course Mth 152 ȷpۏ撴奾ˈassignment #014
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21:30:34 `q001. Note that there are 8 questions in this assignment. {}{} What is the average, or mean value, of the numbers 5, 7, 9, 9, 10, 12, 13, and 15? On the average how 'far' is each number from this mean value?
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RESPONSE --> Mean: 80 / 8 = 10 (5, 3, 1, 1, 0, 2, 3, 5) = 20 20 / 8 =2.5 confidence assessment: 3
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21:31:52 To get the mean value of the numbers, we first note that there are eight numbers. Then we had the numbers and divide by eight. We obtain 5 + 7 + 9 + 9 + 10 + 12 + 13 + 15 = 80. Dividing by 8 we obtain mean = 80 / 8 = 10. The difference between 5 and the mean 10 is 5; the difference between 7 and the mean 10 is 3; the difference between 9 and 10 is 1; the differences between 12, 13 and 15 and the mean 10 are 2, 3 and 5. So we have differences 5, 3, 1, 1, 0, 2, 3 and 5 between the mean and the numbers in the list. The average difference between the mean and the numbers in the list is therefore ave difference = ( 5 + 3 + 1 + 1 + 0 + 2 + 3 + 5 ) / 8 = 20 / 8 = 2.5.
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RESPONSE --> Ok. self critique assessment: 3
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21:33:22 `q002 What is the middle number among the numbers 13, 12, 5, 7, 9, 15, 9, 10, 8?
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RESPONSE --> 5, 7, 8, 9, 9, 10, 12, 13, 15 Median: 9 confidence assessment: 3
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21:34:17 It is easier to answer this question if we place the numbers in ascending order. Listed in ascending order the numbers are 5, 7, 8, 9, 9, 10, 12, 13, and 15. We see that there are 9 numbers in the list. If we remove the first 4 and the last 4 we are left with the middle number. So we remove the numbers 5, 7, 8, 9 and the numbers 10, 12, 13, and 15, which leaves the second '9' as the middle number.
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RESPONSE --> Ok. self critique assessment: 3
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21:34:55 `q003. On a list of 9 numbers, which number will be the one in the middle? Note that the middle number is called the 'median'.
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RESPONSE --> The fifth number. confidence assessment: 3
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21:37:46 If the 9 numbers are put in order, then we can find the middle number by throwing out the first four and the last four numbers on the list. We are left with the fifth number on the list. In general if we have an odd number n of number in an ordered list, we throw out the first (n-1) / 2 and the last (n-1) / 2 numbers, leaving us with the middle number, which is number (n-1)/2 + 1 on the list. So for example if we had 179 numbers on the list, we would throw out the first (179 - 1) / 2 = 178/2 = 89 numbers on the list and the last 89 numbers on the list, leaving us with the 90th number on the list. Note that 90 = (179 - 1) / 2 + 1, illustrating y the middle number in number (n-1)/2 + 1 on the list.
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RESPONSE --> Ok. self critique assessment: 3
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21:38:37 `q004. What is the median (the middle number) among the numbers 5, 7, 9, 9, 10, 12, 13, and 15?
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RESPONSE --> In between the fourth and fifth. Median: 4.5 confidence assessment: 3
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21:38:53 There are 8 numbers on this list. If we remove the smallest then the largest our list becomes 7, 9, 9, 10, 12, 13. If we remove the smallest and the largest from this list we obtain 9, 9, 10, 12. Removing the smallest and the largest from this list we are left with 9 and 10. We are left with two numbers in the middle; we don't have a single 'middle number'. So we do the next-most-sensible thing and average the two numbers to get 9.5. We say that 9.5 is the middle, or median, number.
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RESPONSE --> Ok. self critique assessment: 3
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21:44:48 `q005. We saw that for the numbers 5, 7, 9, 9, 10, 12, 13, and 15, on the average each number is 2.5 units from the average. Are the numbers in the list 48, 48, 49, 50, 51, 53, 54, 55 closer or further that this, on the average, from their mean?
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RESPONSE --> Closer: 2.25 to 2.5 average. confidence assessment: 3
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21:45:51 The mean of the numbers 48, 48, 49, 50, 51, 53, 54, and 55 is (48 + 48 + 49 + 50 + 51 + 53 + 54 + 55) / 8 = 408 / 8 = 51. 48 is 3 units away from the mean 51, 49 is 2 units away from the mean 51, 50 is 1 unit away from the mean 51, and the remaining numbers are 2, 3 and 4 units away from the mean of 51. So on the average the distance of the numbers from the mean is (3 + 3 + 2 + 1 + 0 + 2 + 3 + 4) / 8 = 18 / 8 = 2.25. This list of numbers is a bit closer, on the average, then the first list.
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RESPONSE --> Ok. self critique assessment: 3
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21:50:36 `q006. On a 1-10 rating of a movie, one group gave the ratings 1, 8, 8, 9, 9, 10 while another gave the ratings 7, 7, 8, 8, 9, 10. Find the mean (average) and the median (middle value) of each group's ratings. Which group would you say liked the movie better?
