#$&* course MTH 152 Note that I have a few questions regarding the material before 100% completion. After the questions are squared away I'll send the complete version titled MTH152-QA15 (2). If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: 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, 1.5, .5, .5, 1.5, 3.5. Averaging these deviations we get ave deviation from mean = (3.5 + 1.5 + .5 + .5 + 1.5 + 3.5) / 6 = 1.83 approx.. A simpler measure of the spread is the range, which is the difference 14 - 7 = 7 between the lowest and highest number. Self-critique: - So there are ideally two separate ways to calculate the spread. - 1. Average distance from the mean 2. Range ------------------------------------------------ Self-critique rating: Ok
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Given Solution: 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. Self-critique: - I forgot to note that the range is also equivalent (being that range = difference between highest and lowest number) ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - For the set ( 7, 9, 10, 11, 12, 14) the average deviation is (3.5, 1.5, .5, .5, 1.5, 3.5) - In order to calculate the standard deviation, one must first square the set of deviations. - (12.25, 2.25, .25, .25, 2.25, 12.25) = 29.5 / 6 = 4.917 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 = 5.58, approx.. This average of the squared deviations is not the standard deviation, which will be calculated in the next exercise. Self-critique: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& - Which set is being utilized? In the given problem above it reads “For the distribution 7, 9, 10, 11, 12, 14…” but it appears that the set (7, 8, 9, 12, 13, 14) is being used instead. - In the above solution it lists the squared deviations of the second set (provided in parentheses above) rather than the first set (7, 9, 10, 11, 12, 14). - I can do either set, by switching back and forth between the two and comparing my notes, but perhaps it’s a simple “pilot error” so to speak. ------------------------------------------------ Self-critique rating: See comment above &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - From the last question, the solution was found by squaring the deviations of the set 7, 9, 10 , 11, 12, 14 = (3.5, 1.5, .5, .5, 1.5, 3.5) whose mean was found to be 1.833. - The square deviations were thus (12.25, 2.25, .25, .25, 2.25 12.25) whose average = 4.958 - Sqrt`(4.958) = 2.226656687 = 2.2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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.. Self-critique: - I may have calculated the average squared deviation wrong. I tried again and got 29.5 /6 = 4.91 ------------------------------------------------ Self-critique rating:
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Given Solution: 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. Self-critique: - I’m a bit perplexed. Are we combining the two sets to get get (12.25 + 6.25 + 0.25 + 0.25 + 2.25 + 12.25)? ------------------------------------------------ Self-critique rating: Okay. See comment above.
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Given Solution: 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. Self-critique: ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. Self-critique: ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The maximum possible spread of the distribution would be achieved when half of the numbers are all 7 and the other half are all 14. This would give us the distribution of 7, 7, 7, 14, 14, 14. Each of these six numbers has a deviation of 3.5 from the mean of 10.5. Thus the squared deviation for each number is 12.25. Since there are six numbers in the distribution, the total of the squared deviations must be 6 * 12.25 = 75. Our modified average of the squared distributions will therefore be 75/5 = 15, and the standard deviation will be square root of 15 or approximately 3.9. ********************************************* Question: `q009. The mean of the numbers 1.05, 1.03, 1.06, 1.08, 1.06 is 105.6. On the average by how much do these numbers deviate from the mean? (You would have answered this question in the preceding qa). What is the standard deviation of these numbers? How does the standard deviation compare with the average deviation from the mean? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: