#$&* course mth 152 10:25am12/19/13 Question: `q001. Note that there are 9 questions in this assignment.
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Given Solution: We could calculate the average distance of the numbers from the mean. The mean of these numbers is (7 + 9 + 10 + 11 + 12 + 14) / 6 = 10.5. The deviations from the mean are 3.5, 1.5, .5, .5, 1.5, 3.5. Averaging these deviations we get ave deviation from mean = (3.5 + 1.5 + .5 + .5 + 1.5 + 3.5) / 6 = 1.83 approx.. A simpler measure of the spread is the range, which is the difference 14 - 7 = 7 between the lowest and highest number. Self-critique: ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. Comparing the distributions 7, 9, 10, 11, 12, 14 and 7, 8, 9, 12, 13, 14, which distribution would you say is more spread out? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 14 - 7 = 7. both have the same mean, 10.5. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ------------------------------------------------ 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: The squared deviations are 3.5^2 = 12.25, 2.5^2 = 6.25, .5^2 = .25, and 1.5^2 = 2.25. 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 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The squared deviations are 3.5^2 = 12.25, 2.5^2 = 6.25, .5^2 = .25, and 1.5^2 = 2.25. Since 3.5 and .5 to occur twice each, the average of the squared deviations is (12.25 + 6.25 + 0.25 + 0.25 + 2.25 + 12.25) / 6 = 5.58, approx.. This average of the squared deviations is not the standard deviation, which will be calculated in the next exercise. Self-critique: ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. In the last problem we calculated the average of the squared deviations. Since this average was calculated from the squared deviations, it seems appropriate to now take the square root of our result. The standard deviation is the square root of the average of the squared deviations. Continuing the last problem, what is the standard deviation of the distribution 7, 9, 10, 11, 12, 14? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the squared deviations in the last we came up with 5.67. standard deviation = `sqrt(ave of squared deviations) = `sqrt(5.67) = 2.4, approx.. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (12.25 + 6.25 + 0.25 + 0.25 + 2.25 + 12.25) = 34. divide by 5 (12.25 + 6.25 + 0.25 + 0.25 + 2.25 + 12.25) / 5 = 6.8. `sqrt(6.8) = 2.6 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The total of the squared deviations is (12.25 + 6.25 + 0.25 + 0.25 + 2.25 + 12.25) = 34. When we divide by 5 instead of 6 we get (12.25 + 6.25 + 0.25 + 0.25 + 2.25 + 12.25) / 5 = 6.8. This 'average' of the squared deviations (not really the average but the 'average' we use in calculating the standard deviation) is therefore 6.8, not the 5.67 we obtain before. Thus the standard deviation is std dev = square root of 'average' of squared deviations = `sqrt(6.8) = 2.6, approximately. Note that this value differs slightly from that obtained by doing a true average. Note also that if we are totaling 30 or more squared deviations subtracting the 1 doesn't make much difference, and we just use the regular average of the squared deviations. Self-critique: ------------------------------------------------ Self-critique rating: ********************************************* Question: `q006. We just calculated the standard deviation of the distribution 7, 9, 10, 11, 12, 14. Earlier we noted that the distribution 7, 8, 9, 12, 13, 14 is a bit more spread out than the distribution 7, 9, 10, 11, 12, 14. Calculate the standard deviation of the distribution 7, 8, 9, 12, 13, 14 and determine how much difference the greater spread makes in the standard deviation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: std dev = `sqrt(8.4) = 2.9 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: 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. 8 - 1 = 7 squared deviations equal to 64/7 = 9.1. taking the square root 3.03. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The mean of these numbers is easily found to be 10.5. Note that we have here still another distribution with mean 10.5 and range 7. The deviations from the mean are 3.5, 2.5, 2.5, 2.5, 2.5, 2.5, 2.5, 3.5. The squared deviations are 12.25, 6.25, 6.25, 6.25, 6.25, 6.25, 6.25, 12.25. The sum of these squared deviations is 64. There are 8 numbers in the distribution, so in calculating the modified 'average' use with the standard deviation we will divide the total 64 by 8 - 1 = 7 to get a modified 'average' of the squared deviations equal to 64/7 = 9.1. Taking the square root to get the standard deviation we obtain approximately 3.03. The quickest way to have calculated this standard deviation would be to note that the deviations of 7, 8, 13, and 14 from our previously calculated mean of 10.5 are respectively 3.5, 2.5, 2.5, and 3.5, corresponding to square deviations of 12.25, 6.25, 6.25, and 12.25. Noting that since 8 occurs three times and 13 occurs three times, the total of the squared deviations will be 12.25 + 3 * 6.25 + 3 * 6.25 + 12.25 = 12.25 + 18.75 + 18.75 + 12.25 = 64. The rest of the calculation is done as before. Using multiplication instead of addition to calculate the sum of the repeated numbers is more efficient then doing unnecessary repeated additions. Self-critique: ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. What is the maximum possible standard deviation for a set of six numbers ranging from 7 through 14 and averaging (7 + 14 ) / 2 = 10.5? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 7, 7, 7, 14, 14, 14. Each of these six numbers has a deviation of 3.5 from the mean of 10.5. squared deviation for each number is 12.25 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: "
#$&* course mth 152 10:31am12/19/13 Question: `q Query problem 13.2.10 .3, .4, .3, .8, .7, .9, .2, .1, .5, .9, .6
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Given Solution: `aThe 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q Query problem 13.2.24 more effect from extreme value YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: medium is 2.39 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q Query problem 13.2.30 Salaries 6 @$19k, 8 @ 23k, 2 @ 34.5k, 7 @ 56.9k, 1 @ 145.5k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 28 =$887,000 / 28 = $31,700. medium ($23000 +$28300) / 2 = $25,650. 23,000 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIF 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 $23,000. The mode is 23,000, since this salary occurs more frequently than any other. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q Query problem 13.2.51 mean, med, mode of 0, 1, 3, 14, 14, 15, 16, 16, 17, 17, 18, 18, 18, 19, 20 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 13.73 = 15. The median 16. The mode is 18 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q Query problem 13.2.51 mean, med, mode of 0, 1, 3, 14, 14, 15, 16, 16, 17, 17, 18, 18, 18, 19, 20 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 13.73 = 15. The median 16. The mode is 18 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!