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RESPONSE --> Mean: 45 / 6 = 7.5 Median: 8.5 Mean: 49 / 6 = 8.17 Median: 8 Group one, with the exception of the low scorer, like the movie better. confidence assessment: 3
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21:51:14 The mean of the first list is (1 + 8 + 8 + 9 + 9 + 10) / 6 = 45 / 6 = 7.5. The median is obtained a throwing out the first 2 numbers on the list and the last 2 numbers. This leaves the middle two, which are 8 and 9. The median is therefore 8.5. The mean of the numbers on the second list is (7 + 7 + 8 + 8 + 9 + 10) / 6 = 49 / 6 = 8 .16. The median of this list is found by removing the first 210 the last 2 numbers on the list, leaving the middle two numbers 8 and 8. The median is therefore 8. The first group had the higher median and the lower mean, while the second group had the lower median but the higher mean. Since everyone except one person in the first group scored the movie as 8 or higher, and since everyone in both groups except this one individual scored the movie 7 or higher, it might be reasonable to think that the one anomalous score of 1 is likely the result of something besides the quality of the movie. We might also note that this score is much further from the mean that any of the other scores, giving it significantly more effect on the mean than any other score. We might therefore choose to use the median, which limits the otherwise excessive weight given to this unusually low score when we calculate the mean. In this case we would say that the first group liked the movie better.
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RESPONSE --> Ok. self critique assessment: 3
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21:54:50 `q007. Suppose that in a certain office that ten employees make $700 per pay period, while five make $800 per pay period and the other two make $1000 per pay period. What is the mean pay per period in the office? What is the median?
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RESPONSE --> 10 * 700 + 5 * 800 + 2 * 1000 = 13000 Mean: 13000 / 17 = 764 Median: 700 confidence assessment: 3
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21:55:32 There are a total of 10 + 5 + 2 = 17 employees in the office. The total pay per pay period is 10 * $700 + 5 * $800 + 2 * $1000 = $13,000. The mean pay per period is therefore $13,000 / 17 = $823 approx.. The median pay is obtained by 'throwing out' the lowest 8 and the highest 8 in an ordered list, leaving the ninth salary. Since 10 people make $700 per period, this leaves $700 as the median.
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RESPONSE --> Ok. self critique assessment: 3
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21:56:49 `q008. In the preceding problem ten employees make $700 per pay period, while five make $800 per pay period and the other two make $1000 per pay period; we just found that the mean pay per period was $823. On the average, how much to the individual salaries differ from the mean?
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RESPONSE --> confidence assessment:
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21:56:53 The mean was found in the preceding problem to be $823. The deviation of $700 from the mean is therefore $123, the deviation of $800 from the mean is $23 and the deviation of $1000 from the mean is $177. Since $700 is paid to 10 employees, $800 to five and $1000 to two, the total deviation is 10 *$123 + 5 * $23 + 2 * $177 = $1630. The mean deviation is therefore $1630 / 17 = $96, approx..
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RESPONSE --> self critique assessment:
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~TËPĽnЫ assignment #015 015. range vs ave dev vs std dev Liberal Arts Mathematics II 07-08-2007
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22:01:56 `q001. Note that there are 8 questions in this assignment. {}{}In what ways can you measure how 'spread out' the distribution 7, 9, 10, 11, 12, 14 is?
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RESPONSE --> The average difference from the mean. confidence assessment: 3
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22:07:34 We could calculate the average distance of the numbers from the mean. The mean of these numbers is (7 + 9 + 10 + 11 + 12 + 14) / 6 = 10.5. The deviations from the mean are 3.5, 2.5, .5, .5, 1.5, 3.5. Averaging these deviations we get ave deviation from mean = (3.5 + 2.5 + .5 + .5 + 1.5 + 3.5) / 6 = 2. A simpler measure of the spread is the range, which is the difference 14 - 7 = 7 between the lowest and highest number.
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RESPONSE --> Ok. self critique assessment: 3
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22:25:29 `q002. Comparing the distributions 7, 9, 10, 11, 12, 14 and 7, 8, 9, 12, 13, 14, which distribution would you say is more spread out?
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RESPONSE --> The second group is more spread out from their mean. confidence assessment: 3
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22:27:33 Both distributions have the same range, which is 14 - 7 = 7. Note that both distributions have the same mean, 10.5. However except for the end numbers 7 and 14, the numbers in the second distribution are spread out further from the mean (note that 8 and 9 in the second distribution are further from the mean than are 9 and 10 in the first, and that 12 and 13 in the second distribution are further from the mean that are 11 and 12 from the first). We can easily calculate the average deviation of the second distribution from the mean, and we find that the average deviation is 2.67, which is greater than the average deviation in the first.
