#$&* course MTH 152 1;00 p.m. 11/17/13 016. ``q Query 16
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Given Solution: `a x dev. from mean squared dev. 55 15.4 237.16 63 7.4 54.76 65 5.4 29.16 67 3.4 11.56 68 2.4 5.76 72 1.6 2.56 77 6.6 43.56 83 12.6 158.76 84 13.6 184.96 634 728.08 mean = 634 / 9 = 70.4 std. dev. = `sqrt (728.08 / 8) = 9.54 range = 84 - 55 = 29 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Okay ------------------------------------------------ Self-critique Rating: Okay ********************************************* Question: `q Query probl 13.3.12 freq dist 14,8; 16,12; 18,15; 20,14; 22,10; 24,6; 26,3 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Okay I’m not going to lie the 2 numbers in the frequency distribution has completely thrown me into a state of confusion…. confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Okay this makes a bit more sense to see it wrote out in a table. ------------------------------------------------ Self-critique Rating: Okay ********************************************* Question: `q 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1-1/5^2 = 1 - 1/25 = 25/25 - 1/25 = 24/25 = .96 = 96% confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe formula 1 - 1/k^2 gave you .96. That's the proportion which must under any circumstances lie between mean and 5 std dev from the mean. So the number is .96 n, where n is the number of elements in the sample. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Okay ------------------------------------------------ Self-critique Rating: Okay ********************************************* Question: `q Query probl 13.3.48 mean length of stay 2.7 days, std dev 7.1 days. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I’m lost on this one…I’ve tried to find examples to help me solve it but none really help me… confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aA 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. ** **** Describe your sketch of the distribution of lengths of stay. 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. **** Is your distribution skewed? If so why, and if not why do you think it shouldn't be? STUDENT RESPONSE AND INSTRUCTOR COMMENT: I said no, but I'm not really sure how to determine this if I draw it myself, because 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay the student questions and responses really helped me out! ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!