# Excursions in Modern Mathematics, 7e: 16.7 - 2Copyright © 2010 Pearson Education, Inc. 16 Mathematics of Normal Distributions 16.1Approximately Normal.

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Excursions in Modern Mathematics, 7e: 16.7 - 2Copyright © 2010 Pearson Education, Inc. 16 Mathematics of Normal Distributions 16.1Approximately Normal Distributions of Data 16.2Normal Curves and Normal Distributions 16.3Standardizing Normal Data 16.4The 68-95-99.7 Rule 16.5Normal Curves as Models of Real- Life Data Sets 16.6Distribution of Random Events 16.7Statistical Inference

Excursions in Modern Mathematics, 7e: 16.7 - 3Copyright © 2010 Pearson Education, Inc. Suppose that we have an honest coin and intend to toss it 100 times. We are going to do this just once, and we will let X denote the resulting number of heads. Been there, done that! What’s new now is that we a have a solid understanding of the statistical behavior of the random variable X–it has an approximately normal distribution with mean  = 50 and standard deviation  = 5–and this allows us to make some very reasonable predictions about the possible values of X. Statistical Inference

Excursions in Modern Mathematics, 7e: 16.7 - 4Copyright © 2010 Pearson Education, Inc. For starters, we can predict the chance that X will fall somewhere between 45 and 55 (one standard deviation below and above the mean)–it is 68%. Likewise, we know that the chance that X will fall somewhere between 40 and 60 is 95%, and between 35 and 65 is a whopping 99.7%. What if, instead of tossing the coin 100 times, we were to toss it n times? Statistical Inference

Excursions in Modern Mathematics, 7e: 16.7 - 5Copyright © 2010 Pearson Education, Inc. Not surprisingly, bell-shaped distribution would still be there–only the values of  and  would change. Specifically, for n sufficiently large (typically n ≥ 30), the number of heads in n tosses would be a random variable with an approximately normal distribution with mean  = n/2 heads and standard deviation heads. This is an important fact for which we have coined the name the honest-coin principle. Statistical Inference

Excursions in Modern Mathematics, 7e: 16.7 - 6Copyright © 2010 Pearson Education, Inc. Let X denote the number of heads in n tosses of an honest coin (assume n ≥ 30). Then, X has an approximately normal distribution with mean  = n/2 and standard deviation THE HONEST-COIN PRINCIPLE

Excursions in Modern Mathematics, 7e: 16.7 - 7Copyright © 2010 Pearson Education, Inc. An honest coin is going to be tossed 256 times. Before this is done, we have the opportunity to make some bets. Let’s say that we can make a bet (with even odds) that if the number of heads tossed falls somewhere between 120 and 136, we will win; otherwise, we will lose. Should we make such a bet? Let X denote the number of heads in 256 tosses of an honest coin. Example 16.9Coin-Tossing Experiments: Part 2

Excursions in Modern Mathematics, 7e: 16.7 - 8Copyright © 2010 Pearson Education, Inc. By the honest-coin principle, X is a random variable having a distribution that is approxi- mately normal with mean  = 256/2 = 128 heads and standard deviation heads. The values 120 to 136 are exactly one standard deviation below and above the mean of 128, which means that there is a 68% chance that the number of heads will fall somewhere between 120 and 136. Example 16.9Coin-Tossing Experiments: Part 2

Excursions in Modern Mathematics, 7e: 16.7 - 9Copyright © 2010 Pearson Education, Inc. We should indeed make this bet! A similar calculation tells us that there is a 95% chance that the number of heads will fall somewhere between 112 and 144, and the chance that the number of heads will fall somewhere between 104 and 152 is 99.7%. Example 16.9Coin-Tossing Experiments: Part 2

