course Mth 174 V}MED͌assignment #011
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18:40:12 Query 8.8.2 (3d edition 8.7.2) 8.7.2. Probability and More On Distributions, p. 421 daily catch density function piecewise linear (2,.08) to (6.,24) to (8,.12)
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RESPONSE -->
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18:49:38 what is the mean daily catch?
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RESPONSE --> Mean = Integral (from 2 to 8) of x*p(x) dx = Integral (from 2 to 6): x*(0.04x) dx + Integral (from 6 to 8): x*(-0.06x+0.6) dx = Integral (from 2 to 6): 0.04x^2 dx + Integral (from 6 to 8): -0.06x^2+0.6x dx = (0.04x^3)/3, evaluated from 2 to 6 + [(-0.06x^3)/3 + (0.6x^2)/2], evaluated from 6 to 8 = (0.04/3)(216-8) + [-0.02(512-216) + 0.3(64-36)] Mean = 5.253 tons
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18:49:54 What integral(s) did you perform to compute a mean daily catch?
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RESPONSE --> See previous response.
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19:11:19 What does this integral have to do with the moment integrals calculated in Section 8.3?
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RESPONSE --> The integral for calculating the mean produces the equivalent of the center of mass for the mass functions in Section 8.3. The moment integral is the approximation of the sums of individual moments (density * length approaching 0 * distance). Dividing this by the integral of the density function (which is total mass) gives the center of mass, which is the mean point, at which one-half the moment is on each side. The mean integral in Section 8.8 simply factors out the integral of the density function, which is the total population (equivalent to total mass). Dividing both numerator and denominator by the total population leaves the mean integral divided by 1. Thus, the mean integral is equivalent to the moment integral/mass integral, and is equivalent to the center of mass.
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19:21:18 Query 8.8.13 (3d edition 8.7.13). Probability and More On Distributions, p. 423 cos t, 0
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19:50:44 which function might best represent the probability for the time the next customer walks in?
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RESPONSE --> Function (a) (cos t 0<,=t<=2pi; e^(t-2pi) t>=2pi) is impossible since it produces negative probabilities from pi/2 to 3pi/2. Function (d) is possible, but not likely, since there is no provision for a time greater than 4 minutes. Function (c) is impossible since the cumulative distribution function (from 0 to infinity) is (1/3) (see below). Function (b) shows a decreasing probability over time, which seems reasonable, and has a cumulative distribution function = 1. Therefore, function (b) is the best choice. For (c) P(t) = Integral (0 to infinity): e^(-3t) dt = (-e^(-3t))/3, evaluated from 0 to infinity. As t approaches infinity, the expression approaches 0, when t=0, the expression = -1/3. 0 - (-1/3) = 1/3. For (b), following the same process as above, the integral is -e^(-3t), as t approaches infinity the expression approaches 0, when t=0, the expression = -1. 0 - (-1) = 1.
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19:50:54 for each of the given functions, explain why it is either appropriate or inappropriate to the situation?
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RESPONSE --> See previous response.
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20:00:05 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> The idea of the mean of a function being equivalent to the center of mass is very interesting. Problem #12 (For a normal population of mean 0, show that the fraction of the population within one standard deviation of the mean does not depend on the standard deviation. With a hint to subsitute w for x/sigma.) gave me fits. The only solution I came up with is to take the integral of the normal distribution function from 0 to b (using the upper one standard deviation) and setting it equal to 0.34. Eventually I ended up with 0.34 = (1/(2sqrtpi)) * integral (from 0 to b) of e^(-(1/2)*u^2) du. Since this has no elementary antiderivative, I tried Derive on it and came up with .34 = an expression that contained an ERF in terms of b, with the remainder of the expression being constants. Since sigma drops out of the expression, I conclude that the fraction within one standard deviation is in fact not dependant upon the value of sigma. Is this a correct solution and/or is there a simpler solution?
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