# In no more than 3-4 sentences, respond to the following: What is a “z Score”? Why is it used? What is the value of a z score when analyzing something such as scores on a test? After you comment, pleas

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In no more than 3-4 sentences, respond to the following: What is a “z Score”? Why is it used? What is the value of a z score when analyzing something such as scores on a test? After you comment, please reply to at least one other student’s posting for this Option A.

In no more than 3-4 sentences, respond to the following: What is a “z Score”? Why is it used? What is the value of a z score when analyzing something such as scores on a test? After you comment, pleas

Page 1 Week Three 1. Introduction to Transformed Scores. Raw scores are meaningful only when compared with some preset standard or in relation to a distribution of scores. For example, suppose a friend comes up to you and proudly announces receiving a score of 50 on a statistics exam. Unless you have additional information, a score of 50 is meaningless. To provide a context, you need to know information such as the grade cutoff points, the modal/median/mean score, or how well the rest of the class did. If your friend said 50 is the second highest score, or that it is better than 94% of the others, then you can understand the pride. Context is efficiently con veyed by transforming the particular raw score into a new value, called a transformed score. You can look at a transformed score and immediately know how its raw -score equivalent stands in a distribution of raw scores. Two types of transformed scores are percentile ranks or simply percentiles, and z scores. 2. Percentile Ranks (Percentiles). Transforming raw scores into percentile ranks gives the percentage of raw scores that are less than or equal to that particular raw score. For example, if a raw score of 50 can be transformed to a percentile rank of 94, this means 94% of the raw scores are less than or equal to 50. In other words, a raw score that is at the 94th percentile is one such that 94% of the raw scores are less than or equal to it. Many people confuse percentile ranks and percentiles with conventional “percentage” scores, such as the percentage of points obtained on an exam. For example, a “percentage” score of 60% on an exam merely tells you that you got 60% of the possible points correct. It tells nothing about how you scored compared with other people taking the exam. To transform one or more raw scores into percentile ranks, begin by constructing a cumulative frequency distribution. For e xample, suppose you measure the heights (in inches) of 15 women in your statistics class and summari ze the results as follows. X f cf 60 1 1 61 0 1 62 1 2 63 1 3 64 3 6 65 4 10 66 3 13 67 1 14 68 1 15 Page 2 Since a cf value represents the number of scores less than or equal to a certain raw score, percentile ranks may be calculated by converting the cf value to a percentage. This is actually a two -step process. The first step is to look at the cumulative frequency distribution and find the cf value corresponding to the raw score of interest. The second step is to divide this cf value by N and multiply by 100 to yield the percentile rank. For example, to calculate the percentile rank for a height of 65 inches, first find its corresponding cf of 10. Then divide 10 by N = 15 and multiply by 100. The percentile rank is 66.7: (100 )= 10 15 (100 )= .667 (100 )= 66 .7 The percentile rank of 66.7 means that 66.7% of the women are 65 inches tall or shorter. The percentile rank of 66.7 also tells you that 33.3% of the women are taller than 65 inches. In general, the formula for transforming a raw score into a percentile ra nk is in Formula 1 below : = (100 ) Sometimes you may want to convert a percentile rank to a raw score . For example, suppose you want to determine what height corresponds to the 25th percentile. You need to invert the two steps used above. The first step is to rework Formula 1 and solve for cf. This renders Formula 2 below: = 100 () Then search the cumulative frequency distribution for the raw score corresponding to the calculated cf. This raw score is the answer. Frequently, however, your calculated cf is not specifically listed in the cf column of the cumulative frequency distribution. In such cases, locate the closest listed cumulative frequency value that is larger than the calculated cf. The raw score corresponding to this listed cf value is the answer . For example, to find the raw score for the 25th percentile, use Formula 4.2. Divide the percentile rank of 25 by 100 and multiply by 15 to yield the cf of 3.75: = 100 ()= 25 100 (15 )= .25 (15 )= 3.75 Page 3 Since 3.75 is not listed in the cf column, the next larger cf value listed is 6. The raw score corresponding to the cf value of 6 is 64, and thus 64 is the raw -score equivalent for a percentile rank of 25. The procedures given above for transforming raw scores to percentiles and percentil es to raw scores often involve considerable rounding off. You should therefore not expect to be able to take a raw score and convert it to a percentile and then convert that percentile back to exactly the original raw score. More precise methods involving interpolation are available but are not often used, so these methods are not discussed in here . 