Excursions in Modern Mathematics, 7e: 16.7 - 10Copyright © 2010 Pearson Education, Inc. What happens when the coin being tossed is not an honest coin? Surprisingly, the distribution of the number of heads X in n tosses of such a coin is still approximately normal, as long as the number n is not too small (a good rule of thumb is n ≥ 30). All we need now is a dishonest-coin principle to tell us how to find the mean and the standard deviation. Dishonest Coin

Excursions in Modern Mathematics, 7e: 16.7 - 11Copyright © 2010 Pearson Education, Inc. Let X denote the number of heads in n tosses of a coin (assume n ≥ 30). Let p denote the probability of heads on each toss of the coin. Then, X has an approximately normal distribution with mean  = n P and standard deviation THE DISHONEST-COIN PRINCIPLE

Excursions in Modern Mathematics, 7e: 16.7 - 12Copyright © 2010 Pearson Education, Inc. A coin is rigged so that it comes up heads only 20% of the time (i.e., p = 0.20). The coin is tossed 100 times (n = 100) and X is the number of heads in the 100 tosses. What can we say about X? Example 16.10Coin-Tossing Experiments: Part 3

Excursions in Modern Mathematics, 7e: 16.7 - 13Copyright © 2010 Pearson Education, Inc. According to the dishonest-coin principle, the distribution of the random variable X is approximately normal with mean m = 100  0.20 = 20 and standard deviation Applying the 68-95-99.7 rule with  = 20 and  = 4 gives the following facts: Example 16.10Coin-Tossing Experiments: Part 3

Excursions in Modern Mathematics, 7e: 16.7 - 14Copyright © 2010 Pearson Education, Inc. ■ There is about a 68% chance that X will be somewhere between 16 and 24 (  –  ≤ X ≤  +  ). ■ There is about a 95% chance that X will be somewhere between 12 and 28 (  – 2  ≤ X ≤  + 2  ). ■ The number of heads is almost guaranteed (about 99.7%) to fall somewhere between 8 and 32 (  – 3  ≤ X ≤  + 3  ). Example 16.10Coin-Tossing Experiments: Part 3

Excursions in Modern Mathematics, 7e: 16.7 - 15Copyright © 2010 Pearson Education, Inc. In this example, heads and tails are no longer interchangeable concepts–heads is an outcome with probability p = 0.20 while tails is an outcome with much higher probability (0.8). We can, however, apply the principle equally well to describe the distribution of the number of tails in 100 coin tosses of the same dishonest coin: The distribution for the number of tails is approximately normal with mean  = 100  0.80 = 80 and standard deviation Example 16.10Coin-Tossing Experiments: Part 3

Excursions in Modern Mathematics, 7e: 16.7 - 16Copyright © 2010 Pearson Education, Inc. The dishonest-coin principle is a special version of one of the most important laws in statistics, a law generally known as the central limit theorem. We will now briefly illustrate why the importance of the dishonest-coin principle goes beyond the tossing of coins. Central Limit Theorem

Excursions in Modern Mathematics, 7e: 16.7 - 17Copyright © 2010 Pearson Education, Inc. An assembly line produces 100,000 light bulbs a day, 20% of which generally turn out to be defective. Suppose that we draw a random sample of n = 100 light bulbs. Let X represent the number of defective light bulbs in the sample. What can we say about X? A moment’s reflection will show that, in a sense, this example is completely parallel to Example 16.10–think of selecting defective light bulbs as analogous to tossing heads with a dishonest coin. Example 16.11Sampling for Defective Light Bulbs

Excursions in Modern Mathematics, 7e: 16.7 - 18Copyright © 2010 Pearson Education, Inc. We can use the dishonest-coin principle to infer that the number of defective light bulbs in the sample is a random variable having an approximately normal distribution with a mean of 20 light bulbs and standard deviation of 4 light bulbs. Using these facts, we can draw the following conclusions: Example 16.11Sampling for Defective Light Bulbs