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RESPONSE --> Ok. self critique assessment: 3
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22:42:28 `q003. Another measure of the spread of a distribution is what is called the standard deviation. This quantity is similar in many respects to the average deviation, but this measure of deviation is more appropriate to statistical analysis. To calculate the standard deviation of a distribution of numbers, we begin as before by calculating the mean of the distribution and then use the mean to calculate the deviation of each number from the mean. For the distribution 7, 9, 10, 11, 12, 14 we found that the mean was to 10.5 deviations were 3.5, 2.5, .5, .5, 1.5 and 3.5. To calculate the standard deviation, we first square the deviations to find the squared deviations. We then average the squared deviations. What the you get for the squared deviations, then for the average of the squared deviations, for the given distribution?
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RESPONSE --> (12.25 + 6.25 + 0.25 + 0.25 + 2.25 + 12.25) / 6 = 5.67 confidence assessment: 3
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22:45:36 The squared deviations are 3.5^2 = 12.25, 2.5^2 = 6.25, .5^2 = .25, and 1.5^2 = 2.25. Since 3.5 and .5 to occur twice each, the average of the squared deviations is (12.25 + 6.25 + 0.25 + 0.25 + 2.25 + 12.25) / 6 = 5.67. This average of the squared deviations is not the standard deviation, which will be calculated in the next exercise.
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RESPONSE --> Ok. self critique assessment: 3
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22:50:00 `q004. In the last problem we calculated the average of the squared deviations. Since this average was calculated from the squared deviations, it seems appropriate to now take the square root of our result. The standard deviation is the square root of the average of the squared deviations. Continuing the last problem, what is the standard deviation of the distribution 7, 9, 10, 11, 12, 14?
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RESPONSE --> Sqrt (5.67) = 2.38 confidence assessment: 3
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22:50:59 The standard deviation is a square root of the average of the squared deviations. We calculated the average of the squared deviations in the last exercise, obtaining 5.67. So to get the standard deviation we need only take the square root of this number. We thus find that standard deviation = `sqrt(ave of squared deviations) = `sqrt(5.67) = 2.4, approx..
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RESPONSE --> Ok. self critique assessment: 3
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22:58:46 `q005. The last problem didn't really lie to you, there is one more subtlety in the calculation of the standard deviation. When we calculate the standard deviation for a distribution containing less than about 30 numbers, then in the step where we calculated the average deviation we do something a little bit weird. Instead of dividing the total of the squared deviations by the number of values we totaled, we divide by 1 less than this number. So instead of dividing (12.25 + 6.25 + 0.25 + 0.25 + 2.25 + 12.25) by 6, as we did, we only divide by 5. With this modification, what is the standard deviation?
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RESPONSE --> (12.25 + 6.25 + 0.25 + 0.25 + 2.25 + 12.25) / 5 = 6.8 Sqrt(6.8) = 2.61 confidence assessment: 3
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23:00:07 The total of the squared deviations is (12.25 + 6.25 + 0.25 + 0.25 + 2.25 + 12.25) = 34. When we divide by 5 instead of 6 we get (12.25 + 6.25 + 0.25 + 0.25 + 2.25 + 12.25) / 5 = 6.8. This 'average' of the squared deviations (not really the average but the 'average' we use in calculating the standard deviation) is therefore 6.8, not the 5.67 we obtain before. Thus the standard deviation is std dev = square root of 'average' of squared deviations = `sqrt(6.8) = 2.6, approximately. Note that this value differs slightly from that obtained by doing a true average. Note also that if we are totaling 30 or more squared deviations subtracting the 1 doesn't make much difference, and we just use the regular average of the squared deviations.
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RESPONSE --> Ok. self critique assessment: 3
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23:19:06 `q006. We just calculated the standard deviation of the distribution 7, 9, 10, 11, 12, 14. Earlier we noted that the distribution 7, 8, 9, 12, 13, 14 is a bit more spread out than the distribution 7, 9, 10, 11, 12, 14. Calculate the standard deviation of the distribution 7, 8, 9, 12, 13, 14 and determine how much difference the greater spread makes in the standard deviation.
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RESPONSE --> 42 / 5 = 8.4 Sqrt(8.4) = 2.90 Greater spread increases by around 0.3 confidence assessment: 3
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23:20:55 The mean of the distribution 7, 8, 9, 12, 13, 14 is still 10.5. The deviations are 3.5, 2.5, 1.5, 1.5, 2.5 and 3.5, giving us squared deviations 12.25, 6.25, 2.25, 2.25, 6.25 and 12.25. The total of the squared deviations is 42, and the 'average', as we compute it using division by 5 instead of the six numbers we totaled, is 42/5 = 8.4. The standard deviation is therefore the square root of this 'average', or std dev = `sqrt(8.4) = 2.9, approximately. We see that the greater spread increases are standard deviation by about 0.3 over the previous result.
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RESPONSE --> Ok. self critique assessment: 3
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23:37:40 `q007. What is the standard deviation of the distribution 7, 8, 8, 8, 13, 13, 13, 14. What would be the quickest way to calculate this standard deviation?
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RESPONSE --> 64 / 7 = 9.14 Sqrt(9.14) = 3.02 confidence assessment: 3
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23:42:21 The mean of these numbers is easily found to be 10.5. Note that we have here still another distribution with mean 10.5 and range 7. The deviations from the mean are 3.5, 2.5, 2.5, 2.5, 2.5, 2.5, 2.5, 3.5. The squared deviations are 12.25, 6.25, 6.25, 6.25, 6.25, 6.25, 6.25, 12.25. The sum of these squared deviations is 64. There are 8 numbers in the distribution, so in calculating the modified 'average' use with the standard deviation we will divide the total 64 by 8 - 1 = 7 to get a modified 'average' of the squared deviations equal to 64/7 = 9.1. Taking the square root to get the standard deviation we obtain approximately 3.03. The quickest way to have calculated this standard deviation would be to note that the deviations of 7, 8, 13, and 14 from our previously calculated mean of 10.5 are respectively 3.5, 2.5, 2.5, and 3.5, corresponding to square deviations of 12.25, 6.25, 6.25, and 12.25. Noting that since 8 occurs three times and 13 occurs three times, the total of the squared deviations will be 12.25 + 3 * 6.25 + 3 * 6.25 + 12.25 = 12.25 + 18.75 + 18.75 + 12.25 = 64. The rest of the calculation is done as before. Using multiplication instead of addition to calculate the sum of the repeated numbers is more efficient then doing unnecessary repeated additions.
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RESPONSE --> Ok. self critique assessment: 3
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23:46:35 `q008. What is the maximum possible standard deviation for a set of six numbers ranging from 7 through 14 and averaging (7 + 14 ) / 2 = 10.5?
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RESPONSE --> 75 / 5 = 15 Sqrt(15) = 3.87 confidence assessment: 2
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23:46:59 The maximum possible spread of the distribution would be achieved when half of the numbers are all 7 and the other half are all 14. This would give us the distribution of 7, 7, 7, 14, 14, 14. Each of these six numbers has a deviation of 3.5 from the mean of 10.5. Thus the squared deviation for each number is 12.25. Since there are six numbers in the distribution, the total of the squared deviations must be 6 * 12.25 = 75. Our modified average of the squared distributions will therefore be 75/5 = 15, and the standard deviation will be square root of 15 or approximately 3.9.
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RESPONSE --> Ok. self critique assessment: 3
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~\˟aꦩ assignment #016 016. mean, std dev of freq dist (incl binomial) Liberal Arts Mathematics II 07-08-2007
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00:35:47 `q001. Note that there are 8 questions in this assignment. {}{}When rolling 2 dice a number of times, suppose you get a total of 5 on four different rolls, a total of 6 on seven rolls, a total of 7 on nine rolls, and total of 8 of six rolls a total of 9 on three rolls. What was your mean total per roll of the two dice?
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RESPONSE --> 20 + 42 + 63 + 48 + 27 = 200 200 / 29 = 6.90 confidence assessment: 3
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00:37:22 You obtained four 5's, which total 4 * 5 = 20. You obtained seven 6's, which total 7 * 6 = 42. You obtained nine 7's, which total 9 * 7 = 63. You obtained six 8's, which total 6 * 8 = 48. You obtained three 9's, which total 3 * 9 = 27. The total of all the outcomes is therefore 20 + 42 + 63 + 48 + 27 = 200. Since there are 4 + 7 + 9 + 6 + 3 = 29 outcomes (i.e., four outcomes of 5 plus 7 outcomes of 6, etc.), the mean is therefore 200/29 = 6.7, approximately. This series of calculations can be summarized in a table as follows: Result Frequency Result * frequency 5 4 20 6 7 42 7 9 63 8 6 48 9 3 27 9 3 27 ___ ____ ____ 29 200 mean = 200 / 29 = 6.7
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RESPONSE --> Ok. self critique assessment: 3
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00:54:54 `q002. The preceding problem could have been expressed in the following table: Total Number of Occurrences 5 4 6 7 7 9 8 6 9 3 This table is called a frequency distribution. It expresses each possible result and the number of times each occurs. You found the mean 6.7 of this frequency distribution in the preceding problem. Now find the standard deviation of the distribution.
