Elementary Statistics Thirteenth Edition Chapter 6 Normal Probability

Normal Probability Distributions 6 -1 The Standard Normal Distribution 6 -2 Real Applications of Normal Distributions 6 -3 Sampling Distributions and Estimators 6 -4 The Central Limit Theorem 6 -5 Assessing Normality 6 -6 Normal as Approximation to Binomial Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Key Concept (1 of 2) This section presents a method for using a normal distribution as an approximation to the binomial probability distribution. • Given probabilities p and q (where q = 1 − p) and sample size n, if the conditions of np ≥ 5 and nq ≥ 5 are both satisfied, then probabilities from a binomial probability distribution can be approximated reasonably well by using a normal distribution having these parameters: Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Key Concept (2 of 2) • The binomial probability distribution is discrete (with whole numbers for the random variable x), but the normal approximation is continuous. To compensate, we use a “continuity correction” with a whole number x represented by the interval from x − 0. 5 to x + 0. 5. Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Brief Review of Binomial Probability Distribution (1 of 2) In a previous section we saw that a binomial probability distribution has (1) a fixed number of trials; (2) trials that are independent; (3) trials that are each classified into two categories commonly referred to as success and failure; and (4) trials with the property that the probability of success remains constant. Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Brief Review of Binomial Probability Distribution (2 of 2) Notation n = the fixed number of trials x = the specific number of successes in n trials p = probability of success in one of the n trials q = probability of failure in one of the n trials (so q = 1 − p) Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Rationale for Using a Normal Approximation We saw in a previous section that the sampling distribution of a sample proportion tends to approximate a normal distribution. Also, see the following probability histogram for the binomial distribution with n = 580 and p = 0. 25. (In one of Mendel’s famous hybridization experiments, he expected 25% of his 580 peas to be yellow, but he got 152 yellow peas, for a rate of 26. 2%. ) The bellshape of this graph suggests that we can use a normal distribution to approximate the binomial distribution. Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Normal Distribution as an Approximation to the Binomial Distribution (1 of 3) Requirements 1. The sample is a simple random sample of size n from a population in which the proportion of successes is p, or the sample is the result of conducting n independent trials of a binomial experiment in which the probability of success is p. 2. np ≥ 5 and nq ≥ 5. Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Normal Distribution as an Approximation to the Binomial Distribution (2 of 3) Normal Approximation If the above requirements are satisfied, then the probability distribution of the random variable x can be approximated by a normal distribution with these parameters: Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Normal Distribution as an Approximation to the Binomial Distribution (3 of 3) Continuity Correction When using the normal approximation, adjust the discrete whole number x by using a continuity correction so that any individual value x is represented in the normal distribution by the interval from x − 0. 5 to x + 0. 5. Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Procedure for Using a Normal Distribution to Approximate a Binomial Distribution 1. Check the requirements that np ≥ 5 and nq ≥ 5. 3. Identify the discrete whole number x that is relevant to the binomial probability problem being considered, and represent that value by the region bounded by x − 0. 5 and x + 0. 5. 4. Graph the normal distribution and shade the desired area bounded by x − 0. 5 or x + 0. 5 as appropriate. Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Example: Was Mendel Wrong? (1 of 7) In one of Mendel’s famous hybridization experiments, he expected that among 580 offspring peas, 145 of them (or 25%) would be yellow, but he actually got 152 yellow peas. Assuming that Mendel’s rate of 25% is correct, find the probability of getting 152 or more yellow peas by random chance. That is, given n = 580 and p = 0. 25, find P(at least 152 yellow peas). Is 152 yellow peas significantly high? Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Example: Was Mendel Wrong? (2 of 7) Solution Step 1: Requirement check: With n = 580 and p = 0. 25, we get np = (580)(0. 25) = 145 and nq = (580)(0. 75) = 435, so the requirements that np ≥ 5 and nq ≥ 5 are both satisfied. Step 2: We now find µ and σ needed for the normal distribution. Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Example: Was Mendel Wrong? (3 of 7) Solution Step 3: We want the probability of at least 152 yellow peas, so the discrete whole number relevant to this example is x = 152. We use the continuity correction as we represent the discrete value of 152 in the graph of the normal distribution by the interval between 151. 5 and 152. 5 (as shown in the figure after step 4). Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Example: Was Mendel Wrong? (4 of 7) Solution Step 4: See the bottom portion of the figure, which shows the normal distribution and the area to the right of 151. 5 (representing “ 152 or more” yellow peas). Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Example: Was Mendel Wrong? (5 of 7) Solution Copyright © 2018, 2014, 2012 Pearson