Excursions in Modern Mathematics, 7e: 16.7 - 19Copyright © 2010 Pearson Education, Inc. ■ There is a 68% chance that the number of defective light bulbs in the sample will fall somewhere between 16 and 24. ■ There is a 95% chance that the number of defective light bulbs in the sample will fall somewhere between 12 and 28. ■ The number of defective light bulbs in the sample is practically guaranteed (a 99.7% chance) to fall somewhere between 8 and 32. Example 16.11Sampling for Defective Light Bulbs

Excursions in Modern Mathematics, 7e: 16.7 - 20Copyright © 2010 Pearson Education, Inc. Probably the most important point here is that each of the preceding facts can be rephrased in terms of sampling errors (Chapter 13). For example, say we had 24 defective light bulbs in the sample; in other words, 24% of the sample (24 out of 100) are defective light bulbs. If we use this statistic to estimate the percentage of defective light bulbs overall, then the sampling error would be 4% (because the estimate is 24% and the value of the parameter is 20%). Example 16.11Sampling for Defective Light Bulbs

Excursions in Modern Mathematics, 7e: 16.7 - 21Copyright © 2010 Pearson Education, Inc. By the same token, if we had 16 defective light bulbs in the sample, the sampling error would be –4%. Coincidentally, the standard deviation is  = 4 light bulbs, or 4% of the sample. (We computed it in Example 16.10.) Thus, we can rephrase our previous assertions about sampling errors as follows: Example 16.11Sampling for Defective Light Bulbs

Excursions in Modern Mathematics, 7e: 16.7 - 22Copyright © 2010 Pearson Education, Inc. ■ When estimating the proportion of defective light bulbs coming out of the assembly line by using a sample of 100 light bulbs, there is a 68% chance that the sampling error will fall somewhere between –4% and 4%. Example 16.11Sampling for Defective Light Bulbs

Excursions in Modern Mathematics, 7e: 16.7 - 23Copyright © 2010 Pearson Education, Inc. ■ When estimating the proportion of defective light bulbs coming out of the assembly line by using a sample of 100 light bulbs, there is a 95% chance that the sampling error will fall somewhere between –8% and 8%. Example 16.11Sampling for Defective Light Bulbs

Excursions in Modern Mathematics, 7e: 16.7 - 24Copyright © 2010 Pearson Education, Inc. ■ When estimating the proportion of defective light bulbs coming out of the assembly line by using a sample of 100 light bulbs, there is a 99.7% chance that the sampling error will fall somewhere between –12% and 12%. Example 16.11Sampling for Defective Light Bulbs

Excursions in Modern Mathematics, 7e: 16.7 - 25Copyright © 2010 Pearson Education, Inc. Suppose that we have the same assembly line as in Example 16.11, but this time we are going to take a really big sample of n = 1600 light bulbs. Before we even count the number of defective light bulbs in the sample, let’s see how much mileage we can get out of the dishonest-coin principle. The standard deviation for the distribution of defective light bulbs in the sample is Example 16.12Sampling with Larger Samples

Excursions in Modern Mathematics, 7e: 16.7 - 26Copyright © 2010 Pearson Education, Inc. which just happens to be exactly 1% of the sample (16/1600 = 1%). This means that when we estimate the proportion of defective light bulbs coming out of the assembly line using this sample, we can have some sort of a handle on the sampling error. Example 16.12Sampling with Larger Samples

Excursions in Modern Mathematics, 7e: 16.7 - 27Copyright © 2010 Pearson Education, Inc. ■ We can say with some confidence (68%) that the sampling error will fall somewhere between –1% and 1%. ■ We can say with a lot of confidence (95%) that the sampling error will fall somewhere between –2% and 2%. ■ We can say with tremendous confidence (99.7%) that the sampling error will fall somewhere between –3% and 3%. Example 16.12Sampling with Larger Samples

