Probability HW#5 - 國立臺灣大學

Probability HW#5 Due: May 20, (6 points)ComputeE[X] ifXhas a density function given by(a)f(x) ={14xe x/2x >00otherwise;(b)f(x) ={c(1 x2) 1< x <10otherwise;(c)f(x) ={5x2x >50x (a) Note that limx xne x/2= 0 for alln [X] = xf(x) dx= 014x2e x/2dx= 12x2e x/2 0 0 xe x/2dx= 12x2e x/2 0+ ( 2xe x/2) 0 0 2e x/2dx= 4e x/2 0= 4(b)E[X] = xf(x) dx= 1 1c(x x3) dx= 0 The last equality holds sincex x3is an odd function (c)E[X] = xf(x) dx= 55xdx= 5 lnx| 5= (4 points)The density function ofXis given byf(x) ={a+bx20 x 10otherwiseIfE[X] =35, ,f(x) is a Probability density function. f(x) dx= 10a+bx2dx=ax+b3x3 10=a+b3= 1 And we can computeE[X],E[X] = xf(x) dx= 10ax+bx3dx=a2x2+b4x4 10=a2+b4=35 Hence{3a+b= 310a+ 5b= 12We havea= 3/5 andb= 6 (4 points)The lifetime in hours of an electronic tube is a random variable having aprobability density function given byf(x) =xe x,x 0 Compute the expected lifetime of such a xf(x) dx= 0x2e xdx= x2e x 0 0 2xe xdx= x2e x 0+ ( 2xe x) 0 0 2e xdx= 2e x 0= 2 The expected lifetime of such a tube is (6 points)Suppose that the height, in inches, of a 25-year-old man is a normal randomvariable with paramet}}}}}

The lifetimes of interactive computer chips produced by a certain semicon-ductor manufacturer are normally distributed with parameters µ = 1.4×106 hours and σ = 3 × 105 hours. What is the approximate probability that a batch of 100 chips will contain at least 20 whose lifetimes are less than 1.8× 106. Solution.

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