3. z Scores . z scores provide another way of determining where a particular raw score lies in a distribution. Transforming a raw score into a z score tells you how many standard deviation units that score lies above or below the mean of a distribution. You were doing this at the end of Week Two when th e standard deviation was discus sed in light of a normal distribution. In the example with IQ scores, µ = 100 and σ = 15. Thus an IQ of 115 can be translated to +1σ. from the mean, an IQ of 85 can be translated to -1σ from the mean, an IQ of 130 is equivalent to + 2σ , an IQ of 70 is equivalent to -2σ, and so on. A shortcut way to denote this equivalence is with a z score. A z score tells you how many standard deviation units a raw score lies from the mean. A positive z score indicates that the raw score is above the mean, and a negative z score indicates that the raw score is below the mean. For example, an IQ of 115 is equivalent to a z sc ore of +1, an IQ of 85 is equivalent to a z score of -1, an IQ of 130 is equivalent to a z score of +2, and so on. It is easy to transform raw scores into z scores when the raw score is exactly 0, 1, 2, 3, and so on, standard deviation units above or below the mean, as are the IQs of 55, 70, 85, 100, 115, 130, and 145. But how can you determine the z score for IQs between these whole standard deviation units? For example, what are the z scores for IQs of 106, or 72? The general formula for transforming raw scores into z scores is Formula 3 below . This is the most important formula in Week Three. You will run into it in successive weeks as well. = − µ Although you already know an IQ of 130 is two standard deviations above the mean and is equivalent to a score of 2, following Formula 3 gives the derivation: Page 4 = − µ = 130 − 100 15 = 30 15 = 2 Consider another example. Suppose you measure the weights of 83 women in your statistics class and find µ = 120.29 pounds and σ = 14.75 pounds. A person with a body weight of 109 has what z score? You can use Formula 3 to find that a weight of 109 is .77 s tandard deviation units below the mean: = − µ = 109 − 120 .29 14 .75 = −11 .29 14 .75 = −.77 To convert a z score to its raw -score equivalent, simply solve Formula 4 for X: = µ+ For example, what weight corresponds to a z score of -1.5? Since a z score is the number of standard deviation units a raw score lies away from the mean, the raw -score equivalent for a z of -1.5 lies 1.5 standard deviation units below the mean. Thus the raw score is 1.5 times 14.75 pounds below 120.29, which is 98.16: = µ+ = 120 .29 + (−1.5)(14 .75 ) = 120 .29 + (−22 .13 )= 98 .16 Some people erroneously assume that you can only use z scores with normal or approximately normal distributions. In actuality, z scores may be calculated for distributions of any shape. However, you may not find 68.26% of the scores falling within one standard deviation unit above and below the mean, or 95.44% of the scores within two standard deviation units falling above and below the mean, or 99.74% of the scores falling within three standard deviation unit above and below the mean. z scores are very useful for comparing raw scores from different distributions, as well as for determining at a glance where a score lies in a distribution. For example, suppose you want to know if you are a better swimmer (measured in elapsed time) than a mathematician (measured in problems correct), or whether you are taller (measured in inches) than heavy (measured in pounds), or whether you do better on standa rdi zed achievement tests (SAT score) than on course work (GPA). Since all of these abilities/traits come from different distributions, it is impossible to compare raw scores. For example, if you swim 100 meters in 90 seconds and get 85 out of 100 math problems correct, are you a better swimmer or a better mathemati cian? Or how can you compare your SAT score with your GPA? To make such comparisons, consider first what it means to do well or score high. Scoring high is relative to the other scores in the distribution. To compare the scores in different distributions, first you need to determine how each Page 5 score compares to the rest of the scores in its respective distribution. Since a z score is a common measure of where a score stands in comparison to the rest of the scores in a distribution, it serves your purpose. Suppose you have a GPA of 3.3 and score a 735 on the SAT. To determine whether you did better on the standardi zed achievement test or in your course work, transform each score to a z score. Th e SAT has µ= 500 and σ = 100. Suppose GPA at your school has µ = 2.6 and σ = .5 grade