Excursions in Modern Mathematics, 7e: 16.7 - 28Copyright © 2010 Pearson Education, Inc. In California, school bond measures require a 66.67% vote for approval. Suppose that an important school bond measure is on the ballot in the upcoming election. In the most recent poll of 1200 randomly chosen voters, 744 of the 1200 voters sampled, or 62%, indicated that they would vote for the school bond measure. Let’s assume that the poll was properly conducted and that the 1200 voters sampled represent an unbiased sample of the entire population. Example 16.13Measuring the Margin of Error of a Poll

Excursions in Modern Mathematics, 7e: 16.7 - 29Copyright © 2010 Pearson Education, Inc. What are the chances that the 62% statistic is the result of sampling variability and that the actual vote for the bond measure will be 66.67% or more? Here, we will use a variation of the dishonest- coin principle, with each vote being likened to a coin toss: A vote for the bond measure is equivalent to flipping heads, a vote against the bond measure is equivalent to flipping tails. Example 16.13Measuring the Margin of Error of a Poll

Excursions in Modern Mathematics, 7e: 16.7 - 30Copyright © 2010 Pearson Education, Inc. In this analogy, the probability (p) of “heads” represents the proportion of voters in the population that support the bond measure: If p turns out to be 0.6667 or more, the bond measure will pass. Our problem is that we don’t know p, so how can we use the dishonest-coin principle to estimate the mean and standard deviation of the sampling distribution? Example 16.13Measuring the Margin of Error of a Poll

Excursions in Modern Mathematics, 7e: 16.7 - 31Copyright © 2010 Pearson Education, Inc. We start by letting the 62% (0.62) statistic from the sample serve as an estimate for the actual value of p in the formula for the standard deviation given by the dishonest- coin principle. (Even though we know that this is only a rough estimate for p, it turns out to give us a good estimate for the standard deviation .) Example 16.13Measuring the Margin of Error of a Poll

Excursions in Modern Mathematics, 7e: 16.7 - 32Copyright © 2010 Pearson Education, Inc. Using p = 0.62 and the dishonest-coin principle, we get votes. This number represents the approximate standard deviation for the number of “heads” (i.e., voters who will vote for the school bond measure) in the sample. Example 16.13Measuring the Margin of Error of a Poll

Excursions in Modern Mathematics, 7e: 16.7 - 33Copyright © 2010 Pearson Education, Inc. If we express this number as a percentage of the sample size, we can say that the standard deviation represents approximately 1.4% of the sample (16.8/1200 = 0.014). The standard deviation for the sampling distribution of the proportion of voters in favor of the measure expressed as a percentage of the entire sample is called the standard error. (For our example, we have found above that the standard error is approximately 1.4%.) Example 16.13Measuring the Margin of Error of a Poll

Excursions in Modern Mathematics, 7e: 16.7 - 34Copyright © 2010 Pearson Education, Inc. In sampling and public opinion polls, it is customary to express the information about the population in terms of confidence intervals, which are themselves based on standard errors: A 95% confidence interval is given by two standard errors below and above the statistic obtained from the sample, and a 99.7% confidence interval is given by going three standard errors below and above the sample statistic. Example 16.13Measuring the Margin of Error of a Poll

Excursions in Modern Mathematics, 7e: 16.7 - 35Copyright © 2010 Pearson Education, Inc. For the school bond measure, a 95% confidence interval is 62% plus or minus 2  (1.4%) = 2.8%. This means that we can say with 95% confidence (we would be right approximately 95 out of 100 times) that the actual vote for the bond measure will fall somewhere between 59.2% (62 – 2.8) and 64.8% (62 + 2.8) and thus that the bond measure will lose. Example 16.13Measuring the Margin of Error of a Poll

Excursions in Modern Mathematics, 7e: 16.7 - 36Copyright © 2010 Pearson Education, Inc. Take a 99.7% confidence interval of 62% plus or minus 3  (1.4%) = 4.2%–it is almost certain that the actual vote will turn out somewhere in that range. Even in the most optimistic scenario, the vote will not reach the 66.67% needed to pass the bond measure. Example 16.13Measuring the Margin of Error of a Poll