course Mth 152 稇槮罇肱鳘笊籋瓿┹䦛嚎w晶撟鉧ssignment #015
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09:15:19 `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 mean and the median confidence assessment: 2
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09:16:11 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 --> It is a lot simpler to subtract the lowest from the highest. self critique assessment: 2
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09:22:28 `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 mean for the first set is 63/6 = 10.5 the mean for the second set is 63/6 = 10.5 10.5-7 = 3.5 10.5-9 = 1.5 10.5-10= .5 11-10.5= .5 12-10.5=1.5 14-10.5=3.5 3.5+1.5+.5+.5+1.5+3.5=11/6 = 1.83 10.5-7 = 3.5 10.5-8 = 2.5 10.5-9 = 1.5 12-10.5=1.5 13-10.5=2.5 14-10.5=3.5 3.5+2.5+1.5+1.5+2.5+3.5=15/6= 2.5 confidence assessment: 2
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09:23:40 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 --> I got that they were not the same but they did nave the same range. self critique assessment: 2
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09:30:47 `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 --> 3.5*3.5=12.25 2.5*2.5=6.25 .5*.5=.25 .5*.5=.25 1.5*1.5=2.25 3.5*3.5=12.25 12.25+6.25+.25+.25+2.25+12.25=33.5 6-1= 5 33.5/5=6.7 Square root of 6.7 confidence assessment: 2
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09:31:37 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 think I worked through the square self critique assessment: 2
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09:32:27 `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 --> Square of 6.7 confidence assessment: 2
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09:32:59 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 --> I got 6.7 for the answer self critique assessment: 2
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09:33:56 `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 --> Square of 6.7 confidence assessment: 2
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09:34:43 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 is how I got the answer on the previous problems, I had already subtracted the 1 to get 5 to divide by instead of 6. self critique assessment: 2
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09:43:39 `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 --> 7+8+9+12+13+14= 63/6=10.5 10.5-7 =3.5 10.5-8 =2.5 10.5-9 =1.5 12-10.5=1.5 13-10.5=2.5 14-10.5=3.5 3.5^2=12.25 2.5^2=6.25 1.5^2=2.25 1.5^2=2.25 2.5^2=6.25 3.5^2=12.25 12.25+6.25+2.25+2.25+6.25+12.25=41.5 6-1 = 5 41.5/5 = 8.30 Square of 8.3 approx 2.9 confidence assessment: 2
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09:44:16 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 --> that is what I got and how I worked it. self critique assessment: 2
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09:56:44 `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 --> 7+8+8+8+13+13+14 = 71/7 = 10.14 10.14-7= 3.14 10.14-8= 2.14*3 =6.42 13-10.14=2.86*2=5.72 14-10.14=3.86 3.14+6.42+5.72+3.86=19.14 7-1=6 19.14/6=3.19 confidence assessment: 2
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09:58:49 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 --> I mutiplied instead of adding all the numbers too but I must have done something wrong because I got a different outcome self critique assessment: 2
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10:02:56 `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 --> the numbers would be between 7 and 14 so the maximum deviation would be 14 confidence assessment: 2
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10:05:11 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 wasn't sure how to get the answer but I thought it would have to be between 7 and 14 self critique assessment: 2
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course Mth 152 稇槮罇肱鳘笊籋瓿┹䦛嚎w晶撟鉧ssignment #015
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09:15:19 `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 mean and the median confidence assessment: 2
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09:16:11 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 --> It is a lot simpler to subtract the lowest from the highest. self critique assessment: 2
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09:22:28 `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 mean for the first set is 63/6 = 10.5 the mean for the second set is 63/6 = 10.5 10.5-7 = 3.5 10.5-9 = 1.5 10.5-10= .5 11-10.5= .5 12-10.5=1.5 14-10.5=3.5 3.5+1.5+.5+.5+1.5+3.5=11/6 = 1.83 10.5-7 = 3.5 10.5-8 = 2.5 10.5-9 = 1.5 12-10.5=1.5 13-10.5=2.5 14-10.5=3.5 3.5+2.5+1.5+1.5+2.5+3.5=15/6= 2.5 confidence assessment: 2
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09:23:40 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 --> I got that they were not the same but they did nave the same range. self critique assessment: 2
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09:30:47 `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 --> 3.5*3.5=12.25 2.5*2.5=6.25 .5*.5=.25 .5*.5=.25 1.5*1.5=2.25 3.5*3.5=12.25 12.25+6.25+.25+.25+2.25+12.25=33.5 6-1= 5 33.5/5=6.7 Square root of 6.7 confidence assessment: 2
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09:31:37 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 think I worked through the square self critique assessment: 2
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09:32:27 `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 --> Square of 6.7 confidence assessment: 2
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09:32:59 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 --> I got 6.7 for the answer self critique assessment: 2
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09:33:56 `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 --> Square of 6.7 confidence assessment: 2
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09:34:43 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 is how I got the answer on the previous problems, I had already subtracted the 1 to get 5 to divide by instead of 6. self critique assessment: 2
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09:43:39 `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 --> 7+8+9+12+13+14= 63/6=10.5 10.5-7 =3.5 10.5-8 =2.5 10.5-9 =1.5 12-10.5=1.5 13-10.5=2.5 14-10.5=3.5 3.5^2=12.25 2.5^2=6.25 1.5^2=2.25 1.5^2=2.25 2.5^2=6.25 3.5^2=12.25 12.25+6.25+2.25+2.25+6.25+12.25=41.5 6-1 = 5 41.5/5 = 8.30 Square of 8.3 approx 2.9 confidence assessment: 2
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09:44:16 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 --> that is what I got and how I worked it. self critique assessment: 2
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09:56:44 `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 --> 7+8+8+8+13+13+14 = 71/7 = 10.14 10.14-7= 3.14 10.14-8= 2.14*3 =6.42 13-10.14=2.86*2=5.72 14-10.14=3.86 3.14+6.42+5.72+3.86=19.14 7-1=6 19.14/6=3.19 confidence assessment: 2
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09:58:49 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 --> I mutiplied instead of adding all the numbers too but I must have done something wrong because I got a different outcome self critique assessment: 2
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10:02:56 `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 --> the numbers would be between 7 and 14 so the maximum deviation would be 14 confidence assessment: 2
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10:05:11 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 wasn't sure how to get the answer but I thought it would have to be between 7 and 14 self critique assessment: 2
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