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RESPONSE --> 41.7 / 28 = 1.49 Sqrt(1.49) = 1.22 confidence assessment: 2
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00:57:30 We must calculate the square root of the 'average' of the squared deviation. We calculate the deviation of each result from the mean, then find the squared deviation. To find the total of the squared deviations we must add each squared deviation the number of times which is equal to the number of times the corresponding result occurs. For example, the first result is 5 and it occurs four times. Since the deviation of 5 from the mean 6.7 is 1.7, the squared deviation is 1.7^2 = 2.89. Since 5 occurs four times, the squared deviation 2.89 occurs four times, contributing 4 * 2.89 = 11.6 to the total of the squared deviations. Using a table in the manner of the preceding exercise we obtain Result Freq Result * freq Dev Sq Dev Sq Dev * freq 5 4 20 1.7 2.89 11.6 6 7 42 .7 0.49 3.4 7 9 63 0.3 0.09 0.6 8 6 48 1.3 1.69 10.2 9 3 27 2.3 5.29 15.9 ___ ____ ____ ___ 29 200 41.7 mean = 200 / 29 = 6.7 'ave' squared deviation = 41.7 / (29 - 1) = 1.49 std dev = `sqrt(1.49) = 1.22
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RESPONSE --> Ok. self critique assessment: 3
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01:13:31 `q003. If four coins are flipped, the possible numbers of 'heads' are 0, 1, 2, 3, 4. Suppose that in an experiment we obtain the following frequency distribution: # Heads Number of Occurrences 0 4 1 20 2 22 3 13 4 3 What is the mean number of 'heads' and what is the standard deviation of the number of heads from this mean?
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RESPONSE --> 115 / 62 = 1.85 51 / 62 = .82 Sqrt(.82) = .91 confidence assessment: 2
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01:16:27 Using the table in the manner of the preceding problem we obtain the following: Result Freq Result * freq Dev Sq Dev Sq Dev * freq 0 4 0 1.86 3.5 0 1 20 20 0.86 0.7 14 2 22 44 0.24 0.1 2 3 13 39 1.24 1.5 20 4 3 12 2.24 5.0 15 ___ ____ ____ ___ 62 115 51 mean = 115 / 62 = 1.86 approx. Note that the mean must be calculated before the Dev column is filled in. 'ave' squared deviation = 51 / 62 = .83 std dev = `sqrt(.83) = .91
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RESPONSE --> Ok. self critique assessment: 3
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01:23:32 `q004. If we rolled 2 dice 36 times we would expect the following distribution of totals: Total Number of Occurrences 2 1 3 2 4 3 5 4 6 5 7 6 8 5 9 4 10 3 11 2 12 1 What is the mean of this distribution and what is the standard deviation?
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RESPONSE --> 252 / 36 = 7 230 / 36 = 6.39 Sqrt(6.39) = 2.53 confidence assessment: 2
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01:33:02 Using the table in the manner of the preceding problem we obtain the following: Result Freq Result * freq Dev Sq Dev Sq Dev * freq 2 1 2 5 25 25 3 2 6 4 16 32 4 3 12 3 9 37 5 4 20 2 4 16 6 5 30 1 1 5 7 6 42 0 0 0 8 5 40 1 1 5 9 4 36 2 4 16 10 3 30 3 9 27 11 2 22 4 16 32 12 1 12 5 25 25 ___ ____ ____ ___ 36 252 230 mean = 252 / 36 = 7. Note that the mean must be calculated before the Dev column is filled in. 'ave' squared deviation = 230 / 36 = 6.4 approx. std dev = `sqrt(6.4) = 2.5 approx.
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RESPONSE --> Ok. self critique assessment: 3
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01:52:38 `q005. If we flip n coins, there are C(n, r) ways in which we can get r 'heads' and 2^n possible outcomes. The probability of r 'heads' is therefore C(n, r) / 2^n. If we flip five coins, what is the probability of 0 'heads', of 1 'head', of 2 'heads', of 3 'heads', of 4 'heads', and of 5 'heads'?
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RESPONSE --> C(5,0) = 1 / 32 C(5,1) = 5 / 32 C(5,2) = 10 / 32 C(5,3) = 10 / 32 C(5,4) = 5 / 32 C(5,5) = 1 / 32 confidence assessment: 3
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01:59:32 If we flip 5 coins, then n = 5. To get 0 'heads' we find C(n, r) with n = 5 and r = 0, obtaining C(5,0) = 1 way to get 0 'heads' out of the 2^5 = 32 possibilities for a probability of 1 / 32. To get 1 'heads' we find C(n, r) with n = 5 and r = 1, obtaining C(5,1) = 5 ways to get 1 'heads' out of the 2^5 = 32 possibilities for a probability of 5 / 32. To get 2 'heads' we find C(n, r) with n = 5 and r = 2, obtaining C(5,2) = 10 ways to get 2 'heads' out of the 2^5 = 32 possibilities for a probability of 10 / 32. To get 3 'heads' we find C(n, r) with n = 5 and r = 3, obtaining C(5,3) = 10 ways to get 3 'heads' out of the 2^5 = 32 possibilities for a probability of 10 / 32. To get 4 'heads' we find C(n, r) with n = 5 and r = 4, obtaining C(5,4) = 5 ways to get 4 'heads' out of the 2^5 = 32 possibilities for a probability of 5 / 32. To get 5 'heads' we find C(n, r) with n = 5 and r = 5, obtaining C(5,5) = 1 way to get 5 'heads' out of the 2^5 = 32 possibilities for a probability of 1 / 32.