points. The z score for your SAT score is 2.35: = − µ = 735 − 500 100 = 235 100 = 2.35 The z score for your GPA is 1.4: = − µ = 3.3− 2.6 .5 = .7 .5= 1.4 Since your SAT score is more standard deviations above the mean than your GPA, you are better at taking achievement tests than at getting grades. This type of comparison can be used to compare scores in any two distributions of the same shape. z scores are especially useful in relation to normal and approximately normal distributions. The normal distribution is very useful because it offers a close approximation to many empirical real -life ability/trait distributions. For example, due to the way IQ tests are constructed, for all practical purposes IQ is determined by a large number of random and independent factors, such as genetic contributions and contributions of the environment , and may be thought of as being approximately normally distributed. There are uncountable abilities/traits in the behavioral sciences which are approximately normally distributed. The normal distribution is symmetrical, unimodal, and bell -shaped. It is important to remember that a normal distribution implies only a particular shape and does not imply a specific mean or standard deviation. For example, SAT scores form an approximately normal distribution with µ = 500 and σ = 100, and IQ scores also form an approximately normal distribution, but with µ = 100 and σ = 15. Therefore the “normal distribution” refers to a family of distributions with varying means and standard deviations, but all with the same shape. One useful property of a normal distribution is that a specified percentage of scores falls between the mean and a given number of standard deviation units away from the mean. For example, in Figure 1, 34.13% Page 6 Figure 1 . The symmetrical characteristic of the normal distribution shown in relation to the percentage of scores falling between the mean and the positive and negative z scores of 1, 2, and 3. of the scores lie between the mean and z = +1, 47.72% of the scores lie between the mean and z = +2, and 49.87% of the scores lie between the mean and z = +3. Since the normal distribution is symmetrical, the same percentage of scores falls between the mean and a given number of standard deviations below the mean as between the mean and a given number of standard deviations above the mean. As seen in Figure 1, 34.13% of the scores lie between the mean and z = -1, 47.72% of the scores lie between the mean and z = -2, and 49.87% of the scores lie between the mean and z = -3. In general, the percentage of scores between the mean and any given z score in a normal distribution may be determined from Table z, which can be found at the end of this Week Three pdf file , and in the “Tables ” tab on the home page . To use Table z, first round off to the hundredths place. If there are no digits at the tenths and the hundredths place, add zeros to the right of the decimal as placeholders. Then look down the column labeled z for the digits in the ones and tenths places of your z scor e. This leads you to the proper row. Then find the column labeled with the digit in your z score at the hundredths place. The four – digit value listed at the intersection of the row and column corresponding to your z score is the percentage of scores fallin g between the mean and the z score. For example, to find the percentage of scores between the mean and z = 1.5, place a zero in the hundredths place, making z = 1.50. Look down the leftmost column of Table z for 1.5. Since the digit at the hundredths place is 0, look at the very top row for the column having a 0 at the hundredths place. This is the column labeled .00. The number listed at the intersection of the row and column is 43.32%, which is the percentage of raw scores lying betwee n the mean and z = Page 7 1.5. This is shown pictorially in Figure 2 . When solving problems requiring use of Table z, we urge you to draw a diagram similar to that in Figure 2 so that you can better conceptuali ze what you are looking for. Figure 2. The shaded area is the percentage of scores derived from Table z for z = 1.5. Notice that Table z lists only positive z scores because it represents the right half of the normal distribution. Since the normal distribution is symmetrical, it is not necessary to list the negative z scores. The percentage of scores between the mean and a negative z score is exactly the same as the percentage of scores between the mean and a positive z score of the same magnitude , as seen in Figure 3. . For example, Figure 3 . The shaded area is the percentage of scores that were derived from Table z for z = 2.47. The same percentage holds for z = -2.47, because the normal distribution is symmetrical. Page 8 to find the percentage of scores between the mean and z = -2.47, look up a z value of 2.47 in Table z and find 49.32%. Remember that the symmetry of the normal distribution and demonstrates why Table z may be used to obtain the percentage of scores for both positive and negative z scores. To find the percentage of scores lying above a given z score , you can take further advantage of the symmetrical character of the normal distribution. Exactly 50% of the scores lie above the mean and 50% lie below the mean. The process of finding the percentage of scores above a given z score depends on whether you have a positive or negative z score. For example, to find the percentage above a positive z score such as 2.53, Table z reve als that 49.43% of the scores fall be tween it and the mean. Since 50% of the scores fall above the mean, the percentage above z = 2.53 is found by subtracting: 50% minus 49.43% leaves .57% as the answer. This subtraction process is shown in Figure 4. Figure 4 . To determine the percentage of scores above a positive z score, subtract the percentage found in Table z from 50%. The shaded area under the curve is the desired percentage. To find the percentage of scores above a negative z score , say -.75, look in Table z to determine that 27.34% of the scores lie between the mean and z = -.75. Since 50% of the scores lie above the mean, you can find the percentage above z = -.75 by adding: 27.34% plus 50% sums to the answer of 77.34%. This addition process is shown in Figure 5. Page 9 Figure 5 . To determine the percentage of scores above a negative z score, add the percent found in Table z to 50%. The shaded area under the curve is the desired percent. The same addition and subtraction processes are used to find the percentage of scores lying b elow a given raw score, but in reverse fashion . The addition process is used with positive z scores, whereas the subtraction process is used with negative z scores. For example, to find the percentage of scores below z = 1.20, Table z reveals that 38.49% o f the scores lie between it and the mean. Since 50% of the scores fall below the mean, the percentage below a positive z score is found through adding: 50% plus 38.49% gives 88.49% as the answer (see Figure 6 below ). Figure 6 . To determine the percentage of scores below a positive z score, add the percentage found in Table z to 50%. The shaded area under the curve is the desired percentage. Page 10 Conversely, to find the percentage below a negative z sco re, say z = -1.67, Table z reveals that 45.25% of the scores lie between it and the mean. Since 50% of the scores lie below the mean, you can find the percentage below a negative z score by subtracting: 50% minus 45.25% leaves 4.75% as the answer. This process is shown in Figure 7. Figure 7 . To determine the percentage of scores below a negative z score, subtract the percent found in Table z from 50%. The shaded area under the curve is the desired percent. The addition and subtraction processes used above are also used to find the percentage of scores between two z score s. Whether you add or subtract, however, depends on whether the two z scores are on opposite sides of the mean, or on the same side of the mean. When the two z scores are on opposite sides of the mean, one z will be negative and the other will be positive. For example, suppose you want to find the percentage of scores between the z scores of -1.5 and 2.5. First find the percentage of scores between the mean and z = -1.5 from Table z, which is 43.32%. Next, fin d the percent between the mean and z = 2.5, which is 49.38%. Then the total percentage of scores between two z scores is found by adding: 43.32% plus 49.38% yields 92.70% as the answer. Figure 8 illustrates this process. Page 11 Figure 8 . To determine the percentage of scores lying between z scores on opposite sides of the mean, add the percentages found in Table z. The shaded area is the desired percentage. When the two z scores are on the same side of the mean, both will be positive or both will be negative. For example, to find the percentage of scores between z scores of 2.0 and 3.0, look in Table z to find that 47.72% falls between the mean and z = 2.0, and 49.87% falls between the mean and z = 3.0. The percentage of scores between the two z scores is the difference between the larger percentage and the smaller percentage, found by subtracting the smaller percentage from the larger: 49.87% minus 47.72% leaves 2.15% as the answer. Figure 9 represents this process. The same technique i s used to determine the percentage of scores between two negative z scores. Figure 9. To determine the percentage of scores between z scores lying on the same side of the mean, subtract the smaller percentage found in Table z from the larger percentage. The shaded area represents the desired percentage. Page 12 4. Practical Applications of z Scores . Transforming raw scores into z scores and using the information in Table z has many applications when you can assume the distribution to be approximately normally distributed. Several examples follow. Suppose you are interested in finding the percentage of college women who have body weights between 100 and 130 pounds. If you h ave an empirical frequency