course Mth 152 }֡ifassignment #015
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09:51:04 `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 --> Find the range: 14-7=7 or find the standard deviation: mean=10.5 7-10.5=(-3.5)^2=12.25 9-10.5=(-1.5)^2=2.25 11-10.5=.5^2=.25 12-10.5=1.5^2=2.25 14-10.5=3.5^2=12.25 standard deviation: (12.25+2.25+.25+2.25+12.25)/6-1 = 5.85 confidence assessment: 3
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09:52:10 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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15:50:20 `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 --> since the first set of numbers has an avg deviation of 2 and the second has an avg deviation of 2.5, the second set would be more spread out confidence assessment: 3
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15:52:24 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 I see where I divided by 6 instead of 6-1 self critique assessment: 3
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16:13:59 `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 --> squared deviations: 12.25, 6.25,.25,.25,2.25,12.25(I did not get 2.5 for one of my deviations, however I did get two 1.5s, but I used your deviations for my answers) avg. of squared dev.: 33.5/6-1=6.7 confidence assessment: 3
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16:15:12 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 --> I see I went ahead and did standard deviation. self critique assessment: 3
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16:20:44 `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 --> I took 6.7 which is the sum of the squared dev. divided by 6-1. I then took the square root of 6.7 and got 2.588435821=2.59 confidence assessment: 2
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16:24:36 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 --> I thought I was suppoded to divide by 5, not 6? I must have been doing ""sample standard deviation"". self critique assessment: 2
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16:26:15 `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 --> ok I was using the book pg. 750. confidence assessment: 3
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16:28:30 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 --> That was my answer on the previous question. I understand. self critique assessment: 3
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16:35:31 `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 --> I got the squared deviations: 12.25+6.25+2.25+2.25+6.25+12.25=41.5/5=8.3 'sqrt(8.3)=approximately 2.89 confidence assessment: 3
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16:36:25 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, I've got it. self critique assessment: 3
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16:42:54 `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 --> I used previous answers of squared deviations and added them: 12.25+18.75(sum for three 8s)+18.75(sum for three 13s)+12.25=62/7=8.857142 then took the square root and got 2.98 confidence assessment: 2
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16:46:44 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 --> The sum of the squared deviations is really 62, I hope it is ok that I used that instead of the 64 that you wrote in the explanation. self critique assessment: 3
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10:55:53 `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 --> I guessed and checked to arrive at 10.5 avg, using these numbers I determined the answer to be approximately 2.43 confidence assessment:
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10:58:07 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 --> I pushed cance of the ""rating"" box and it deleted my response as well as the question! self critique assessment: 3
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