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RESPONSE --> Ok. self critique assessment: 3
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02:03:27 `q006. The preceding problem yielded probabilities 1/32, 5/32, 10/32, 10/32, 5/32 and 1/32. On 5 flips, then, we the expected values of the different numbers of 'heads' would give us the following distribution: : # Heads Number of Occurrences 0 1 1 5 2 10 3 10 4 5 5 1 Find the mean and standard deviation of this distribution.
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RESPONSE --> 80 / 32 = 2.5 40 / 32 = 1.25 Sqrt(1.25) = 1.12 confidence assessment: 3
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02:03:51 Using the table in the manner of the preceding problem we obtain the following: Result Freq Result * freq Dev Sq Dev Sq Dev * freq 0 1 0 2.5 6.25 6.25 1 5 5 1.5 2.25 11.25 2 10 20 0.5 0.25 2.50 3 10 30 0.5 0.25 2.50 4 5 20 1.5 2.25 12.25 5 1 5 2.5 6.25 6.25 ___ ____ ____ ___ 32 80 32.00 mean = 80 / 32 = 2.5. Note that the mean must be calculated before the Dev column is filled in. 'ave' squared deviation = 40 / 32 = 1.25. Thus std dev = `sqrt(1.25) = 1.12 approx.
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RESPONSE --> Ok. self critique assessment: 3
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02:15:01 `q007. Suppose that p is the probability of success and q the probability of failure on a coin flip, on a roll of a die, or on any other action that must either succeed or fail. If a trial consists of that action repeated n times, then the average number of successes on a large number of trials is expected to be n * p. For large values of n, the standard deviation of the number of successes is expected to be very close to `sqrt( n * p * q ). For values of n which are small but not too small, the standard deviation will still be close to this number but not as close as for large n. If the action is a coin flip and 'success' is defined as 'heads', then what is the value of p and what is the value of q? For this interpretation in terms of coin flips, if n = 5 then what is n * p and what does it mean to say that the average number of successes will be n * p? In terms of the same interpretation, what is the value of `sqrt(n * p * q) and what does it mean to say that the standard deviation of the number of successes will be `sqrt( n * p * q)? How does this result compare with the result you obtained on the preceding problem?
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RESPONSE --> N * P = 5 * .5 = 2.5 Sqrt(5*.5^2) = 1.12 confidence assessment: 2
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02:17:59 We first identify the quantities p and q for a coin flip. Success is 'heads', which for a fair coin occurs with probability .5. Failure therefore has probability 1 - .5 = .5. Now if n = 5, n * p = 5 * .5 = 2.5, which represents the mean number of 'heads' on 5 flips. The idea that the mean number of occurrences of some outcome with probability p in n repetitions is n * p should by now be familiar (e.g., from basic probability and from the idea of expected values). For n = 5, we have `sqrt(n * p * q) = `sqrt(5 * .5 * .5) = `sqrt(1.25) = 1.12, approx.. In the preceding problem we found that the standard deviation expected on five flips of a coin should be exactly 1. This differs from the estimate `sqrt(n * p * q) by a little over 10%, which is a fairly small difference.
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RESPONSE --> Ok. self critique assessment: 3
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02:27:53 `q008. Suppose that p is the probability of success and q the probability of failure on a coin flip, on a roll of a die, or on any other action that must either succeed or fail. If a trial consists of that action repeated n times, then the average number of successes on a large number of trials is expected to be n * p. The standard deviation of the number of successes is expected to be `sqrt( n * p * q ). If the action is a roll of a single die and a success is defined as rolling a 6, then what is the probability of a success and what is the probability of a failure? If n = 12 that means that we count the number of 6's rolled in 12 consecutive rolls of the die, or alternatively that we count the number of 6's when 12 dice are rolled. How many 6's do we expect to roll on an average roll of 12 dice? What do we expect is the standard deviation the number of 6's on rolls of 12 dice?
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RESPONSE --> Sqrt(12*1/6*5/6) = 1.29 confidence assessment: 3
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02:29:27 We first identify the quantities p and q for a rolling a die. Success is defined in this problem as getting a 6, which for a fair die occurs with probability 1/6. Failure therefore has probability 1 - 1/6 = 5/6. Now if n = 12, n * p = 12 * 1/6 = 2, which represents the mean number of 6's expected on 12 rolls. This is the result we would expect. For n = 12, we have `sqrt(n * p * q) = `sqrt(12 * 1/6 * 5/6) = `sqrt(1.66) = 1.3, approx..