distribution, you could simply count the women with w e ig h t s be tw e e n 1 00 a nd 130 po un d s to s olv e t hi s pr ob le m. Bu t e v e n if you do not have an empirical frequency distribution of the weights, you can still solve this problem because you can assume the weights of college women are approximately normally distributed with µ= 120 pounds and σ = 15 pounds. You already know how to find the percentage of sc ore s be tw een a ny tw o g iven z score s in a n app roxima tely norma l distribution, but in this problem you are only given raw scores. The solution to problems like this involves two major steps. The first step is to transform the ra w sc o re s in to z sc ore s , so y ou c a n u se Ta bl e z. T he se c ond ste p is to determine the relevant percentages from Table z that are appropriate to the problem. For this problem, the first step is to transform the scores of 100 and 130 pounds to z scores: For 100 pounds : = − µ = 100 − 120 15 = −20 15 = −1.33 For 130 pounds: z = −µ = 130 −120 15 = 10 15 = .67 In the second step look in Table z for z = -1.33 and find 40.82%, and find 24.86% for z = .67. Since you are finding the percentage of scores between z’s lying on opposite sides of the mean, you must add the percentages to determine that 65.68% of college women have body weights between 100 and 130 pounds. Figure 10 illustrates this problem. Page 13 Figure 10 . To determine the percentage of raw scores lying between two raw scores on opposite sides of the mean, transform the raw scores into z scores and add the percentages found in Table z. The shaded area under the curve is the percentage of college women weighing between 100 and 130 pounds . Now suppose you are interested in finding the percentage of women having weights less than 90 pounds. First transform the raw score to a z score: For 90 pounds: Second, from Table z find 47.72%, the percentage of scores between the mean a z of -2. The percentage of weights below a z = -2 is 50% minus 47.72% or 2.28%. Figure 11 below illustrates this problem. Figure 11. To determine the percentage of scores lying below a given raw score, transform the raw score into a z and find the percentage in Table z. The shaded area under the curve is the percentage of college women weighing less than 90 pounds. Assume you are interested in finding the percentage of women weighing between 95 and 105 pounds . First transform the raw score to a z score: For 9 5 pounds: = −µ = 95 −120 15 = −25 15 = − 1.67 For 105 pounds: = −µ = 105 −120 15 = −15 15 = −1 = − µ = 90 − 120 15 = −30 15 = −2 Page 14 Second, from Table z, 45.25% is the percentage of scores between the mean and z = 1.67, and 34.13% is the percentage between the mean and z = -1. Since you want to find the percentage of scores between z’s lying on the same side of the mean, subtract the smaller percentage from the larger, leaving 11.12%. Figure 12 illustrates this problem. Figure 12 . To determine the percentage of scores between two raw scores lying on the same side of the mean, tran sform the raw scores into z’s and subtract the smaller percentage found in Table z from the larger percentage. The shaded area under the curve is the percentage of college women weighing between 95 and 105 pounds . Numerous problems arise where you must solve the inverse of the above sort of problem: you are given a percentage of raw scores and asked to solve for the raw score that cuts off that percentage . For example, suppose you are interested in starting an intel lectual club that will only admit people with IQs in the top 10% of the population. What is the lowest IQ you will admit to the club? If you do not have an empirical frequency distribution, you can still proceed if you know that IQ is approximately normall y distributed with µ = 100 and σ = 15. You can reverse the process you used to solve for a percentage in an approximately normal distribution. First, determine the percentage that needs to be looked up in the body of Table z, and find the z score that corresponds to it. Second, convert the z score to its raw -score equivalent. In this problem, if you erroneously look up 10% in the body of Table z, you will find a z score such that 10% of the scores lie between it and the mean, and 40% lie abov e it. This is not what you want. Remember, Table z only gives the percentage of scores between the mean and a given z score . In this example, you want to find a z such that 10% of the scores remain above z. Draw a diagram such as Figure 13 to help you avoid any confusion Page 15 Figure 13 . Always draw a diagram to determine the percentage to look up in Table z. In this example, although you want to find the IQ which cuts off the top 10% of the people, do not look up 10% in Table z. Table z gives only the percentage of scores between the mean and a specified z score. Thus, the shaded area under the curve is the percentage to look up to find its corresponding z score. as to which percentage to look up in Table z. Since 50% of the scores lie above