course Mth 152 ?????-??qM???·??assignment #016
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10:15:20 `questionNumber 160000 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 dev. sq. dev. -15.44 238.3936 -7.44 55.3536 -5.44 29.5936 -3.44 11.8336 -2.44 5.9536 1.56 2.4336 6.56 43.0336 12.56 157.7536 13.56 183.8736 total 728.2224/9-1=91.0278 sqrt91.0278=9.54 approx.
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10:15:46 `questionNumber 160000 ** 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 --> I get it!
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10:22:13 `questionNumber 160000 **** 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=1296/68=19.06 dev. sq. dev. sq. dev.*freq. -5.06 25.6036 204.8288 -3.06 9.3636 112.3632 -1.06 1.1236 16.854 .94 .8836 12.3704 2.94 8.6436 86.436 4.94 24.4036 146.4216 6.94 48.1636 144.4908 total 723.7648 723.7648/68-1=10.8025 sqrt10.8025=3.29 approx.
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10:22:43 `questionNumber 160000 ** 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 -->
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10:22:48 `questionNumber 160000 ** 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 --> ok
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10:25:44 `questionNumber 160000 **** 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/k^2= 1-1/25=24/25=.96 or 96% of the items
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10:26:20 `questionNumber 160000 ** 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 --> ok
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10:36:37 `questionNumber 160000 query probl 13.3.48 mean length of stay 2.7 days, std dev 7.1 days.
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RESPONSE --> the graph would be skewed to the right
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10:37:58 `questionNumber 160000 ** A sketch of a normal distribution will be a normal, or ?ell-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 ?ails 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, ?unched up·on one side of the mean 2.7 and more spread out for larger values. **
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RESPONSE --> ok
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10:40:20 `questionNumber 160000 **** Describe your sketch of the distribution of lengths of stay.
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RESPONSE --> it is very skewed,it will be more spread out for the larger numbers and more bunched up on the other side
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10:40:58 `questionNumber 160000 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 --> ok
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10:42:27 `questionNumber 160000 **** Is your distribution skewed? If so why, and if not why do you think it shouldn't be?
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RESPONSE --> I'm not sure!
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10:45:12 `questionNumber 160000 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 put in my first answer that the graph would be skewed to the right I just wasn't sure how to explain it
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