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RESPONSE --> Ok. self critique assessment: 3
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ymydžəӽ assignment #014 014. `query 14 Liberal Arts Mathematics II 07-09-2007
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17:13:16 query problem 13.1.6 freq dist 35 IQ scores class width 5
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RESPONSE --> 91:95 1/50 = .02 96:100 3/50 = .06 101:105 5/50 = .1 106:110 7/50 = .14 111:115 12/50 = .24 116:120 9/50 = .18 121:125 8/50 = .16 126:130 3/50 = .06 131:135 2/50 = .04
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17:16:37 ** These numbers might or might not be completely accurate, but the calculation of the relative frequencies as percents follows from the tallies: 91-95 1 1/50 = 2% 96-100 3 3/50 = 6% 101-105 5 5/50 = 10% 106-110 7 7/50 = 14% 111-115 12 12/50 = 24% 116-120 9 9/50 = 18% 121-125 8 8/50 = 16% 126-130 3 3/50 = 6% 131-135 2 2/50 = 4% **
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RESPONSE --> Ok.
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17:27:53 query problem 13.1.10 stem and leaf for yards gained by 44 rushers
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RESPONSE --> 0: 3, 7 1: 2, 6, 8, 9 2: 2, 4, 5, 8, 9, 9 3: 0, 2, 3, 3, 6, 6, 7, 9 4: 0, 1, 2, 3, 3, 5, 6, 9 5: 1, 4, 5, 8 6: 0, 2, 7 7: 3, 3, 9 8: 6, 8 9: 4 10: 2 11: 2 12: 3
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17:28:26 ** In order we get the following: 0 3 7 1 2 6 8 9 2 2 4 5 8 9 9 3 0 2 3 3 6 6 7 9 4 0 1 2 3 3 5 6 9 5 1 4 5 8 6 0 2 7 7 3 3 9 8 6 8 9 4 10 2 11 2 12 3 **
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RESPONSE --> Ok.
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17:32:55 query problem 13.1.35 empirical probability distribution for letters of alphabet
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RESPONSE --> .08 / .385 = .208 .13 / .385 = .338 .065 / .385 = .169 .08 / .385 = .208 .03 / .385 = .078
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17:34:12 ** The probabilities for A, E, I, O, U are: .08. .13, .065, .08, .03. The sum of these probabilities is .385 The probabilities of the letters given a vowel: .08 / .385 = .208 .13 / .385 = .338 .065 / .385 = .169 .08 / .385 = .208 .03 / .385 = .078 These probabilities total 1. **
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RESPONSE --> Ok.
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W`}D`w assignment #015 015. `query 15 Liberal Arts Mathematics II 07-09-2007
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17:36:54 query problem 13.2.10 .3, .4, .3, .8, .7, .9, .2, .1, .5, .9, .6
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RESPONSE --> Mean: .518 Median: .5 Mode: .3, .9
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17:37:40 ** The numbers, in order, are .1, .2, .3, .3, .4, .5, .6, .7, .8, .9, .9 The mean, obtained by adding the 11 numbers then dividing by 11, is .518. The median occurs at position (n + 1 ) / 2 = 6 in the ordered list. This number is .5. Note that there are five numbers before .5 and five numbers after .5. The maximum number of times a number repeats in this distribution is 2. So there are two modes (and we say that the distribution is bimodal). The modes are .3 and .9. **
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RESPONSE --> Ok.
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17:40:12 **** query problem 13.2.24 more effect from extreme value
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RESPONSE --> The median is least affected.
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17:41:18 ** The mean is drastically affected by the error; correcting the error changes the mean by about 3 units. The median number, however, simply shifts 1 position, changing from 2.28 to 2.39. **
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RESPONSE --> Ok.
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17:50:33 **** query problem 13.2.30 Salaries 6 @$19k, 8 @ 23k, 2 @ 34.5k, 7 @ 56.9k, 1 @ 145.5k.
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RESPONSE --> 6 * 19500 = 117000 8 * 23000 = 184000 4 * 28300 = 113200 2 * 34500 = 69000 7 * 36900 = 258300 1 * 145500 = 145500 Mean: 887000 / 28 = 31679 Median: (23000 + 28300) / 2 = 25650 Mode: 23000
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17:51:02 ** IF THERE ARE 28 EMPLOYEES: The totals paid for each salary level are: 6 * $19,500 = $117,000 8 * $23,000 = $184,000 4 * $28,300 = $113,200 2 * $34,500 = $69,000 7 * $36,900 = $258,300 1 * $145,500 = $145,500 The grand total paid in salaries to the 28 employees is therefore $887,000, giving an average of $887,000 / 28 = $31,700. The median occurs at position (n + 1) / 2 = (28 + 1) / 2 = 14.5. Since the 14 th salaray on a list ordered from least to greatest is $23,000 and the 15 th is $28300 the median is ($23000 +$28300) / 2 = $25,650. The mode is 23,000, since this salary occurs more frequently than any other. IF THERE ARE 24 EMPLOYEES: The totals paid for each salary level are: $19,000 * 6 = $114,000 $23,000 * 8 = $184,000 $34,500 * 2 = $69,000 $56,900 * 7 = $398,300 $145,500 * 1 = $145,500 Adding these gives a grand total, which is divided by the number 24 of employees to obtain the mean $37,950. The median occurs at position (n + 1) / 2 = (24 + 1) / 2 = 12.5. Since the $23000 salary covers positions 7 thru 14 in an ordered lise of salaries the median is $23,000. The mode is 23,000, since this salary occurs more frequently than any other.
