 successes, inclusive. The midpoint to the left of 270 is 269.5 and the midpoint to the right of 310 is 310.5. So, is the normal distribution probability. 2. The probability of getting at least 158 successes We are looking at. So, making the continuity correction, we subtract 0.5 and we get for the normal distribution probability. 3. The probability of getting fewer than 63 successes. We are looking at. So, making the continuity correction, we subtract 0.5 and we get for the normal distribution probability. 3
Approximating Binomial Probabilities GUIDELINES Using Normal Distribution to Approximate Binomial Probabilities IN WORDS 1. Verify that the binomial distribution applies. 2. Determine whether you can use a normal distribution to approximate x, the binomial variable. IN SYMBOLS Specify n, p, and q. Is Is 3. Find the mean μ and the standard deviation σ for the distribution. 4. Apply thee appropriate continuity correction. Shade the corresponding area under the curve. Add 0.5 to (or subtract 0.5 from) the binomial probability. 5. Find the corresponding z scores. 6. Find the probability. Use the Standard Normal Table 4
In a survey of 8 to 18 year old heavy media users in the United States, 47% said they get fair or poor grades (C's or below). You randomly select forty five 8 to 18 year old heavy media users in the United States and ask them whether they get fair or poor grades. What is the probability that fewer than 20 of them respond yes? (fewer than 20 means 19 or less) From example 1, we have μ = 21.15 and σ = 3.35 to approximate the binomial distribution. Since we are looking for values to the left of 19, we add 0.5 to the right hand value of 19 to get 19.5. or on the calculator 2nd VARS normcdf ENTER lower: 10000 upper: 19.5 μ = 21.15 σ = 3.35 Enter Enter Enter The probability that fewer than 20 8 to 18 year olds respond yes is approximately 0.3111 or about 31.11%. 5
EXAMPLE 4 Fifty eight percent of adults say they never wear a helmet when riding a bicycle. You randomly select 200 adults in the United States and aask them whether they wear a helmet when riding a bicycle. What is the probability that at least 120 adults will say they never wear a helmet when riding a bicycle? n = 200 p = 0.58 q = 0.42 Both are greater than 5, so the binomial variable x is approximately normally distributed, with Since we are looking for values greater than 120, we subtract 0.5 from 120 to get 119.5. Try It Yourself 4 In the above example, what is the probability that at most 100 adults will say they never wear a helmet when riding a bicycle? (a) Find the corresponding z score (b) Find the probability. 6
EXAMPLE 5 A study of National Football League (NFL) retirees, ages 50 and older, found 62.4% have arthritis. You randomly select 75 NFL retirees who are at least 50 years old and ask them if they have arthritis. What is the probability that exactly 48 will say yes. The binomial variable x is approximately normally distributed. Since we want 48 exactly, we must go 0.5 above and 0.5 below for the continuity correction. 2nd VARS normalcdf lower: 47.5 upper: 48.5 μ = 46.8 σ = 4.19 Enter Enter Enter 0.09112 7
Try It Yourself 5 In a study, it was found that 32.0% of all men in the United States ages 50 and older have arthritis. You randomly select 75 men in the United States who are at least 50 years old and ask them whether they have arthritis. What is the probability that exactly 15 will say yes? (a) Find np and qp. (b) Find μ and σ. (c) Apply the continuity correction (d) Use the calculator to find P(x = 15). P. 281 1 29 odds 8