the mean in a normal distribution and you want to cut off the upper 10%, you can determine by subtraction that 40% lies between the mean and the desired z score. Thus 40% is the proper percentage to look up in the body of Table z. The closest val ue to 40% you can find in Table z is 39.97, which has a corresponding z = 1.28. The second step is to convert this z score to its raw – score equivalent using Formula 4: = µ+ = 100 + (1.28 )(15 )= 119 .2 Therefore, the cutoff score for your club is 119.2. Similarly, you may wish to start an “Average Joe’s Club,” comprised only of people with IQs in the middle 50% of the population as measured with IQ. Only 25% above the mean and 25% below the mean will be admitted. The first step is to find the closest value corresponding to 25% in the body of Table z, which is 24.86%. The corresponding z = .67. Since about 25% of the IQs fall between the mean and z = .67, then, by symmetry, roughly 25% of the IQs will fall between the mean and z = -.67. Thus, the qualifying IQs for your club are the raw -score equivalents of z’s equal to -.67 and .67. For z = -.67: X= µ + zσ = 100 + ( -.67)(15) = 90 For z = .67: ‘X = µ + zσ = 100 + (.67)(15) = 110 Page 16 Therefore, anyone with an IQ from 90 to 110 will be eligible to be an “Average Joe.” This is illustrated in Figure 14. Figure 14 . Finding the IQ qualifications of the “Average Joe’s Club,” in which only the middle 50% are eligible. To summari ze, you have learned how to solve two general types of practical problems in approximately normal distributions. In one you are given raw scores or z scores and asked to calculate a percentage. In the other you are given some percentage and asked to calcula te raw -score cutoffs. Table z is used in solving both types of problems when you can assume an approximately normal distribution. The general strategy for solving each of these types of problems is summari zed below : One final problem is a spinoff of the preceding types of problems. In the new type of proble m, you want to find the frequency of score s in addition to the percentage. If a particular problem asks for the frequency of scores instead of the percentage, you can calculate the frequency provided tha t y ou know N. For e xa mple , you pre viously found tha t 11.12% of women weigh between 95 and 105 pounds. If you have to find the number of women in a class of 83 who have weights between 95 and 105 pounds, you can determine the answer by taking 11.12% of N = 83. This is equivalent to multiplying 83 by .1112, which gives the answer of 9.23 women. When you want to find a Page 17 percentage of N, first convert the percentage to a proportion 1 by moving the decimal point two places to the left. For example, 11.12% means 11.12/100, which is .1112. Then multiply N by this decimal. Similarly, you can determine the raw score that cuts off a certain number of the population by first converting th e frequency to a percentage. For example, if you want to find the weight that cuts off the five lightest women in your class, first convert the frequency of 5 to a proportion by dividing by N and then move the decimal point two places to the right to obtain a percentage: 5 83 = .0602 = 6.02% Then look up the closest value to 43.98% (50% – 6.02%) in Table z to find z = -1.55. The raw -score equivalent for z = -1.55 is 76.75 pounds. Thus 76.75 pounds is the weight separating the five lightest women from the rest. This process is illustrated in Figure 15. Figure 15 . In problems dealing with score frequencies instead of percentages, first convert the frequency to a percentage. The shaded area under the curve represents the five college women weighing less than or equal to 76.75 pounds. 1 A proportion is a fraction of 1 . In contrast, a percent is a number out of 100 . A percent can be converted to a proportion by dividing by 100 (moving the decimal point two places to the left). A proportion may be converted to a percent by multiplying by 100 (moving the decimal point two places to the right). Page 18 SUMMARY OF KEY CONCEPTS and FORMULAS for WEEK 3 1. A transformed score provides context, telling where its raw -score equivalent falls in a distribution of scores. 2. Percentile rank , or percentile, refers to the percentage of scores that are less than or equal to its raw -score equivalent. Raw scores are transformed to percentile ranks by making a cumulative frequency distribution and then calculating: = (100 ) 3. A z score refers to the number of standard deviation units away from the mean that its raw -score equivalent lies. z scores are useful in comparing raw scores in different distributions of the same shape. Raw scores are transformed to z scores by : = − µ To convert a z score to its raw -score equivalent, use the formula below: = µ+ 4. The normal distribution is a theoretical distribution and is defined mathematically. It approximates many empirical distributions of abilities/ traits. Approximately normal distributions arise when an ability /trait is determined by the sum of many random, independent factors. 