Unit 16 Normal Distributions

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1 Unit 16 Normal Distributions Objectives: To obtain relative frequencies (probabilities) and percentiles with a population having a normal distribution While there are many different types of distributions that a population may have (see Figures 15-1a to 15-1f), normal distributions are of particular importance in statistics One of the primary reasons normal distributions are so important is because, as we have seen previously, the sampling distribution of x with simple random sampling takes on the properties of a normal distribution, with a sufficiently large sample size n Figure 15-2 gave us a somewhat detailed description of a normal distribution in terms of one, two, and three standard deviations away from the mean We have seen that a normal distribution is symmetric and bell-shaped, and that practically all of the items in a normally distributed population are within three standard deviations of the mean We now want to describe normal distributions in a much more detailed manner than merely in terms of one, two, and three standard deviations away from the mean Table A2 in the appendix provides areas under a normal density curve in terms of z-scores A normal curve described entirely in terms of z-scores is called a standard normal curve By converting raw scores to z-scores, we can use Table A2 to find many different areas under any normal density curve These areas under a normal density curve can be interpreted as relative frequencies or as probabilities In addition to using Table A2, statistical software packages, spreadsheets, programmable calculators, etc can also be used to find areas under a normal density curve The figure at the top of the Table A2 indicates that you can use this table directly to find the area under a normal curve above any value greater than the mean µ Such an area is found by obtaining the z-score of a value greater than the mean and reading the corresponding area in the body of the table The z-scores in the table are all to two decimal place accuracy, with the z-scores to the first decimal place displayed as the row labels, and the column labels providing the second decimal place The areas in the body of the table are all given to four decimal place accuracy To illustrate the use of Table A2, we shall consider several examples involving the population of weights of oranges from a particular grove Let us suppose that the weights of oranges from the grove have a normal distribution with mean µ = 781 oz and standard deviation σ = oz We shall consider finding the percentage of orange weights that lie in a given range, or, in other words, the probability that one randomly 107

3 The probability that one randomly selected orange weighs less than 1066 oz is = (or 9846%) As another illustration, we shall have you find the percentage, or relative frequency, of oranges that weigh more than 653 oz, which can also be interpreted as the probability that one randomly selected orange has a weight more than 653 oz Find the z-score for 653 oz, and label the value 653 oz on the horizontal axis in Figure 16-4 Then, shade the desired area under the normal curve, and use Table A2 to obtain the desired probability (You should find that the z-score is 097, and that the probability that one randomly selected orange weighs more than 653 oz is (or 8340%)) If the desired area under the standard normal curve is in between two given values, then we need to read Table A2 twice Let us obtain the percentage, or relative frequency, of oranges that weigh between 6 and 8 oz, which can also be interpreted as the probability that one randomly selected orange has a weight between 6 and 8 oz To find this probability, we first use µ = 781 and σ = to find the z-score of 6 oz to be = 137, and to find the z-score of 8 oz to be = The proportion of shaded area in Figure 16-5 is the desired probability The unshaded area below 6 oz in Figure 16-5 is a mirror image of the shaded area in the figure at the top of Table A2; also, the unshaded area above 8 oz in Figure 16-5 corresponds exactly to the shaded area in the figure at the top of Table A2 We find the total unshaded area by adding the entry of Table A2 in the row labeled 13 and the column labeled 007 to the entry of Table A2 in the row labeled 01 and the column labeled 004; we then obtain the desired area by subtracting this unshaded area from 1 The probability that one randomly selected orange weighs between 6 and 8 oz is 1 ( ) = (or 4704%) In the illustration just completed, the desired area under the standard normal curve was in between two values, where one was below µ, and the other was above µ We shall now consider instances where the desired area under the standard normal curve is in between two values, where either both values are greater than µ, or both values are less than µ First, we obtain the percentage, or relative frequency, of oranges that weigh between 55 and 65 oz, which can also be interpreted as the probability that one randomly selected orange has a weight between 55 and 65 oz To find this probability, we first use µ = 781 and σ = to find the z-score of 55 oz to be = 175, 109

4 and to find the z-score of 65 oz to be = 099 The proportion of shaded area in Figure 16-6 is the desired probability The unshaded area below 55 oz in Figure 16-6 is a mirror image of the shaded area in the figure at the top of Table A2; also, if the shaded and unshaded areas below 65 oz are combined together in Figure 16-6, this combined area is a mirror image of the shaded area in the figure at the top of Table A2 We find the desired area by subtracting the entry of Table A2 in the row labeled 17 and the column labeled 005 from the entry of Table A2 in the row labeled 09 and the column labeled 009 The probability that one randomly selected orange weighs between 55 and 65 oz is = (or 1210%) Now, we shall let you find the percentage, or relative frequency, of oranges that weigh between 854 and 932 oz, which can also be interpreted as the probability that one randomly selected orange has a weight between 854 and 932 oz Find the z-score for each of 854 and 932 oz, and label the values for 854 and 932 oz on the horizontal axis in Figure 16-7 Then, shade the desired area under the normal curve, and use Table A2 to obtain the desired probability (You should find that the z-scores are +055 and +114, and that the probability that one randomly selected orange weighs between 854 and 932 oz is (or 1641%)) Since practically all of the area under a normal curve is within three standard deviations of the mean, there are no z-scores in Table A2 below 309 or above +309 When one orange weight is randomly selected from the population with mean µ = 781 oz and standard deviation σ = oz, we would consider the probability of observing a weight greater than 20 oz to be practically zero (0), because the z-score of 20 oz is +923 Of course, common sense might suggest to us that the likelihood of finding an orange weighing more than 20 oz is extremely small! In a similar fashion, we would conclude that practically all oranges weigh more than 2 oz (for which the z-score is 440), or in other words, the probability of selecting an orange weighing more than 2 oz is one (1) It is sometimes desirable to obtain percentiles and percentile ranks from a density curve We define the pth percentile to be the value with p% of the total area below it and (100 p)% of the total area above it; also, the percentage of area below a given value x is defined to be the percentile rank of x To illustrate the use of Table B2 in obtaining percentiles and percentile ranks with a normally distributed population, we shall return to the population of orange weights having a normal distribution with mean µ = 781 oz and standard deviation σ = oz Finding the percentile rank for a value x from a normally distributed population simply amounts to calculating the area under the normal density curve which lies below x For instance, earlier we found the percentage of oranges weighing less than 698 oz to be 2643%; we can then say that the percentile rank of an orange weighing 698 oz is 2643 or 26 Often, the percent sign (%) is omitted when stating a percentile rank, 110