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RESPONSE --> Ok.
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17:55:23 **** query problem 13.2.51 mean, med, mode of 0, 1, 3, 14, 14, 15, 16, 16, 17, 17, 18, 18, 18, 19, 20
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RESPONSE --> Mean: (206) / 15 = 13.73 Median: 16 Mode: 18
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17:56:11 ** The mean is 13.73, obtained by adding together all the numbers and dividing by n = 15. The median is in position (n+1) / 2 = (15+1)/2 = 8 on the ordered list; the 8 th number is 16. The mode is 18, which is the only number occurring as many as 3 times. **
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RESPONSE --> Ok.
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17:56:32 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> None
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ڃD assignment #016 016. `query 16 Liberal Arts Mathematics II 07-09-2007
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18:04:00 query probl 13.3.6 range, std dev of {67, 83, 55, 68, 77, 63, 84, 72, 65}
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RESPONSE --> Mean: 634 / 9 = 70.44 Sqrt(91.01) = 9.54 Range = 29
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18:04:13 ** x dev. from mean squared dev. 55 15.4 237.16 63 7.4 54.76 65 5.4 29.16 67 3.4 11.56 68 2.4 5.76 72 1.6 2.56 77 6.6 43.56 83 12.6 158.76 84 13.6 184.96 634 728.08 mean = 634 / 9 = 70.4 std. dev. = `sqrt (728.08 / 8) = 9.54 range = 84 - 55 = 29 **
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RESPONSE --> Ok.
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18:06:39 **** query probl 13.3.12 freq dist 14,8; 16,12; 18,15; 20,14; 22,10; 24,6; 26,3
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RESPONSE --> Value Freq Value * Freq Dev^2 * F 14 8 112 204.80 16 12 192 112.32 18 15 270 16.80 20 14 280 12.32 22 10 220 86.40 24 6 144 146.40 26 3 78 144.48 723.5 / 68 = 10.64 Sqrt(10.64) = 3.26
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18:06:53 ** Value freq Value * Freq Dev^2 * F 14 8 112 204.80 16 12 192 112.32 18 15 270 16.80 20 14 280 12.32 22 10 220 86.40 24 6 144 146.40 26 3 78 144.48 Total 68 1296 723.52 Total squared dev is 723.5 so ave squared dev is 723.5 / 68 = 10.6, approx. Std dev is sqrt(ave squared dev) = sqrt(10.6) = 3.3 approx. **
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RESPONSE --> Ok.
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18:08:02 **** query probl 13.3.18 chebyshev for z=5 What is the least possible number of elements of a sample which lie within 5 standard deviations of the mean?
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RESPONSE --> .96N
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18:08:17 ** The formula 1 - 1/k^2 gave you .96. That's the proportion which must under any circumstances lie between mean and 5 std dev from the mean. So the number is .96 n, where n is the number of elements in the sample. **
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RESPONSE --> Ok.
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18:11:13 query probl 13.3.48 mean length of stay 2.7 days, std dev 7.1 days.
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RESPONSE -->
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18:11:46 **** Describe your sketch of the distribution of lengths of stay.
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RESPONSE -->
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18:12:16 2.7 is in the center with each number within 7.1 of the right or left of 2.7 and each additonal number on the left or right within 7.1 of each other. I see the curve as not being skewed.
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RESPONSE --> Ok.
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18:13:26 **** Is your distribution skewed? If so why, and if not why do you think it shouldn't be?
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RESPONSE --> Skewed to the right since you can't have negative values.
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18:13:38 STUDENT RESPONSE AND INSTRUCTOR COMMENT: I said no, but I'm not really sure how to determine this if I draw it myself, becasue I place all the numbers at an equal distance from each other. It just hit me while typing this that it must be skewed to the right because I can't really have negative days stay in the hospital. I think I have confused myself ** You didn't confuse yourself. That's exactly the point. You can't stay fewer than 0 days; since even 1 standard deviation is way below 0 the deviations must be primarily to the right of the mean. So the distribution must be skewed significantly to the right. GENERAL SUGGESTION: In general to understand the graphs of various distributions, try to understand in terms first of the bell-shaped curve with max height at the mean, dropping to about 60% height at a distance of 1 std dev from the mean and to about 14% at 2 std dev from the mean. Then understand that this distribution can be distorted, or skewed, as in this problem. This occurs when most of the distribution lies close to the mean on one side, with a smaller part of the distribution spread out further from the mean on the other. **
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RESPONSE --> Ok.
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