course Mth 152 ’ÐûŽ~ÖЋM°zùQÔ«SÍé›ñyassignment #016
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20:52:05 query probl 13.3.6 range, std dev of {67, 83, 55, 68, 77, 63, 84, 72, 65}
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RESPONSE --> Range = 84- 55 = 29 Mean = 634/9 = 70 Std dev = 225+49+25+9+4+49+169+196 = 730 / 8 = 91.25 sqrt = 9.55
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20:53:05 ** 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 --> Correct. I rounded to 70 instead of leaving it 70.4
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21:29:07 **** 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 --> range - 26-14=12 mean 140/7=20 std dev = sqrt 18.6 = 4.31
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21:31:03 ** 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 --> I see where I went wrong I squared the wrong set of numbers.
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21:36:16 **** 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 --> 1-1/5^2 = 25/25- 1/25 = 24/25 = .96 = 96%
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21:36:43 ** 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 --> Correct
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21:40:58 query probl 13.3.48 mean length of stay 2.7 days, std dev 7.1 days.
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RESPONSE --> I cannot find this problem in 13.3. I am not sure with the information given what I am solving for.
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21:45:08 ** A sketch of a normal distribution will be a normal, or ‘bell-shaped’ curve with its peak at the mean, dropping to about 60% of peak value at 1 std dev from the mean and to about 14% of peak value at 2 std dev from the mean. For a normal distribution with mean 2.7 and std dev. 7.1 one std dev from the mean occurs at 2.7 + 7.1 = 9.8 and at 2.7 – 7.1 = -4.4; two std dev from the mean occurs at 16.9 and -11.5. The corresponding normal curve cannot represent length of stay, since length of stay must not be less than zero. As a result we obtain a curve which ‘tails off’ for large values of x, but whose area is concentrated mostly between 0 and 2.7. This curve is not symmetric like the bell curve but is very skewed, ‘bunched up’ on one side of the mean 2.7 and more spread out for larger values. **
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RESPONSE --> I see where this problems come from. Pg 814 For furthur thought. I see that we need to use the formula of sk=3*(mean - median) / standard deviation.
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21:59:18 **** Describe your sketch of the distribution of lengths of stay.
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RESPONSE --> To sketch this and use the formaula we would have to have the median also. If the shape is non symetric then it would be skewed to the left or right but with the largest concentration of numbers being between 0 and 2.7 then the sketch would be skewed to the right.
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22:01:36 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 --> If the center number is 7.1 to the left or right of 2.7 then would some of the numbers be to the negative of 0?
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22:04:49 **** Is your distribution skewed? If so why, and if not why do you think it shouldn't be?
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RESPONSE --> I do not understand how to sketch the distribution to see if it is skewed or not but it would seem to me that if the patients day starts at 0 and goes up to 7.1 days as deviation but the largest group being btween 0 and 2.7 it would be skewed to the right.
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22:06:22 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 --> I see so it would be skewed to the right.
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¯Š†ÁÔóªÚÙÜÀÐT¯Tîåõ„±Ã assignment #017 017. `query 17 Liberal Arts Mathematics II 03-19-2009
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23:06:03 **** query problem 13.4.12 z score for KG's rebounds (.4 from bottom range 10-13)
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RESPONSE --> I do not see this problm in the book I have or one that resembles this. I know to find the z score we would take the number subtract the mean from it and then divide by the standard deviation.
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23:07:33 ** The z score for KG is his total number of rebounds minus the mean average number of rebounds for all the players and then divided by the standard deviation. In KG' s case: z = (489 - 538.2) / 38.8 = -1.3 **
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RESPONSE --> I do not know where we get the numbers from but this is the correct concept that I explained on the formula.
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23:11:06 query problem 13.4.30 midquartile same as median? (Q1+Q3)/2
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RESPONSE --> The midquartile is not always the same as the median. It is the same as the median if there is an odd numbers of items but it is the same as the mean if there is an even numbers of items.
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23:11:39 ** If the median is the actual number in the middle, the it's not necessarily equal to the mean of the first and third quartile. There are different ways to see this. For example suppose that in a large set of numbers, the median number is at least 2 greater than the next smaller number and 2 smaller than the next greater number. Then if all the other numbers stay the same, but the median is increased or decreased by 1, it's still in the middle, so it's still the median. Since all the other numbers stay the same, the first and third quartiles are the same as before, so (Q1 + Q3) / 2 is still the same as before. However the median has changed. So if the median was equal to (Q1 + Q3) / 2, it isn't any more. And if it is now, it wasn't before. In either case we see that the median is not necessarliy equal to the midquartile. To be even more specific, the median of the set {1, 3, 5, 7, 9, 11, 13} is 7. The median of the set {1, 3, 5, 8, 9, 11, 13} is 8. The midquartile of both sets is the same, so for at least one of the two sets (namely the second, as you can verify for yourself) the median and the midquartile are different. **
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RESPONSE --> Correct
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