5 since a percentile rank is understood to be a percentage As another illustration, recall that we have previously found that the percentage of oranges weighing less than 1066 oz is 9846%; we can then say that the percentile rank of an orange weighing 1066 oz must be 9846 or 98 Finding the pth percentile of a normally distributed population amounts to finding the value below which lies p percent of the area under the normal density curve Since a normal curve is symmetric, the 50th percentile (ie, the median) is equal to the mean µ We determine other percentiles by using Table A2 to find the z-score of the desired percentile and converting this z-score to a raw score To illustrate, we shall obtain the 90th percentile of the orange weights and the 30th percentile of the orange weights To find the 90th percentile, we first realize that the 90th percentile must be greater than the median (which is the mean µ), implying that the z-score of the 90th percentile must be positive We then use Table A2 to determine the desired z-score, that is, the z-score of the orange weight below which lie 90% of all the orange weights and above which lie 10% of all the orange weights Since the areas listed in Table A2 are the areas under the normal curve above a positive z-score, by searching for the area closest to 010 (which is 01003), we find that the desired z-score is +128 In Figure 16-8, label the z-score +128 on the horizontal axis; if you draw a vertical line through the graph at the location of the z-score +128, you should be able to see that roughly 90% of the area under the normal curve lies below this z-score, and 10% of the area under the normal curve lies above this z-score Recall that we use x = µ + zσ to convert a z-score to a raw score Convert the z-score +128 to a raw score to obtain the 90th percentile of the orange weights (You should find that the 90th percentile is approximately 950 oz) To find the 30th percentile of the orange weights, we must first realize that the 30th percentile must be smaller than the median (which is the mean µ), implying that the z-score of the 30th percentile must be negative We then use Table A2 to determine the desired z-score, that is, the z-score of the orange weight below which lie 30% of all the orange weights and above which lie 70% of all the orange weights The areas listed in Table A2 are the areas under the normal curve above a positive z-score, but since normal distributions are symmetric, the areas listed can also be treated as the areas under the normal curve below a negative z-score By searching for the area closest to 030 (which is 03015), we find that the desired z-score is 052 In Figure 16-9, label the z-score 052 on the horizontal axis; if you draw a vertical line through the graph at the location of the z-score 052, you should be able to see that roughly 30% of the area under the normal curve lies below this z-score, and 70% of the area under the normal curve lies above this z-score Use x = µ + zσ to convert the z-score 052 to a raw score to obtain the 30th percentile of the orange weights (You should find that the 30th percentile is approximately 712 oz) We have now illustrated the use of Table A2 in obtaining desired areas under the standard normal curve and in obtaining percentiles Now that we have thoroughly discussed normal distributions, we once again recall how we have previously observed that the sampling distribution of x with simple random samples of sufficiently large size n, takes on the properties of a normal distribution; consequently, the sampling distribution of x can be approximated using a normal distribution, and we shall very shortly begin making heavy use of this fact 111

6 Self-Test Problem 16-1 Suppose the right-hand grip strength for men between the ages of 20 and 40 is normally distributed with mean 863 lbs and standard deviation 78 lbs Draw a sketch illustrating the probability that one randomly selected male between the ages of 20 and 40 will have a right-hand grip strength (a) over 90 lbs, and find this probability; (b) under 96 lbs, and find this probability; (c) over 70 lbs, and find this probability; (d) under 75 lbs, and find this probability; (e) between 85 and 100 lbs, and find this probability; (f) between 88 and 95 lbs, and find this probability; (g) between 75 and 82 lbs, and find this probability (h) Draw a sketch illustrating the probability that the right-hand grip strength for one randomly selected male is within 4 lbs of the population mean, and find this probability (i) Find the percentile rank for a male whose right-hand grip strength is 90 lbs (j) Find the quartiles for the distribution of right-hand grip strengths Answers to Self-Test Problems 16-1 (a) or 3192% (b) or 8925% (c) or 9817% (d) or 735% (e) or 5283% (f) or 2815% (g) or 2177% (h) or 3900% (i) 6808 or 68 (j) The quartiles are approximately 811, 863, and 915 lbs Summary A normal curve described entirely in terms of z-scores is called a standard normal curve Tables of standard normal probabilities (eg, Table A2) provide a detailed description of a normal density curve in terms of z-scores Using such a table together with the fact that a normal distribution is symmetric, we can obtain areas under a normal density curve, which can be interpreted as relative frequencies or as probabilities, and we can obtain percentile ranks and percentiles Statistical software packages, spreadsheets, programmable calculators, etc are often able to supply the same information as a table of standard normal probabilities Since the sampling distribution of x with simple random samples of sufficiently large size n, takes on the properties of a normal distribution, the sampling distribution of x can be approximated using a normal distribution 112

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