5. There are an infinite number of normal distributions, each with a different mean and standard deviation. However, they are all the same shape and thus share the property that the same percentage of scores occurs between the mean and a given z score. These percentages are listed in Table z. 6. Table z can be used to calc ulate the approximate percentage of scores between the mean and a given z score in empirical distributions that can be approximated by the normal distribution. Page 19 Practice Problems for Week 3 1. Use the data in the frequency distribution below to make the following calculations: a. Calculate the percentile rank of a raw score of 4. b. Calculate the percentile rank of a raw score of 9. c. Calculate the raw score corresponding to the 25th percentile. d. Calculate the raw score corresponding to the 50th percen tile. e. Calculate the raw score corresponding to the 75th percentile. X f 0 1 1 0 2 1 3 2 4 1 5 6 6 7 7 12 8 10 9 5 10 5 2. Bu zz A. Field commutes to his statistics class by flying his helicopter to the campus landing pad. The numbers of minutes that it has taken Bu zz to fly to campus each class session for the past six weeks are recorded below: 8 14 14 12 10 8 9 9 10 7 11 12 11 10 9 15 10 11 a. Calculate the percentile rank of a raw score of 8. b. Calculate the percentile rank of a raw score of 13. c. Calculate the raw score corresponding to the 50th percentile. d. Calculate the raw score corresponding to the 90th percentile. 3. Given a normal distribution whose mean is 50 and standard deviation is 6.5, transform the following raw scores to their corresponding z scores: 57 43.5 75 64.1 57.5 4. Given a normal distribution whose mean is 18.5 and standard deviation is 3, calculate the raw scores corresponding to the following z scores: Page 20 1.5 3.25 — 1.5 — .67 2.33 5. Given a normal distribution whose mean is 75 and standard deviation is 20, make the calculations below: a. Calculate the percentage of scores between the mean and the following raw scores: 95 45 84 37 77.6 28.25 b. Calculate the percentage of scores above each of the following raw scores: 95 45 84 37 77.6 28.25 c. Calculate the percentage of scores below each of the following raw scores: 95 45 84 37 77.6 28.25 d. Calculate the percentage of scores between the following raw scores: 28.25 to 37 45 to 84 77.6 to 95 6. Given a normally distributed population of 400 scores whose mean is 500 and standard deviation is 50, make the calculations below: a. Calculate the frequency of scores between the mean and the following raw scores: 380 590 415 515 646 b. Calculate the frequency of scores above each of the following raw scores: 380 590 415 515 646 c. Calculate the frequency of scores below each of the following raw scores: 380 590 415 515 646 d. Calculate the frequency of scores between the following ra w scores: 380 to 415 415 to 590 515 to 646 Page 21 7. Persons applying to the Ph.D. program in computer science are required to take the GRE (Graduate Record Examination) aptitude test as part of their admission requirements. Applicants are not considered for ad mission if their quantitative ability scores fall below the “cutoff score. Given that the quantitative scores of all persons taking the test is normally distributed with a mean of 530 and a standard deviation of 128, make the calculations below: a. Determ ine the quantitative ability score that cuts off the top 10 percent. This score is the Ph.D. program’s cutoff score. b. College Joe obtained a quantitative ability score of 720. Does he have a chance to be admitted to the Ph.D. program? c. College Jane a lso took the GRE test and applied to the Ph.D. program. However, all she knows is that her z score on quantitative ability is 2.13. Does she have a chance to be admitted to the Ph.D. program? d. Assuming that 7200 people took the GRE test, determine the quantitative ability score that cuts off the top 600 people. e. When College Joe took the GRE test he obtained a verbal ability score of 740. Given that the verbal ability scores of all persons taking the test are normally distributed with a mean of 510 a nd a standard deviation of 157, determine whether College Joe was better in verbal ability or in quantitative ability. Page 22 ANSWERS TO PROBLEMS In the answers below where it says : use Formula 4.1, use Formula 1 where is says to use Formula 4.2, use Formula 2 where is says to use Formula 4. 3, use Formula 3 and where is says to use Formula 4. 4, use Formula 4 Page 23 Page 24 b. For problems concerning the percentage of scores falling above a given raw score, the strategy is to proceed as in (a). Then, if z is positive, subtract the percentage found in Table z from 50%. If z is negative, add the percentage found in Table z to 50% . Draw diagrams to make things clearer. For 95: Since z = 1, then subtract: 50% -34.13% = 15.87% Page 25 Page 26 Page 27 Page 28 Page 29 Page 30

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