Example: Was Mendel Wrong? (6 of 7) Solution If using Table A-2, we must first find the z score using x = 151. 5, µ = 145 and σ = 10. 4283 as follows: Using Table A-2, we find that z = 0. 62 corresponds to a cumulative left area of 0. 7324, so the shaded region in the bottom portion of the previous figure is 1 − 0. 7324 = 0. 2676. (The result of 0. 2665 from technology is more accurate. ) Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Example: Was Mendel Wrong? (7 of 7) Interpretation Mendel’s result of 152 yellow peas is greater than the 145 yellow peas he expected with his theory of hybrids, but with P(152 or more yellow peas) = 0. 2665, we see that 152 yellow peas is not significantly high. That is a result that could easily occur with a true rate of 25% for yellow peas. This experiment does not contradict Mendel’s theory. Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Continuity Correction • Continuity Correction – When we use the normal distribution (which is a continuous probability distribution) as an approximation to the binomial distribution (which is discrete), a continuity correction is made to a discrete whole number x in the binomial distribution by representing the discrete whole number x by the interval from x − 0. 5 to x + 0. 5 (that is, adding and subtracting 0. 5). Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Continuity Correction Examples The previous example used a continuity correction when the discrete value of 152 was represented in the normal distribution by the area between 151. 5 and 152. 5. Because we wanted the probability of “ 152 or more” yellow peas, we used the area to the right of 151. 5. Here are other uses of the continuity correction: Statement About the Discrete Value Area of the Continuous Normal Distribution At least 152 (includes 152 and above) To the right of 151. 5 More than 152 (doesn’t include 152) To the right of 152. 5 At most 152 (includes 152 and below) To the left of 152. 5 Fewer than 152 (doesn’t include 152) To the left of 151. 5 Exactly 152 Between 151. 5 and 152. 5 Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Example: Exactly 252 Yellow Peas (1 of 4) Using the same information from the previous example, find the probability of exactly 152 yellow peas among the 580 offspring peas. That is, given n = 580 and assuming that p = 0. 25, find P(exactly 152 yellow peas). Is this result useful for determining whether 152 yellow peas is significantly high? Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Example: Exactly 252 Yellow Peas (2 of 4) Solution The figure shows the normal distribution with µ = 145 and σ = 10. 4283. The shaded area approximates the probability of exactly 152 yellow peas. That region is the vertical strip between 151. 5 and 152. 5, as shown. We can find that area by using the same methods introduced in an earlier section. Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Example: Exactly 252 Yellow Peas (3 of 4) Solution Using Table A-2, we convert 151. 5 and 152. 5 to z = 0. 62 and z = 0. 72, which yield cumulative left areas of 0. 7324 and 0. 7642. Because they are both cumulative left areas, the shaded region in the figure is 0. 7642 − 0. 7324 = 0. 0318. The probability of exactly 152 yellow peas is 0. 0318. Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Example: Exactly 252 Yellow Peas (4 of 4) Interpretation In an earlier section we saw that x successes among n trials is significantly high if the probability of x or more successes is unlikely with a probability of 0. 05 or less. In determining whether Mendel’s result of 152 yellow peas contradicts his theory that 25% of the offspring should be yellow peas, we should consider the probability of 152 or more yellow peas, not the probability of exactly 152 peas. The result of 0. 0305 is not the relevant probability; the relevant probability is 0. 2665 found in the first example. In general, the relevant result is the probability of getting a result at least as extreme as the one obtained. Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved

Technology for Binomial Probabilities (1 of 2) This topic of using a normal distribution to approximate a binomial distribution was once quite important, but we can now use technology to find binomial probabilities that were once beyond our capabilities. For example, see the accompanying Statdisk display showing that for the first example, the probability of 152 or more yellow peas is 0. 2650, and for the second example, the probability of exactly 152 yellow peas is 0. 0301, so there is no real need to use a normal approximation. However, there are cases where we need to use a normal approximation, and a future section uses a normal approximation to a binomial distribution for an important statistical method introduced in that section. Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved