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Unformatted text preview: Homework #4 Due Mon, 9/18/6 ActivStats: b1) quantitative, measured on the same cases, direction, form, linear, strength b2) negative, linear, moderate b3) null, linear, weak b4) positive, linear, strong b5) positive, bent, moderate c) Correlation = 0.85 d) association, linear, outliers, sign, magnitude, 1.0, units Ch.4 Problems: 3a) see attached
x =2 3b)
x =0 x =1 .09375(4  x 2 )dx = 0.5 .09375(4  x 2 )dx = 0.6875 3c)
x =1 x =6 9a)
x =.5 x= .15e .15( x .5) dx = 0.562 .15e .15( x .5) dx = 0.438 .15e .15( x .5) dx = 0.071 9b)
x =6 x =6 9c)
x =5 x F ( x) = 90 y 8 (1  y )dy =  x 9 (9 x  10)
0 15a) see attached for graphs... 0 x<0 F ( x) =  x (9 x  10) 0 x 1 0 1< x
9 15b) F (.5) = (.5)9 (9(.5)  10) = 0.0107 15c) P (.25 <M X P (.25 X M .5) = F (.5)  F (.25) = 0.0107  (.25)9 (9(.25)  10) = 0.0107 .5) = F (.5)  F (.25) = 0.0107  (.25)9 (9(.25)  10) = 0.0107 15d) see attached X = x x8 (1  x)dx =0.818 90
15e)
0 1 X = 1 ( x  0.818) 2 x8 (1  x)dx = 0.111 90
0 11 3 X = x [1  (10  x) 2 ]dx = 10 4 21) 9 A = ( X ) 2 = (10) 2 = 314.16 23) = 1.8(120) + 32 = 248 = 1.8(2) = 3.60 26a) 0.9850  0.5 = 0.485 26b) 0.8413 0.5 = 0.3413 27a) 2.14 27b) 0.291 + 0.5 = 0.791 corresponds to 0.81 27d) 0.668/2 + 0.5 = 0.834 corresponds to 0.97 29b) 1.34 .25  .3 = 32a) 1  1  (.833) = 1  .203 = .797 .06 .1  .3 = 32b) (3.33) = 0.0004 .06 55a) (61)! = 120
x 55b)
0 x 2 e  x dx = 1.329 5 1 56a) .238 56b) .238 20 80 80 = 2 = 2 57a) 20 80 = =4 2 20 = =5 4 20 = = 57b) P ( X < 24) = F (24 / 4;5) = F (6;5) = 0.715 59a) 1 59b) 2 59c) F (4;1) = 0.982 p = 1  e x e x = 1  p  x = ln(1  p ) 64) ln(1  p ) x= ln(1  0.5) ln(0.5)1 ln(2) median =  = = 116a) r ( x ) = e x e x =  x = 1  (1  e  x ) e a 1  ( x / ) a 1  ( x / ) x e x e 1 a 1 x 116b) r ( x ) = = = x = 1  (1  e  ( x / ) ) e ( x / ) can be any real number without affecting the direction (+ or ) of the slope. If < 1 , then r(x) will decrease with x (due to the negative exponent). If > 1 , then r(x) will increase with x (due to the positive exponent). If = 1 , then r(x) will be constant (due to the 0 exponent which eliminates x). 2 2 0 r sin r sin r 2 x  A  = Ar = 1 A = 1 A 1  cos = A  dx r 0 2 2 r r 0 r r Q1) 1 2 x f ( x) strain =  cos 0 x 1 r r r r y 2 1 2 f ( y ) stress =  cos 1 r 0 y r r See attached for graphs... Q2) A random variable X numerically maps the outcomes of random experiments. The random variable itself does not have a value. Instead, it is a function that describes the set of possible outcomes in an experiment. The random variable could be used to create a probability distribution function, which describes the probabilities for various values of x. These functions take many different forms that each have their own properties. Taking the integral of a pdf can provide a cumulative distribution function, which quickly outputs the probability that the result is less than a chosen value. Q3) Friday 23pm is the only one that fits my schedule. ...
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This note was uploaded on 03/31/2008 for the course CEE 3040 taught by Professor Stedinger during the Fall '08 term at Cornell University (Engineering School).
 Fall '08
 Stedinger

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